Summary of Some Reactions and Reaction Mechanisms

ELIMINATION VERSUS SUBSTITUTION IN HALOGENOALKANES

The factors that decide whether halogenoalkanes undergo elimination reactions or nucleophilic substitution when they react with hydroxide ions from,

say, sodium hydroxide or potassium hydroxide.Details for each of these types of reaction are given elsewhere, and you will find links to them from this page.

The reactions

Both reactions involve heating the halogenoalkane under reflux with sodium or potassium hydroxide solution.

Nucleophilic substitution

The hydroxide ions present are good nucleophiles, and one possibility is a replacement of the halogen atom by an -OH group to give an alcohol via a nucleophilic

substitution reaction.

Elimination

Halogenoalkanes also undergo elimination reactions in the presence of sodium or potassium hydroxide.

The 2-bromopropane has reacted to give an alkene - propene.

Notice that a hydrogen atom has been removed from one of the end carbon atoms together with the bromine from the centre one. In all simple elimination reactions

the things being removed are on adjacent carbon atoms, and a double bond is set up between those carbons.

What decides whether you get substitution or elimination?

The reagents you are using are the same for both substitution or elimination - the halogenoalkane and either sodium or potassium hydroxide solution. In all cases,

you will get a mixture of both reactions happening - some substitution and some elimination. What you get most of depends on a number of factors.

The type of halogenoalkane

This is the most important factor.

 type of halogenoalkane substitution or elimination? primary mainly substitution secondary both substitution and elimination tertiary mainly elimination

For example, whatever you do with tertiary halogenoalkanes, you will tend to get mainly the elimination reaction, whereas with primary ones you will tend to get mainly

substitution. However, you can influence things to some extent by changing the conditions.

The solvent

The proportion of water to ethanol in the solvent matters.

• Water encourages substitution.

• Ethanol encourages elimination.

The temperature

Higher temperatures encourage elimination.

Concentration of the sodium or potassium hydroxide solution

Higher concentrations favour elimination.

In summary

For a given halogenoalkane, to favour elimination rather than substitution, use:

• heat

• a concentrated solution of sodium or potassium hydroxide

• pure ethanol as the solvent

The role of the hydroxide ions

The role of the hydroxide ion in a substitution reaction

In the substitution reaction between a halogenoalkane and OH- ions, the hydroxide ions are acting as nucleophiles. For example, one of the lone pairs on the oxygen can attack the slightly positive carbon. This leads on to the loss of the bromine as a bromide ion, and the -OH group becoming attached in its place.

The role of the hydroxide ion in an elimination reaction

Hydroxide ions have a very strong tendency to combine with hydrogen ions to make water - in other words, the OH- ion is a very strong base. In an elimination reaction, the hydroxide ion hits one of the hydrogen atoms in the CH3 group and pulls it off. This leads to a cascade of electron pair movements resulting in the formation of a carbon-carbon double bond, and the loss of the bromine as Br-.

THE MECHANISM FOR THE ESTERIFICATION REACTION

This page looks in detail at the mechanism for the formation of esters from carboxylic acids and alcohols in the presence of concentrated sulphuric acid acting as the catalyst. It uses the formation of ethyl ethanoate from ethanoic acid and ethanol as a typical example.

The mechanism for the formation of ethyl ethanoate

A reminder of the facts

Ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid as a catalyst to produce the ester, ethyl ethanoate. The reaction is slow and reversible. To reduce the chances of the reverse reaction happening, the ester is distilled off as soon as it is formed.

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The mechanism

All the steps in the mechanism below are shown as one-way reactions because it makes the mechanism look less confusing. The reverse reaction is actually done sufficiently differently that it affects the way the mechanism is written. You will find a link to the hydrolysis of esters further down the page if you are interested.

Step 1

In the first step, the ethanoic acid takes a proton (a hydrogen ion) from the concentrated sulphuric acid. The proton becomes attached to one of the lone pairs on the oxygen which is double-bonded to the carbon.

The transfer of the proton to the oxygen gives it a positive charge, but it is actually misleading to draw the structure in this way (although nearly everybody does!). he positive charge is delocalised over the whole of the right-hand end of the ion, with a fair amount of positiveness on the carbon atom. In other words, you can think of an electron pair shifting to give this structure:

You could also imagine another electron pair shift producing a third structure:

So which of these is the correct structure of the ion formed? None of them! The truth lies somewhere in between all of them. One way of writing the delocalised structure of the ion is like this:

The double headed arrows are telling you that each of the individual structures makes a contribution to the real structure of the ion. They don't mean that the bonds are flipping back and forth between one structure and another. The various structures are known as resonance structures or canonical forms.There will be some degree of positive charge on both of the oxygen atoms, and also on the carbon atom. Each of the bonds between the carbon and the two oxygens will be the same - somewhere between a single bond and a double bond.For the purposes of the rest of this discussion, we are going to use the structure where the positive charge is on the carbon atom.

Step 2

The positive charge on the carbon atom is attacked by one of the lone pairs on the oxygen of the ethanol molecule.

Note:  You could work out precisely why that particular oxygen carries the positive charge on the right-hand side. On the other hand, you could realise that there has to be a positive charge somewhere (because you started with one), and that particular oxygen doesn't look right - it has too many bonds. Put the charge on there! That's a quick rough-and-ready reasoning which works every time I use it!

Step 3

What happens next is that a proton (a hydrogen ion) gets transferred from the bottom oxygen atom to one of the others. It gets picked off by one of the other substances in the mixture (for example, by attaching to a lone pair on an unreacted ethanol molecule), and then dumped back onto one of the oxygens more or less at random.

The net effect is:

Step 4

Now a molecule of water is lost from the ion.

The product ion has been drawn in a shape to reflect the product which we are finally getting quite close to! The structure for the latest ion is just like the one we discussed at length back in step 1. The positive charge is actually delocalised all over that end of the ion, and there will also be contributions from structures where the charge is on the either of the oxygens:

It is easier to follow what is happening if we keep going with the structure with the charge on the carbon.

Step 5

The hydrogen is removed from the oxygen by reaction with the hydrogen sulphate ion which was formed way back in the first step.

And there we are! The ester has been formed, and the sulphuric acid catalyst has been regenerated. This page looks in detail at the mechanism for the formation of esters from carboxylic acids and alcohols in the presence of concentrated sulphuric acid acting as the catalyst. It uses the formation of ethyl ethanoate from ethanoic acid and ethanol as a typical example.

The mechanism for the formation of ethyl ethanoate

A reminder of the facts

Ethanoic acid reacts with ethanol in the presence of concentrated sulphuric acid as a catalyst to produce the ester, ethyl ethanoate. The reaction is slow and reversible. To reduce the chances of the reverse reaction happening, the ester is distilled off as soon as it is formed.

The mechanism

All the steps in the mechanism below are shown as one-way reactions because it makes the mechanism look less confusing. The reverse reaction is actually done sufficiently differently that it affects the way the mechanism is written. You will find a link to the hydrolysis of esters further down the page if you are interested.

Step 1

In the first step, the ethanoic acid takes a proton (a hydrogen ion) from the concentrated sulphuric acid. The proton becomes attached to one of the lone pairs on the oxygen which is double-bonded to the carbon.

The transfer of the proton to the oxygen gives it a positive charge, but it is actually misleading to draw the structure in this way (although nearly everybody does!). The positive charge is delocalised over the whole of the right-hand end of the ion, with a fair amount of positiveness on the carbon atom. In other words, you can think of an electron pair shifting to give this structure:

You could also imagine another electron pair shift producing a third structure:

So which of these is the correct structure of the ion formed? None of them! The truth lies somewhere in between all of them. One way of writing the delocalised structure of the ion is like this:

The double headed arrows are telling you that each of the individual structures makes a contribution to the real structure of the ion. They don't mean that the bonds are flipping back and forth between one structure and another. The various structures are known as resonance structures or canonical forms.There will be some degree of positive charge on both of the oxygen atoms, and also on the carbon atom. Each of the bonds between the carbon and the two oxygens will be the same - somewhere between a single bond and a double bond.

NUCLEOPHILIC ADDITION / ELIMINATION IN THE REACTION BETWEEN ACYL CHLORIDES AND AMMONIA

The facts and a simple, uncluttered mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and ammonia. If you want the mechanism explained to you in detail, there is a link at the bottom of the page. Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want.

The reaction between ethanoyl chloride and ammonia

The facts Ethanoyl chloride reacts violently with a cold concentrated solution of ammonia. A white solid product is formed which is a mixture of ethanamide (an amide) and ammonium chloride.

Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ammonia to give ammonium chloride.

The mechanism

The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ammonia.

The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.

That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways: It might be removed by a chloride ion, producing HCl (which would immediately react with excess ammonia to give ammonium chloride as above) . . .

. . . or it might be removed directly by an ammonia molecule.

The ammonium ion, together with the chloride ion already there, makes up the ammonium chloride formed in the reaction.

THE ELIMINATION REACTIONS PRODUCING ALKENES FROM SIMPLE HALOGENOALKANES

ELIMINATION FROM UNSYMMETRIC HALOGENOALKANES

The elimination from unsymmetric halogenoalkanes such as 2-bromobutane.

2-bromobutane is an unsymmetric halogenoalkane in the sense that it has a CH3 group one side of the C-Br bond and a CH2CH3 group the other. You have to be careful with compounds like this because of the possibility of more than one elimination product depending on where the hydrogen is removed from. The basic facts and mechanisms for these reactions are exactly the same as with simple halogenoalkanes like 2-bromopropane. This page only deals with the extra problems created by the possibility of more than one elimination product.

Background to the mechanism

You will remember that elimination happens when a hydroxide ion (from, for example, sodium hydroxide) acts as a base and removes a hydrogen as a hydrogen ion from the halogenoalkane. For example, in the simple case of elimination from 2-bromopropane:

The hydroxide ion removes a hydrogen from one of the carbon atoms next door to the carbon-bromine bond, and the various electron shifts then lead to the formation of the alkene - in this case, propene. With an unsymmetric halogenoalkane like 2-bromobutane, there are several hydrogens which might possibly get removed. You need to think about each of these possibilities.

Where does the hydrogen get removed from?

The hydrogen has to be removed from a carbon atom adjacent to the carbon-bromine bond. If an OH- ion hit one of the hydrogens on the right-hand CH3 group in the 2-bromobutane (as we've drawn it), there's nowhere for the reaction to go.

To make room for the electron pair to form a double bond between the carbons, you would have to expel a hydrogen from the CH2 group as a hydride ion, H-. That is energetically much too difficult, and so this reaction doesn't happen. That still leaves the possibility of removing a hydrogen either from the left-hand CH3 or from the CH2 group.

If it was removed from the CH3 group:

The product is but-1-ene, CH2=CHCH2CH3.

If it was removed from the CH2 group:

This time the product is but-2-ene, CH3CH=CHCH3.In fact the situation is even more complicated than it looks, because but-2-ene exhibits geometric isomerism. You get a mixture of two isomers formed - cis-but-2-ene and trans-but-2-ene.

Which isomer gets formed is just a matter of chance. For the purposes of the rest of this discussion, we are going to use the structure where the positive charge is on the carbon atom.

Step 2

The positive charge on the carbon atom is attacked by one of the lone pairs on the oxygen of the ethanol molecule.

Note:  You could work out precisely why that particular oxygen carries the positive charge on the right-hand side. On the other hand, you could realise that there has to be a positive charge somewhere (because you started with one), and that particular oxygen doesn't look right - it has too many bonds. Put the charge on there!

That's a quick rough-and-ready reasoning which works every time I use it!

Step 3

What happens next is that a proton (a hydrogen ion) gets transferred from the bottom oxygen atom to one of the others. It gets picked off by one of the other substances in the mixture (for example, by attaching to a lone pair on an unreacted ethanol molecule), and then dumped back onto one of the oxygens more or less at random.The net effect is:

Step 4

Now a molecule of water is lost from the ion.

The product ion has been drawn in a shape to reflect the product which we are finally getting quite close to!The structure for the latest ion is just like the one we discusssed at length back in step 1. The positive charge is actually delocalised all over that end of the ion, and there will also be contributions from structures where the charge is on the either of the oxygens:

It is easier to follow what is happening if we keep going with the structure with the charge on the carbon.

Step 5

The hydrogen is removed from the oxygen by reaction with the hydrogen sulphate ion which was formed way back in the first step.

And there we are! The ester has been formed, and the sulphuric acid catalyst has been regenerated.

The elimination reaction involving 2-bromopropane and hydroxide ions

The facts

2-bromopropane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Heating under reflux involves heating with a condenser placed vertically in the flask to avoid loss of volatile liquids. Propene is formed and, because this is a gas, it passes through the condenser and can be collected.

Everything else present (including anything formed in the alternative substitution reaction) will be trapped in the flask.

The mechanism

In elimination reactions, the hydroxide ion acts as a base - removing a hydrogen as a hydrogen ion from the carbon atom next door to the one holding the bromine.The resulting re-arrangement of the electrons expels the bromine as a bromide ion and produces propene.

Geometric isomerism:  Isomerism is where you can draw more than one arrangement of the atoms for a given molecular formula. Geometric isomerism is a special case of this involving molecules which have restricted rotation around one of the bonds - in this case, a carbon-carbon double bond. The C=C bond could only rotate if enough energy is put in to break the pi bond. Effectively, except at high temperatures, the C=C bond is "locked".

In the case of but-2-ene, the two CH3 groups will either both be locked on one side of the C=C (to give the cis isomer), or on opposite sides (to give the trans one).

Beware!  It is easy to miss geometric isomers in an exam. Always draw alkenes with the correct 120° bond angles around the C=C bond as shown in the diagrams for the cis and trans isomers above. If you take a short cut and write but-2-ene as CH3CH=CHCH3, you will almost certainly miss the fact that cis and trans forms are possible. This is a rich source of questions in an exam. You could easily throw away marks if you miss these possibilities.

Nucleophilic Aliphatic Substitution Reactions

An Overview

### Introduction

In our discussion of acid-base chemistry we alluded to the similarities between the reactions of NaOH with HCl on the one hand and CH3Cl on the other. Scheme 1 compares the two.

Scheme 1

One and the Same

The reaction of HCl with NaOH is a specific example of a more general type of reaction known as a nucleophilic substitution. The reaction of CH3Cl with NaOH exemplifies a type of nucleophilic substitution called nucleophilic aliphatic substitution, where the word aliphatic indicates that the substitution occurs at an sp3 hybridized carbon atom. The study of nucleophilic aliphatic substitution reactions has provided chemists with detailed insights into the nature of chemical reactivity. Scheme 2 presents a generalized description of nucleophilic aliphatic substitution reactions.

Scheme 2

Nucleophilic Aliphatic Substitution

Before we begin our discussion of the details of Scheme 2, let's clarify some terms.

### Definitions

• nucleophile A nucleophile is a Lewis base, i.e. an electron pair donor. Any reagent that contains an atom with at least one lone pair of electrons is a potential nucleophile. Common examples include halide ions such as -:I and -:Br, hydroxide ion , -:OH, water, H2O:, and ammonia, :NH3. Note that X may represent a single atom or a polyatomic group. Note, too, that the central atom of a nucleophile may have a formal charge of 0 or -1.

• electrophile An electrophile is a Lewis acid, i.e. an electron pair acceptor. Any reagent that contains an atom which has a formal or a partial positive charge is a potential electrophile.

• leaving group A leaving group is any atom or polyatomic group that is replaced by a nucleophile during a nucleophilic aliphatic substitution reaction. Leaving groups are generally conjugate bases of strong acids. Common examples include halide ions such as -:I and -:Br, water, H2O:, and carboxylate ions such as trifluoroacetate, CF3CO2:-. The central atom of the leaving group is always an electronegative atom, most commonly halogen or oxygen.

• substituent A substituent is any atom or polyatomic group attached to the electrophilic center of the substrate.

There are two features of Scheme 2 that merit further comment.

First, notice that the reaction is depicted as an equilibrium process. In most nucleophilic aliphatic substitution reactions, the value of the equilibrium constant is very large, i.e. the reaction is essentially irreversible. The rules that we developed for assessing the equilibrium constant of any acid-base reaction will serve as a useful guide in estimating the equilibrium constant for nucleophilic aliphatic substitution reactions as well.

Second, since the reaction is an equilibrium process, there is no fundamental difference between a nucleophile and a leaving group. It all depends on your perspective; the reagent that acts as a nucleophile in the forward direction assumes the role of a leaving group in the reverse direction.

Having seen the features of nucleophilic aliphatic substitution reactions in broad outline, we will now examine the process in more detail. Specifically, we will look at the effect of each of the following parameters on the rates of nucleophilic aliphatic substitution reactions:

• the substituent(s) connected to the substrate

• the leaving group

• the nucleophile

• the solvent

### Introduction-

Organometallic reagents are compounds which contains carbon-metal bonds.

For the purposes of the discussion that follows, the only compounds we will consider will be ones where M = Li or Mg. When M= Li, the organometallic reagent is called an organolithium reagent. When M = Mg, it is called a Grignard reagent. Historically Grignard reagents were developed before organolithium reagents. In recent years, however, organolithium reagents have taken over the key role that Grignard reagents played as the most versatile source of nucleophilic carbon.

The nucleophilic character of organometallic reagents stems from the fact that the C-M bond is polarized in such a way that the carbon atom is negative while the metal atom is positive.

As the picture above indicates, the carbon-metal bond has "ionic character". In fact, it is useful to think about Grignard reagents and organolithium reagents as sources of negatively charged carbon atoms, i.e. carbanions. Since carbon is not a very electronegative element, it is very reactive when it bears a negative or partial negative charge.

### Organometallic Reagents as Bases

In order to appreciate just how reactive carbanions are, consider the series of anions and their conjugate acids shown in Figure 1.

Figure 1

A Comparison of Acid/Base Strengths (The pK values are approximate.)

Ignoring for the moment the different ways in which chemists write methane and ammonia on the one hand and water and hydrogen fluoride on the other (wierd, huh?), recall that pKa values are a measure of the acidity of a compound. Furthermore, the pKa scale is logarithmic or exponential. Thus, hydrogen fluoride is 1013 times more acidic than water. Similarly, water is 1022 more acidic than ammonia. Knowing the relative acidities of the compounds in Figure 1 means you also know their relative basicities: the amide anion is 1022 more basic than hydroxide ion, which is 1013 times more basic than fluoride ion.

Clearly then, methane is the weakest acid of the four conjugate acids shown in Figure 1, while the methide anion is the strongest base, and, by extension, the best nucleophile. The trend in base strength exhibited by the four anions in Figure 1 is attributed to the difference in nuclear charge of the central atom in each ion: Carbon has 6 protons attracting the lone pair of electrons, nitrogen has 7, oxygen 8, and fluorine 9.

Reagents such as phenylmagnesium bromide and methyl lithium are among the strongest bases there are. Consequently they will deprotonate compounds such as amines, alcohols, and carboxylic acids. Figure 2 presents one reaction that is representative of each of these situations.

Figure 2

Acid-Base Reactions of Organometallic Reagents

The equilibrium constant for each of these reactions is very large. These reactions all occur extremely rapidly, sometimes explosively! The extreme reactivity of organometallic reagents towards O-H and N-H groups makes these groups incompatible with such strong bases.

#### Organometallic Reagents as Nucleophiles

Whether an organometallic reagent is classified as a base or a nucleophile depends on whether it forms a bond with a hydrogen atom or a carbon atom. If a reactant contains an electrophilic carbon and does not contain O-H or N-H groups, then an organometallic reagent will act as a nucleophile towards that electrophilic carbon atom. The most common source of electrophilic carbon is the carbonyl group, especially the carbonyl group of aldehydes and ketones. Equations 3 and 4 illustrate the nucleophilic reactivity of phenylmagnesium bromide and methyl lithium towards a simple aldehyde and a simple ketone, respectively.

As these equations emphasize, each of these reactions leads to the formation of a C-C bond. This is the basis of synthetic organic chemistry! A fundamental principle that guides the development of logical approaches to the preparation of new molecules is that carbon-carbon bond formation requires the interaction of molecules which contain carbon atoms of opposite polarity. We have already seen that the C-O bond in a carbonyl group has a bond dipole with the carbon atom being electron deficient. In Equation 4 the carbonyl carbon of propiophenone is highlighted in blue to emphasize its electrophilic character, while the nucleophilic nature of the carbon in methyl lithium is stressed by colouring it red.

Reaction Mechanism

The reaction of an organometallic reagent with an aldehyde or ketone embodies the most fundamental reaction of the carbonyl group: nucleophilic addition. The mechanism of the reaction involves two steps: 1. addition of the organometallic reagent to the carbonyl carbon to form a tetrahedral intermediate 2. protonation of the the resulting alkoxide ion. The second step occurs during the work-up of the reaction. Figure 3 summarizes these two steps.

Figure 3

The Mechanism of Nucleophilic Addition to a Carbonyl Group

### Examples

In the following reactions, the carbon-carbon bond that is formed is indicated in red. Equation 5 describes a Grignard reaction that was used in the first total synthesis of the hormone progesterone. The carbon atoms in the product of reaction 5 are numbered to match their positions in progesterone.

An aromatic Grignard reagent played a key role in the synthesis of monensin, a polyether antibiotic, as shown in Equation 6.

Finally, Equation 7 illustrates three of the final steps in the synthesis of a terpene called D2-8-epicedrene.

The middle step involves the nucleophilic addition of methyl lithium to the carbonyl carbon atom of a ketone. The final step in the synthesis entails dehydration of the 3o alcohol formed in the second step. How would you accomplish this?

## NUCLEOPHILIC ADDITION / ELIMINATION IN THE REACTION BETWEEN ACYL CHLORIDES AND AMMONIA

### Introduction and Review

As part of our discussion of organometallic reagents we compared the base strengths of methanide ion, amide ion, hydroxide ion, and fluoride ion. Figure 1 reiterates that comparison.

Figure 1

A Comparison of Acid/Base Strengths (The pK values are approximate.)

We also considered how it was possible to estimate the equilibrium constant for an acid-base reaction by comaparing the pKa values of the two acids involved in the reaction. Equation 1 presents another example. What is the approximate equilibrium constant for this reaction?

Within limits, it is possible to extend the use of pKa values to other reactions. Consider the addition of methyl lithium to formaldehyde as shown in Equation 2.

In this reaction, the negative charge starts out on a carbon atom and ends up on an oxygen. In terms of electronegativity, this is a favorable change. The alkoxide ion is a more stable species than the methanide ion. In other words, the alkoxide ion is a weaker base than the methanide ion. This means the equilibrium constant for reaction 2 will be greater than 1, i.e. at equilibrium there will be more product than reactant. How much more? To make a numerical estimate, compare the pKa values of the conjugate acids of the methanide and ethoxide ions in reaction 2; they are 50 and 16, respectively. Since methane is 1034 times less acidic than ethanol, methanide ion is 1034 times more basic than ethoxide ion. Hence the equilibrium constant for equation 2 is approximately 1034.

Now consider the reactions shown in Equations 3,4, and 5.

Estimate the equilibrium constant for each of these reactions. Is the addition of amide ion to a carbonyl group favorable? What about the addition of hydroxide ion and fluoride (and by extension, any halide ion)?

As this discussion indicates, the addition of anionic nucleophiles to the carbonyl group of aldehydes and ketones becomes less favorable as the nucleophilic atom changes from C to N to O to F. In fact, when the nucleophilic atom is F, the equilibrium constant is so small that it is safe to say that fluoride ion, and by extension, any halide ion, does not add to the carbonyl group of aldehydes or ketones.

Up to this point we have considered the relative nucleophilicities of anionic nucleophiles. The emphasis on the word anionic is necessitated by the fact, unlike ammonia, water, and hydrogen fluoride, methane does not have a lone pair of electrons on the central atom. In other words, methane is not nucleophilic. The methanide ion is. But ammonia and water, and by extension, amines and alcohols, can act as neutral nucleophiles. And under the right conditions they do.

### Water and Alcohols as Nucleophiles

When formaldehde is added to water, the equilibrium shown in Equation 6 is established rapidly. An aqueous solution of formaldehyde is called formalin. It contains virtually no free formaldehyde, i.e. Keq >>> 1.

This reaction is analogous to the hydration of an alkene. If you recall, the mechanism of the acid catalysed hydration of an alkene involves protonation of the pi bond to produce an intermediate carbocation. Acid is required because the C-C bond is not polar. There is no C-C bond dipole to exert a Coulombic attraction toward a water molecule. Protonation produces a positive charge on one of the carbon atoms, which then attracts a lone pair of electrons on the oxygen atom of a water molecule. Since the pi bond in formaldehyde is polarized, water will be attracted to the positive end of the C-O bond dipole without the need for an acid.

The same is true of other aldehydes and ketones. But, as the bulk of the substituents attached to the carbonyl carbon increases, access to that carbon becomes more difficult and the rate of addition becomes slower. More importantly, the value of Keq becomes smaller.

In the same way that acid catalyses the addition of water to an alkene, it will also increase the rate of addition of water (and alcohols) to an aldehyde or ketone. Figure 2 compares the two processes.

Figure 2

Addition of Water to C=O and C=C Bonds

The reactions of aldehydes and ketones with alcohols parallel their reactions with water. Figure 3 illustrates both processes.

Figure 3

### Introduction

We have seen that alcohols undergo acid catalysed nucleophilic addition to aldehydes and ketones to produce hemiacetals, acetals, hemiketals, and ketals as summarized in Figure 1.

Figure 1

Nucleophilic Addition of Alcohols to Aldehydes and Ketones

Amines undergo similar acid catalysed nucleophilic addition reactions. The following discussion is limited to the reactions of primary amines with aldehydes and ketones. Primary amines are amines in which the nitrogen atom is bonded to 1 carbon atom. Figure 2 presents several examples of primary amines.

Figure 2

Three Primary Amines

More important than the number of carbons attached to the nitrogen is the number of hydrogens. The reason for this becomes apparent when we consider, step-by-step, what happenes upon addition of a primary amine to an aldehyde or ketone. As a simple example, consider the reaction of methylamine with acetaldehyde. This reaction could be performed by dissolving acetaldehyde and methylamine in aqueous acid. Under those reaction conditions, the equilibria shown in Equations 1 and 2 would be established.

The trick here is to adjust the pH of the solution so that some of the aldehyde will be protonated while some of the amine is unprotonated. Protonating the aldehyde makes the carbonyl carbon more electrophilic, thus increasing its reactivity toward the nucleophilic nitrogen of an unprotonated methylamine. Figure 3 outlines the complete reaction.

Figure 3

Nucleophilic Addition of Methylamine to Acetaldehyde

The ammonium ion 1 enclosed in the box in Step 2 of the process is analogous to the oxonium ion produced in the reaction of acetaldehyde with methanol. The resonance contributor 2 highlighted in Step 4 parallels that generated during the formation of acetaldehyde dimethylacetal. While these two structures are very similar, the products they yield are very different, as Figure 4 indicates.

Figure 4

Alternative Fates for Similar Structures

The difference stems from the fact that in intermediate 2 the nitrogen atom has an exchangeable H attached to it while the oxygen atom in intermediate 3 does not. Formation of the carbon-nitrogen double bond by deprotonation of the nitrogen atom is simply the most likely fate of intermediate 2. Since intermediate 3 does not have a comparable pathway available to it, an alternative reaction occurs: the electron deficient carbon gains a pair of electrons by forming a bond with another methanol molecule.

### Examples

Equation 1 outlines the reaction of a cyclic ketone with a type of amine called a hydrazine. (In this equation TBS represents an OH protecting group while Ar stands for an aromatic ring.) Although hydrazines are technically not primary amines, they possess the more essential feature required for the formation of imines: two hydrogen atoms attached to the terminal nitrogen atom. (What happens to those two hydrogens and the oxygen atom of the carbonyl group?)

Reaction 1 constituted an early step in the first synthesis of taxol. The product of reaction 1 is a special type of imine called a hydrazone.

Before the development of modern spectroscopic techniques, hydrazones and related compounds such as oximes and semicarbazides played important roles in the characterization of the structures of aldehydes and ketones. Equations 2-4 illustrate the formation of one compound of each type.

Imines are formed as intermediates in the Strecker synthesis of amino acids. This reaction sequence begins with the reaction of an aldehyde with ammonia to produce an imine. The imine then reacts with cyanide ion to form an a-aminonitrile. Hydrolysis of the nitrile group yields the amino acid. The overall sequence is outlined in Figure 5 for the amino acid phenylalanine.

Figure 5

The Strecker Synthesis of Phenylalanine

As a final example, Figure 6 depicts the formation of an imine from the reaction of pyridoxal-5'-phosphate with the amino group of the aspartic acid.

Figure 6

An Imine Intermediate in a Biochemical Decarboxylation

Pyridoxal-5'-phosphate is the coenzyme form of vitamin B6. It is involved in a variety of important biochemical transformations. In the present case, the imine intermediate undergoes loss of carbon dioxide, followed by a series of proton transfers, to produce another amino acid, alanine. Note that the reaction sequence is catalytic in that pyridoxal-5'-phosphate is regenerated.

#### Introduction-

In most stable organic compounds bonds to carbon are either non-polar (C-C) or they have a bond dipole in which the carbon atom is electron deficient (C-X). Figure 1 summarizes the normal state of affairs.

Figure 1

Bond Polarity in C-C and C-X Bonds

The basic idea behind synthetic organic chemistry is simple: mix a compound which contains an electron rich carbon with one that contains an electron deficient carbon and Coulomb's Law will do the rest. The problem lies in making compounds that contain an electron rich carbon atom, i.e. a nucleophilic carbon. One of the most common ways to do so is to convert an alkyl halide into an organometallic compound. The Wittig reaction offers another approach.

#### The Wittig Reaction-

The Wittig reaction is a one-flask, 3-step sequence that converts aldhydes and ketones into alkenes. The three steps are:

1. Reaction of an alkyl halide with a tertiary phosphine to produce a phosphonium salt.

2. Deprotonation of the phosphonium salt to produce an ylide.

3. Nucleophilic addition of the ylide to an aldehyde or ketone.

Figure 2 illustrates each of these steps.

Figure 2

The Three Steps of the Wittig Reaction

The first step of the sequence involves an Sn2 reaction in which the phosphorous displaces the bromine from the methyl bromide. (As such, it is subject to all the usual limitations of the Sn2 mechanism.) The resulting phosphonium salt generally precipitates from the reaction mixture as a white solid. The positive charge on the phosphorous atom of this salt pulls electron density away from the C-H bonds of the methyl group, making those hydrogens more acidic. The pKa of the methyl protons in methyl triphenylphosphonium bromide is approximately 15. Addition of a strong base, in this case n-butyl lithium, deprotonates the methyl group. The carbanion that is produced is called an ylide. The negatively charged carbon gains stabilization by donating electron density into a vacant d orbital on the phosphorous atom:

Here's a photo of a simple ylide:

As the phosphonium salt reacts with the n-butyl lithium, it disappears and an orange solution is formed. Addition of the ketone to this solution, followed by a brief period of reflux, produces the alkene along with a white precipitate of triphenylphosphine oxide.

While the resonance structure on the right above suggests that the carbon atom is neutral, the structure on the left indicates that the electron density on the carbon is high, i.e. the carbon is nucleophilic. The ability of the phosphorous to accomodate the negative charge that develops when the carbon is deprotonated is what makes phosphonium salts suitable precursors of nucleophilic carbon. Ammonium salts cannot afford comparable resonance stabilization, and hence are not viable sources of nucleophilic carbon atoms. This difference is discussed in more detail in Phosphines vs. Amines.

The value of the Wittig reaction lies in its generality. It works well with aliphatic and aromatic aldehydes and ketones. Furthermore, these compounds may contain other functional groups such as alcohols and esters which are not compatible with Grignard reagents.

Is it possible to accomplish the synthesis outlined in Figure 2 using Grignard chemistry? Reaction of cyclohexanone with methyl magnesium bromide would produce 1-methylcyclohexanol. But dehydration of this alcohol using concentrated sulfuric or phosphoric acid may produce 1-methylcyclohexene and/or methylenecyclohexane. Being a trisubstituted alkene, 1-methylcyclohexene is more stable than methylenecyclohexane, and it is the preferred product. Figure 3 summarizes this alternative.

Figure 3

An Alternative to the Wittig Reaction

### Summary-

The Wittig reaction converts aldehydes and ketones into alkenes.

### Examples-

Scheme 1 describes two steps in the total synthesis of monensin.

Scheme 1

The desired cis alkene was formed in approximately 70% yield along with about 20% of the undesired trans isomer. Note the use of the cyclic ketal as a protecting group during the last step of the sequence.

A similar strategy was utilized during the total synthesis of racemic progesterone as outlined in Scheme 2.

Scheme 2

The Wittig reaction generally gives a mixture of cis and trans isomers. In this instance, the cis/trans mixture was treated with additional phenyl lithium to isomerize the cis alkene to the more stable trans configuration. (The methanol is merely a source of protons.)

An interesting example of the Wittig reaction comes from a synthesis of leukotriene A4 as shown in Equation 1.

The starting material in this reaction is a cyclic hemiacetal. Under the reaction conditions it exists in equilibrium with a small amount of the corresponding hydroxy aldehyde. The squiggly line between the carboethoxy group and the double bond indicates that the product is a mixture of the cis and trans isomers.

NUCLEOPHILIC ADDITION / ELIMINATION IN THE REACTION BETWEEN ACYL CHLORIDES AND AMINES

This page gives you the facts and a simple, uncluttered mechanism for the nucleophilic addition / elimination reaction between acyl chlorides (acid chlorides) and amines. If you want the mechanism explained to you in detail, there is a link at the bottom of the page.Ethanoyl chloride is taken as a typical acyl chloride. Any other acyl chloride will behave in the same way. Simply replace the CH3 group in what follows by anything else you want.

Similarly, ethylamine is taken as a typical amine. Any other amine will behave in the same way. Replacing the CH3CH2 group by any other hydrocarbon group won't affect the mechanism in any way.

The reaction between ethanoyl chloride and ethylamine

The facts

Ethanoyl chloride reacts violently with a cold concentrated solution of ethylamine. A white solid product is formed which is a mixture of N-ethylethanamide (an N-substituted amide) and ethylammonium chloride.

Notice that, unlike the reactions between ethanoyl chloride and water or ethanol, hydrogen chloride isn't produced - at least, not in any quantity. Any hydrogen chloride formed would immediately react with excess ethylamine to give ethylammonium chloride.

The mechanism

The first stage (the addition stage of the reaction) involves a nucleophilic attack on the fairly positive carbon atom by the lone pair on the nitrogen atom in the ethylamine.

The second stage (the elimination stage) happens in two steps. In the first, the carbon-oxygen double bond reforms and a chloride ion is pushed off.

` `

That is followed by removal of a hydrogen ion from the nitrogen. This might happen in one of two ways: it might be removed by a chloride ion, producing HCl (which would immediately react with excess ethylamine to give ethylammonium chloride as above) . . .

and

. . . or it might be removed directly by an ethylamine molecule.

The ethylammonium ion, together with the chloride ion already there, makes up the ethylammonium chloride formed in the reaction.

` `
`THE NUCLEOPHILIC SUBSTITUTION REACTIONS BETWEEN HALOGENOALKANES AND AMMONIA`

This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and ammonia to produce primary amines. If you want the mechanisms explained to you in detail, there is a link at the bottom of the page. If you are interested in further substitution reactions, you will also find a link to a separate page dealing with these.

The reaction of primary halogenoalkanes with ammonia

Important!  If you aren't sure about the difference between primary, secondary and tertiary halogenoalkanes

The facts

The halogenoalkane is heated with a concentrated solution of ammonia in ethanol. The reaction is carried out in a sealed tube. You couldn't heat this mixture under reflux, because the ammonia would simply escape up the condenser as a gas.

We'll talk about the reaction using 1-bromoethane as a typical primary halogenoalkane.

The reaction happens in two stages. In the first stage, a salt is formed - in this case, ethylammonium bromide. This is just like ammonium bromide, except that one of the hydrogens in the ammonium ion is replaced by an ethyl group.

There is then the possibility of a reversible reaction between this salt and excess ammonia in the mixture.

The ammonia removes a hydrogen ion from the ethylammonium ion to leave a primary amine - ethylamine.The more ammonia there is in the mixture, the more the forward reaction is favoured.

Note:  You will find considerable disagreement in textbooks and other sources about the exact nature of the products in this reaction. Some of the information you'll come across is simply wrong!

The mechanism

The mechanism involves two steps. The first is a simple nucleophilic substitution reaction:

Because the mechanism involves collision between two species in this slow step of the reaction, it is known as an SN2 reaction.

Note:  Unless your syllabus specifically mentions SN2 by name, you can just call it nucleophilic substitution.

In the second step of the reaction an ammonia molecule may remove one of the hydrogens on the -NH3+. An ammonium ion is formed, together with a primary amine - in this case, ethylamine.

This reaction is, however, reversible. Your product will therefore contain a mixture of ethylammonium ions, ammonia, ethylamine and ammonium ions. Your major product will only be ethylamine if the ammonia is present in very large excess.

Unfortunately the reaction doesn't stop here. Ethylamine is a good nucleophile, and goes on to attack unused bromoethane. This gets so complicated that it is dealt with on a separate page. You will find a link at the bottom of this page.

The reaction of tertiary halogenoalkanes with ammonia

The facts

The facts of the reactions are exactly the same as with primary halogenoalkanes. The halogenoalkane is heated in a sealed tube with a solution of ammonia in ethanol.

For example:

Followed by:

The mechanism

This mechanism involves an initial ionisation of the halogenoalkane:

followed by a very rapid attack by the ammonia on the carbocation (carbonium ion) formed:

This is again an example of nucleophilic substitution.

This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction.

There is a second stage exactly as with primary halogenoalkanes. An ammonia molecule removes a hydrogen ion from the -NH3+ group in a reversible reaction. An ammonium ion is formed, together with an amine.

It is very unlikely that any of the current UK-based syllabuses for 16 - 18 year olds will ask you about this. In the extremely unlikely event that you will ever need it, secondary halogenoalkanes use both an SN2 mechanism and an SN1.

Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to.

THE REACTION BETWEEN METHANE AND BROMINE

This page gives you the facts and a simple, uncluttered mechanism for the free radical substitution reaction between methane and bromine. If you want the mechanism explained to you in detail, there is a link at the bottom of the page.

The facts

This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction - a reaction brought about by light.

Note:  These reactions are sometimes described as examples of photocatalysis - reactions catalysed by light. It is better to use the term "photochemical" and keep the keep the word "catalysis" for reactions speeded up by actual substances rather than light.

CH4  +  Br2CH3Br  +  HBr

The organic product is bromomethane.

One of the hydrogen atoms in the methane has been replaced by a bromine atom, so this is a substitution reaction. However, the reaction doesn't stop there, and all the hydrogens in the methane can in turn be replaced by bromine atoms. Multiple substitution is dealt with on a separate page, and you will find a link to that at the bottom of this page.

Warning!  Check your self at this point. If your  want  to know about the free radical substitution reaction between methane and chlorine as well as this one, don't waste time trying to learn both mechanisms.

The two mechanisms are identical. You just need to learn one of them. If you are asked for the other one, all you need to do is to write bromine, say, instead of chlorine.

In writing the bromine mechanisms on these pages, that's exactly what I've done! If you read both chlorine and bromine versions, you'll find them boringly repetitive!

The mechanism

The mechanism involves a chain reaction. During a chain reaction, for every reactive species you start off with, a new one is generated at the end - and this keeps the process going.

Species:  a useful word which is used in chemistry to mean any sort of particle you want it to mean. It covers molecules, ions, atoms, or (in this case) free radicals.

The over-all process is known as free radical substitution, or as a free radical chain reaction.

Chain initiation
The chain is initiated (started) by UV light breaking a bromine molecule into free radicals.

Br22Br
Chain propagation reactions
These are the reactions which keep the chain going.

CH4  +  BrCH3  +  HBr

CH3  +  Br2CH3Br  +  Br

Chain termination reactions
These are reactions which remove free radicals from the system without replacing them by new ones.

2BrBr2

CH3  +  BrCH3Br

CH3  +  CH3 CH3CH3

THE NUCLEOPHILIC SUBSTITUTION REACTIONS BETWEEN HALOGENOALKANES AND CYANIDE IONS

This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and cyanide ions (from, for example, potassium cyanide). If you want the mechanisms explained to you in detail, there is a link at the bottom of the page.

Important!  If you aren't sure about the difference between primary, secondary and tertiary halogenoalkanes, it is essential that you follow this link before you go on.

The facts

If a halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by a -CN group and a nitrile is produced. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture.

The solvent is important. If water is present you tend to get substitution by -OH instead of -CN.

Note:  A solution of potassium cyanide in water is quite alkaline, and contains significant amounts of hydroxide ions. These react with the halogenoalkane.

For example, using 1-bromopropane as a typical primary halogenoalkane:

You could write the full equation rather than the ionic one, but it slightly obscures what's going on:

The bromine (or other halogen) in the halogenoalkane is simply replaced by a -CN group - hence a substitution reaction. In this example, butanenitrile is formed.

Note:  When you are naming nitriles, you have to remember to include the carbon in the -CN group when you count the longest chain. In this example, there are 4 carbons in the longest chain - hence butanenitrile.

The mechanism

Here is the mechanism for the reaction involving bromoethane:

This is an example ofnucleophilic substitution.

Because the mechanism involves collision between two species in the slow step (in this case, the only step) of the reaction, it is known as an SN2 reaction.

Note:  Unless your ask a specifically mentions SN2 by name, you can just call it nucleophilic substitution.

f your examiners want you to show the transition state, draw the mechanism like this:

The facts

The facts of the reaction are exactly the same as with primary halogenoalkanes. If the halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol, the halogen is replaced by -CN, and a nitrile is produced.

For example:

Or if you want the full equation rather than the ionic one:

The mechanism

This mechanism involves an initial ionisation of the halogenoalkane:

followed by a very rapid attack by the cyanide ion on the carbocation (carbonium ion) formed:

This is again an example of nucleophilic substitution.

This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an SN1 reaction.

The facts

The facts of the reaction are exactly the same as with primary or tertiary halogenoalkanes. The halogenoalkane is heated under reflux with a solution of sodium or potassium cyanide in ethanol.

For example:

The mechanism

Secondary halogenoalkanes use both SN2 and SN1 mechanisms. For example, the SN2 mechanism is:

Should you need it, the two stages of the SN1 mechanism are:

Reactions of Arenes. Electrophilic Aromatic Substitution

Substituent Effects

Here is a table that shows the effect of substituents on a benzene ring have on both the rate and orientation of electrophilic aromatic substitution reactions.

Study Tip:

This is a VERY important table ! It is worth knowing.... your best understanding will come if you learn HOW it works.
It's application goes way beyond electrophilic aromatic substitution reactions.

Key concepts to review ?  Resonance and electronegativity

These effects are a combination of RESONANCE and INDUCTIVE effects
The effects are also important in other reactions and properties (e.g. acidity of the substituted benzoic acids).

Here are some general pointers for recognising the substituent effects:

• The H atom is the standard and is regarded as having no effect.

• Activating groups increase the rate

• Deactivating groups decrease the rate

• EDG = electron donating group

• EDG can be recognised by lone pairs on the atom adjacent to the π system, eg: -OCH3

• except -R, -Ar or -vinyl (hyperconjugation, π electrons)

• EWG = electron withdrawing group

• EWG can be recognised either by the atom adjacent to the π system having several bonds to more electronegative atoms, or

• having a formal +ve or δ +ve charge, eg: -CO2R, -NO2
• EDG / activating groups direct ortho / para

• EWG / deactivating groups direct meta

• except halogens (-X) which are deactivating BUT direct ortho / para

• EDG add electron density to the π system making it more nucleophilic

• EWG remove electron density from the π system making it less nucleophilic.

There are two main electronic effects that substituents can exert:

RESONANCE effects are those that occur through the pi system and can be represented by resonance structures. These can be either electron donating (e.g. -OCH3) where pi electrons are pushed toward the arene or electron withdrawing (e.g. -C=O) where pi electrons are drawn away from the arene.

In certain cases, molecules can be represent by more than one reasonable Lewis structure that differ only in the location of π electrons.
Electrons in σ bonds have a fixed location and so they are said to be localised.
In contrast,  π electrons that can be drawn in different locations are said to be delocalised.
Collectively these Lewis diagrams are then known as resonance structures or resonance contributors or resonance canonicals.
The "real" structure has characteristics of each of the contributors, and is often represented as the resonance hybrid (think of a hybrid breed which is a mixed breed).  In a way, the resonance hybrid is a mixture of the contributors.

(note that a resonance hybrid cannot normally be written as an individual Lewis diagram !).

You should be able to draw all reasonable resonance structures for a given organic molecule.

The best way to "derive" resonance structures is by learning to "push" curly arrows and starting from a reasonable Lewis structure.

INDUCTIVE effects are those that occur through the sigma system due to electronegativity effects.  These too can be either electron donating electron donating (e.g. -Me) where sigma electrons are pushed toward the arene or electron withdrawing (e.g. -CF3, +NR3) where sigma electrons are drawn away from the arene.

• Electronegativity is defined as the ability of an atom to attract electrons towards itself.

• It is one of the most important properties for rationalising and predicting reactivity etc.

• The partial periodic table below has the Pauling electronegativities of some key elements.

 H 2.1 He Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 Ne Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.0 Ar K 0.8 Ca 1.0 Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br 2.8 Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I 2.5 Xe
• Electronegativity increases left to right across a row in the periodic table   e.g. C < N < O < F
(as you move left to right nuclear charge increases so there is a greater attraction for electrons)

• Electronegativity decreases as you move down a group in the periodic table  e.g. F > Cl > Br > I
(each step down a group increases the atomic radii as a "new shell" of electrons are added and the nuclear charge is further shielded by the core electrons, both factors decrease the attraction for electrons)

• F is the most electronegative element

• Metals, e.g. Li and Mg, are less electronegative than C  (i.e. metals are electropositive compared to C)

A simplified approach to understanding substituent effects is provided, based on the "isolated molecule approach".  The text (as do most others) uses the more rigourous approach of drawing the resonance structures for each of the intermediate carbocations formed by attack at each of the o-, m-  and p- positions and looking at how the initial substituent influences the stability of the system.

We are going to break down the types of substituents into various subgroups based on the structural features of the substituent immediately adjacent to the aromatic ring:

• type 1 = substituents with lone pairs (e.g. -OCH3, -NH2) on the atoms adjacent to the pisystem.

• type 2 = substituents that are CH systems (i.e. -alkyl, -vinyl or -aryl).

• type 3 = substituents that are C=C systems (i.e. -vinyl or -aryl).

• type 4 = substituents with pi bonds to electronegative atoms (e.g. -C=O, -CF3, -NO2)

• type 5 = substituents with several bonds to electronegative atoms (e.g.  -CF3)

• type 6 = substituents that are halogens systems (i.e. -F, -Cl, -Br, -I)

Resonance

In chemistry is a tool used to represent and model certain types of non-classical molecular structures. Resonance is a key component of valence bond theory and arises when no single conventional model using only even number of electrons shared exclusively by two atoms can actually represent the observed molecule. There are two closely related but useful-to-distinguish meanings given to the term resonance. One of these has to do with diagrammatic representation of molecules using Lewis structures while the other has to do with the mathematical description of a molecule using valence bond theory. In both cases, resonance involves representing or modeling the structure of a molecule as an intermediate, average (a resonance hybrid) between several simpler but incorrect structures.

History

The concept of resonance was introduced by Linus Pauling in 1928. The term "resonance" came from the analogy between the quantum mechanical treatment of the H2 molecule and a classical system consisting of two coupled oscillators. In the classical system, the coupling produces two modes, one of which is lower in frequency than either of the uncoupled vibrations; quantum-mechanically, this lower frequency is interpreted as a lower energy. The alternative term mesomerism popular in German and French publications with the same meaning was introduced by Christopher Ingold in 1938 but did not catch on in the English literature. The current concept of mesomeric effect has taken on a related but different meaning. The double headed arrow was introduced by the German chemist Arndt (also responsible for the Arndt-Eistert synthesis) who preferred the German phrase zwischenstufe or intermediate phase.

Due to confusion with the physical meaning of the word resonance, as no elements actually appear to be resonating, it has been suggested that the term resonance be abandoned in favor of delocalization.[2] Resonance energy would become delocalization energy and a resonance structure becomes a contributing structure. The double headed arrows would be replaced by commas.

Examples

Scheme 2. Examples of resonance ozone, benzene and the allyl cation

The ozone molecule is represented by two resonance structures in the top of scheme 2. In reality the two terminal oxygen atoms are equivalent and the hybrid structure is drawn on the right with a charge of -1/2 on both oxygen atoms and partial double bonds.

The concept of benzene as a hybrid of two conventional structures (middle scheme 2) was a major breakthrough in chemistry made by Kekulé, and the two forms of the ring which together represent the total resonance of the system are called Kekulé structures. In the hybrid structure on the right the circle replaces three double bonds.

Resonance as a diagrammatic tool

Scheme 1. Resonance structures of benzene

A single Lewis structure often cannot represent the true electronic structure of a molecule. While one can only show an integral number of covalent bonds between two and only two atoms using these diagrams, one often finds that the experimentally deduced or calculated (from Quantum mechanics) structure of the molecule does not match any of the possible Lewis structures but rather has properties in some sense intermediate to these. Resonance structures are then employed to approximate the true electronic structure. Take the example of benzene (shown above, right). In a Lewis diagram, two carbons can be connected by one or two covalent bonds, but in the observed benzene molecule the bond lengths are longer than double bonds yet shorter than single bonds. More importantly, they are all equivalent, a fact no Lewis structure can explain. Therefore one calls the two Lewis structures canonical, contributing or resonating structures and the real molecule is considered to be their average, called a resonance hybrid. Resonance structures of the same molecule are connected with a double-headed arrow.

This form of resonance is simply a way of representing the structure graphically. It is only a notation and does not represent a real phenomenon. The individual resonance structures do not exist in reality: the molecule does not inter-convert between them. Instead, the molecule exists in a single unchanging state, intermediate between the resonance structures and only partially described by any one of them. This sharply distinguishes resonance from tautomerism. When it is said that a molecule is stabilized by resonance or that amides are less basic because the lone pair on nitrogen is involved in resonance with the carbonyl pi electron, no phenomenon is implied. What is meant is simply that the molecule behaves differently from what we expect by looking at its Lewis structure because the structure diagrammed does not represent the actual structure of the molecule. From this viewpoint, the terminology treating resonance as something that 'happens' is perhaps an unfortunate historical burden.

It is also not correct to say that resonance occurs because electrons "flow", "circulate", or change their place within the molecules. Such a behavior would produce a magnetic field, an effect that is not observed in reality. However, a phenomenon of this sort may be induced by the application of an external magnetic field perpendicular to the plane of an aromatic ring: this causes the appearance of an opposing magnetic field, demonstrating that that the delocalised pi electrons are truly flowing. The applied magnetic field induces a current density ("ring current") of circulating electrons in the pi system; this current in turn induces a magnetic field. A common manifestation of this effect is the large chemical shift observed in the NMR spectrum of aromatic structures.

A vector Analogy

An accurate analogy of resonance is given by the algebra of vectors. A vector r is written in component form as xi+yj+zk where x, y, and z are components and i, j, and k are the standard orthogonal Cartesian unit vectors. Just as the real vector r is neither i, nor j, nor k, but rather a combination of all three, a resonance hybrid is a conceptual combination of resonance structures. x, y, and z have no independent existence; they are considered only as a decomposition of r into easier-to-handle components, as is the case with resonance structures. In fact this analogy is very close to the reality, as will be made clear in the following section.

True nature of Resonance

Though resonance is often introduced in such a diagrammatic form in elementary chemistry, it actually has a deeper significance in the mathematical formalism of valence bond theory (VB). When a molecule cannot be represented by the standard tools of valence bond theory (promotion, hybridisation, orbital overlap, sigma and pi bond formation) because no single structure predicted by VB can account for all the properties of the molecule, one invokes the concept of resonance.

Valence bond theory gives us a model for benzene where each carbon atom makes two sigma bonds with its neighbouring carbon atoms and one with a hydrogen atom. But since carbon is tetravalent, it has the ability to form one more bond. In VB it can form this extra bond with either of the neighbouring carbon atoms, giving rise to the familiar Kekulé ring structure. But this cannot account for all carbon-carbon bond lengths being equal in benzene. A solution is to write the actual wavefunction of the molecule as a linear superposition of the two possible Kekulé structures (or rather the wavefunctions representing these structures), creating a wavefunction that is neither of its components but rather a superposition of them, just as in the vector analogy above (which is formally equivalent to this situation).

In benzene both Kekulé structures have equal weight, but this need not be the case. In general, the superposition is written with undetermined constant coefficients, which are then variationally optimized to find the lowest possible energy for the given set of basis wavefunctions. This is taken to be the best approximation that can be made to the real structure, though a better one may be made with addition of more structures.

In molecular orbital theory, the main alternative to VB, resonance often (but not always) translates to a delocalization of electrons in pi orbitals (which are a separate concept from pi bonds in VB). For example, in benzene, the MO model gives us 6 pi electrons completely delocalised over all 6 carbon atoms, thus contributing something like half-bonds. This MO interpretation has inspired the picture of the benzene ring as a hexagon with a circle inside. Often when describing benzene the VB picture and the MO picture are intermixed, talking both about sigma 'bonds' (strictly a concept from VB) and 'delocalized' pi electrons (strictly a concept from MO).

Resonance Energy

Resonance hybrids are always more stable than any of the canonical structures would be, if they existed[1]. The delocalization of the electrons lowers the orbital energies, imparting this stability. The resonance in benzene gives rise to the property of aromaticity. The gain in stability of the resonance hybrid over the most stable of the (non-existent) canonical structures is called the resonance energy. A canonical structure that is lower in energy makes a relatively greater contribution to the resonance hybrid, or the actual picture of the molecule. In fact, resonance energy, and consequently stability, increase with the number of canonical structures possible, especially when these (non-existent) structures are equal in energy. Resonance energy of a conjugated system can be 'measured' by heat of hydrogenation of the molecule. Consider the example of benzene. The energy required to hydrogenate an isolated pi-bond is around 28.6 kcal/mol (120 kJ/mol). Thus, according to the VB picture of benzene (having three pi-bonds), the complete hydrogenation of benzene should require 85.8 kcal/mol (360 kJ/mol). However, the experimental heat of hydrogenation of benzene is around 49.8 kcal/mol (210 kJ/mol). The difference of 36 kcal/mol (150 kJ/mol) can be looked upon as a measure of resonance energy. One must bear in mind again that resonance structures have no physical existence. So, even though the term 'resonance energy' is quite meaningless, it offers an insight into how different the VB picture of a molecule is from the actual molecule itself. The resonance energy can be used to calculate electronegativities on the Pauling scale.

Writing Resonance Structures

1.   Position of nuclei must be the same in all structures, otherwise they would be isomers with real existence.

2.   Total number of electrons and thus total charge must be constant.

3.   When separating charge (giving rise to ions), usually structures where negative charges are on less electronegative elements have little contribution, but this may not be true if additional bonds are gained.

4.   Resonance hybrids can not be made to have lower energy than the actual molecules.

Reactive Intermediates

Often, reactive intermediates such as carbocations and free radicals have more delocalised structure than their parent reactants, giving rise to unexpected products. The classical example is allylic rearrangement. When 1 mole of HCl adds to 1 mole of 1,3-butadiene, in addition to the ordinarily expected product 3-chloro-1-butene, we also find 1-chloro-2-butene. Isotope labelling experiments have shown that what happens here is that the additional double bond shifts from 1,2 position to 2,3 position in some of the product. This and other evidence (such as NMR in superacid solutions) shows that the intermediate carbocation must have a highly delocalised structure, different from its mostly classical (delocalisation exists but is small) parent molecule. This cation (an allylic cation) can be represented using resonance, as shown above.

This observation of greater delocalisation in less stable molecules is quite general. The excited states of conjugated dienes are stabilised more by conjugation than their ground states, causing them to become organic dyes.

An well-studied example of delocalisation that does not involve pi electrons (hyperconjugation) can be observed in the non-classical ion norbornyl cation. Other examples are diborane and methanium (CH5+). These are known as 3-center-2-electron bonds and are represented either by resonance structures involving rearrangement of sigma electrons or by a special notation, a Y that has the three nuclei at its three points.

# Structure and Bonding

## Basic Principles

What are molecules made from? They are made from atoms, which are themselves made from nuclei and electrons. These building blocks carry an electrical charge: nuclei are positively charged, and electrons are negatively charged. The nuclei themselves are made up of (positively charged) protons and (neutral) neutrons. This is all summarised on the following picture:

Different types of atom have different numbers of protons, neutrons, and electrons. For example, carbon atoms have 6 protons, 6 neutrons, and 6 electrons.

Charged species interact with each other: like charges (+ and + or - and -) repel each other, opposite charges (- and +) attract each other. This well-known principle from physics is summarised by Coulomb's law:

(Here, F is the force between the two charges; ε0 is a constant (not important here), q1 and q2 are the values of the charges involved, and r is the distance between them.)

This force between charged species is central to all of chemistry, and in particular to all the types of bonding we will discuss.

First, it explains how atoms hold together: the negatively charged electrons are attracted to the positively charged nucleus more than they are repelled by the other electrons.

There is a fine balance between the attractive force holding the electrons close to the nucleus, and the repulsive force which tends to keep electrons away from each other. The result of this competition between attractive and repulsive charge-charge interactions is what explains the detailed structure of atoms. The electrons in atoms tend to form into concentric shells. For the hydrogen atom, with just one electron and one proton (Z = 1), the electron sits in the first shell, as shown here:

The nucleus is shown in purple. Also shown is the structure of the Helium atom, with two protons, two neutrons (all shown together as the purple nucleus), and 2 electrons (i.e., Z = 2). Both electrons sit in the first shell.

For elements with more electrons, there is no more room in the first shell, and so a second shell is occupied. This is shown below for carbon (Z = 6) and oxygen (Z = 8).

Above 10 electrons, the second shell contains eight electrons and is full. For the elements beyond (starting with sodium, Z = 11), the last electrons therefore sit in the third shell, as shown here for sodium and chlorine (Z = 17):

This structural description leads naturally to an important property of atoms, the octet rule: atoms have a strong tendency to lose, gain, or share electrons if this leads to them having a complete shell of electrons around them. In other words, atoms prefer to have a total of 2, 10, or 18 electrons around them.

# Structure and Bonding in Chemistry

## Ionic Bonds

Elements in the first few columns of the periodic table have a few more electrons than predicted by the octet rule: they therefore lose the electrons in the outermost shells fairly easily. For example, the alkali metals (group I), such as sodium (Na) or potassium (K), which have, respectively, 11 and 19 electrons, easily lose one electron to form monopositive ions, Na+ and K+. These ions have 10 and 18 electrons, respectively, so they are quite stable according to the octet rule.

Elements in the last few columns of the periodic table have one, two or three fewer electrons than predicted by the octet rule: they therefore gain electrons fairly easily. For example, the halogens (group VII), such as fluorine (F) or chlorine (Cl), which have, respectively, 9 and 17 electrons, easily gain one electron to form mononegative ions, F- or Cl-. These ions have 10 and 18 electrons, respectively.

Likewise, elements in group II form doubly positive ions such as Mg++ or Ca++, and elements in group VI form doubly negative ions such as O-- or S--. All these ions obey the octet rule and so are fairly stable.

Now, imagine what will happen when one sodium atom meets one chlorine atom: the sodium atom will lose one electron to give Na+, and the chlorine atom will gain that electron to give Cl-. This can be represented schematically in the following way:

The resulting ions, which have Opposite charges, will be attracted to one another, and will draw closer, until they "touch". This happens when the inner shell of electrons on the sodium ion (shown in blue) starts to overlap with the outer shell of electrons on the chloride anion (shown in green). This pair of ions looks something like this:

It is possible to determine where the valence electrons are situated in this pair of ions. They are almost entirely situated on the chlorine atom, as expected: the sodium atom has lost its only valence (3s) electron, whereas chlorine has gained an electron and has the 3s23p6 valence configuration. The blue transparent surface on this picture encloses the region of space where the valence electrons spend most of the time:

NaCl, or sodium chloride, is however more complicated than this! This is because charge-charge interaction occurs in all directions. Once an Na+ cation has attracted a Cl- anion in one direction, it can attract another in a different direction. So two pairs of ions such as above can come together to form a species with four ions in total, all placed so as to interact favourably with ions of opposite charge:

Here, too, all the valence electrons sit on the chlorine atoms:

And this need not stop here... The next step is to get 8 ions, 4 each of sodium and chlorine:

The stable form of sodium chloride involves a very large number of NaCl units arranged in a lattice (or regular arrangement) millions of atoms across. Because the lattice is rigid, this means that one gets a solid: the ions do not move much one with respect to another. Also, because atoms are so small, even a small crystal of salt will have billions of sodium chloride units in it! The ions are arranged so that each positive (sodium) ion is close to many negative (chloride) ions, as shown on the following picture:

Can you count how many ions each sodium is next to? And how many ions each chlorine is next to? These pairs of ions in close contact are shown with lines joining them. These lines illustrate the strong ionic bonding between ions of opposite charge which are next to each other. However, you should remember that these close contacts are not the same as covalent bonds - there is no pair of electrons shared between the two atoms which are connected by the two lines. Also, there is some ionic bonding between ions which are further away from each other - ions of opposite charge always attract each other, however far they are from each other. Nevertheless, the force holding them together is largest when they are close together. The lines connecting ions in this lattice (and others below) are there to make it easier to detect the pairs of ions in close contact with each other.

Remember - atoms are very small! The distance between a sodium ion and its nearest chloride ion neighbours is about 3 ten-millionths of a millimeter. Imagine a cubic grain of salt with edges which are 3 tenths of a millimeter long. That means there will be a line of about a million ions along each edge. And the grain will contain one billion billion ions in total. If each ion was replaced by a ping-pong ball (roughly 3 centimeters in diameter), each edge would be one hundred million times longer. Instead of being 3 tenths of a millimeter wide, this "grain" would be roughly thirty kilometers (or twenty miles) wide!! Enough to cover most of London...

All ionic compounds adopt a similar three-dimensional structure in which the ions are close to many ions of the opposite charge. There are however several ways of doing this. Caesium chloride (CsCl), for example, adopts a different structure to that of NaCl, as shown on the following picture:

Can you count how many ions each caesium ion (pink) is in close contact with? And how many ions each chloride ion (green) is close to?

As another example, let us consider a salt with a divalent (doubly positive) ion, for example calcium fluorite, CaF2. This adopts the structure shown below (the calcium atoms are shown as large grey spheres, the fluorine atoms are smaller and orange):

Can you count how many fluoride ions each calcium is in close contact with? And how many caesium ions each fluoride ion is close to?

Experienced chemists can often predict the structure that a given ionic species will adopt, based on the nature of the ions involved. This means that it is often possible to design ionic compounds having certain well-defined and desirable properties. As an example, chemists have been able to make high-temperature superconductors, such as the complicated ionic compound, YBa2Cu3O4. This solid conducts electricity with no resistance at all at low temperature (below ca. -100 degrees centigrade). Previous superconductors only had this property at much lower temperatures. The lack of resistance makes superconductors very useful in a number of technological applications - e.g. in designing high-speed trains that levitate above the track!

The repeating structure of this solid is shown below (oxygen is large and red, barium large and yellow-ish, yttrium small and pink, and copper small and blue). Notice how many oxygen ions surround each barium and yttrium ion.

## Ionic Bonds - Conclusions

Ionic bonds form between elements which readily lose electrons and others which readily gain electrons. Because the interaction between charges as given by Coulomb's law is the same in all directions, ionic compounds do not form molecules. Instead, periodic lattices with billions of ions form, in which each ion is surrounded by many ions of opposite charge. Therefore, ionic compounds are almost always solids at room temperature. By careful consideration of the properties of each ion, it is possible to design ionic solids with certain well-defined and desirable properties, like superconductors.

# Structure and Bonding in Chemistry

## Covalent Bonds

In the previous page, we saw how atoms could achieve a complete shell of electrons by losing or gaining one or more electrons, to form ions. There is another way atoms can satisfy the octet rule: they can share electrons. For example, two hydrogen atoms can share their electrons, as shown below. Because each of the shared electrons then "belongs" to both atoms, both atoms then a fulled shell, with two electrons. The pair of shared electrons is symbolised by the heavy line between the atoms.

In terms of charge-charge interactions, what happens is that the shared electrons are located between the two bonded atoms. The force attracting them to both nuclei is stronger than the repulsive force between nuclei.

The methane (CH4) molecule illustrates a more complex example. Each of the 4 electrons in the outermost ("valence") shell of carbon is shared with one hydrogen. In turn, each of the hydrogens also shares one electron with carbon. Overall, carbon "owns" 10 electrons - satisfying the octet rule - and each hydrogen has 2. This is shown here:

When a molecule of methane is studied experimentally, it is found that the four hydrogens spread out evenly around the carbon atom, leading to the three-dimensional structure shown here:

As you would expect given that the electrons are shared, if we plot the region where the electrons sit, this is not localised on one atom, as it was for the ionic compounds, but is all over the molecule:

### Covalent Bonding and Isomers

As we have seen above, atoms can share electrons with others to form chemical bonds. This can also take place between two carbon atoms, to form a molecule such as ethane (C2H6):

When we add two more carbon atoms and 4 more hydrogens, to make butane (C4H10), an interesting situation arises: There are two different ways of bonding the carbons together, to form two different molecules, or isomers!!! These are shown below. For one of the isomers, the first carbon is bonded to three hydrogens, and to the second carbon, which is itself bonded to another two hydrogens and to the third carbon, which is itself bonded to the fourth carbon. In the other isomer, one of the carbons forms a bond to all three carbon atoms:

Larger compounds can also be formed, and they will have even more isomers! For example, this compound with 8 carbons is called isooctane, and is one of the main components of petrol for cars:

Can you check that the formula for this compound is C8H18? Can you sketch another compound with the same formula?

Because covalent bonds can be formed in many different ways, it is possible to write down, and to make, many different molecules. Many of these are natural compounds, made by living animals or plants within their cells. This example shows one such molecule, cholesterol (C27H46O), which can contribute to heart disease in people whose diet is too rich in fats:

Note that in this structure, two neighbouring carbon atoms appear to form only three bonds, which would go against the octet rule. In fact, these atoms bond by sharing two electrons each (a total of four electrons). In this way, they complete their electron shell like the others. This situation is referred to as a double bond, and is shown in the pop-up window as a thicker stick between those two atoms (Can you find this bond? Check that all other carbon atoms do form four bonds).

Other compounds are synthetic, they are made by chemists. Chemists can also make the natural compounds, starting from only simple things like methane and water. The "natural" molecules made in this way are identical to the "real" natural compounds! Other synthetic molecules do not exist in nature. They can have desirable properties, for example, many medicines are made in this way. An example of a "small" medicine molecule is aspirin, C9H8O4, shown below. In this molecule, two bonds between carbon and oxygen are double bonds, and are shown as thicker sticks in the model.

### Properties of Covalent Molecules

The covalent bonds between atoms in a given molecule are very strong, as strong as ionic bonds. However, unlike ionic bonds, there is a limit to the number of covalent bonds to other atoms that a given atom can form. For example, carbon can make four bonds - not more. Oxygen can form two bonds. As a result, once each atom has made all the bonds it can make, as in all the molecules shown above, the atoms can no longer interact with other ones. For this reason, two covalent molecules barely stick together. Light molecules are therefore gases, such as methane or ethane, above, hydrogen, H2, nitrogen, N2 (the main component of the air we breathe, etc. Heavier molecules, such as e.g. the isooctane molecule, are liquids at room temperature, and others still, such as cholesterol, are solids.

### Covalent Solids

As well as the solids just referred to, formed by piling lots of covalent molecules together, and relying on their slight "stickiness" to hold the solid together, one can also form solids entirely bound together by covalent bonds. An excellent example is diamond, which is pure carbon, with each carbon atom bonding to four others, to form a huge "molecule" containing many millions of millions of atoms. This shows a part of a diamond molecule:

In diamond, all the carbon atoms share one electron with each of their four neighbouring carbon atoms. There is another form in which pure carbon can be formed: graphite. This is the main component of the "lead" in pencils. Here, instead of each carbon having four neighbours, it only has three. Each carbon shares one electron with two of its neighbours, and 2 electrons with the third neighbour. In this way, one C-C bond out of three is a double bond. The atoms all bond together in planes, and the planes stack on top of each other as shown:

In graphite, the C-C bonds in the planes are very strong, but the force between the different planes is quite weak, and they can slip over one another. This explains the "soft" feel of graphite, and the fact that it is used as a lubricant, for example in motor oil.

### Other "Big" Covalent Molecules

In solids like diamond and graphite, the different atoms all bond to one another to form one very large molecule. The atoms are bonded to each other in all directions in diamond, and in two directions (within the planes) in graphite, with no bonding in the other direction. Some important covalent molecules involve atoms bonding to each other repeatedly along just one direction, with no bonds in the others. These are called polymers, and one simple example if polyethene (also called polythene, or polyethylene). The structure of polythene is shown here (the dangling bonds at each end indicate how the bonding should really continue for thousands of atoms on each side):

Polythene is what most plastic bags are made of. Other polymers include molecules such as nylon, teflon (these, like polythene, are man-made), or cellulose (the stuff that makes wood hard), a biological polymer.

## Covalent Bonds - Conclusions

Covalent bonds involve sharing electrons between atoms. The shared electrons "belong" to both atoms in the bond. Each atom forms the right number of bonds, such that they have filled shells. There is lots of flexibility in terms of which atom bonds to which other ones. This means that many isomeric molecules can be formed, and Nature as well as chemists are skilled at designing and making new molecules with desirable properties. In most cases, only a small number of atoms are bonded together to make a molecule, and there is no bonding between atoms in one molecule and other atoms in other molecules. This means that molecules are only very slightly "sticky" between themselves, and covalent compounds are either gases, or liquids, or sometimes solids. In some cases, bonding occurs to form large molecules with thousands or millions of atoms, and these can be solids.

# Structure and Bonding in Chemistry

## Other Types of Bonding

In the previous page, we have learned about the two most important types of bonding: ionic bonding and covalent bonding. Both of these are ultimately driven by the desire that atoms have to be surrounded by a complete shell of electrons. They achieve this by respectively either gaining or losing, or sharing, one or more electrons.

There are other principles which can lead to atoms bonding to each other, and we will examine here two important cases: metallic bonding, and hydrogen bonding.

### Metallic Bonding

Metals are well known to be solids (except for Mercury!). The bonds between metals can loosely be described as covalent bonds (due to sharing electrons), except that the metal atoms do not just share electrons with 1, 2, 3 or 4 neighbours, as in covalent bonding, but with many atoms. The structure of the metal is determined by the fact that each atom tries to be as close to as many other atoms as possible. This is shown here for one typical metal structure (adopted, for instance, by iron at some temperatures):

Can you count how many neighbours each iron atom is bonded to? Contrast this with the structure of diamond seen previously.

Because the electrons are shared with all the neighbours, it is quite easy for the electrons in metals to move around. If each "shared" electron shifts one atom to the right or left, this leads to a net shift of charge. This occurs quite easily in metals, but much less so in ionic solids, or covalent ones, where the electrons are rigidly associated with either a particular atom or ion, or a particular pair of atoms. It is because electrons can move around so easily inside metals that the latter conduct electricity.

### Hydrogen Bonding

In covalent bonds, the electrons are shared, so that each atom gets a filled shell. When the distribution of electrons in molecules is considered in detail, it becomes apparent that the "sharing" is not always perfectly "fair": often, one of the atoms gets "more" of the shared electrons than the other does.

This occurs, in particular, when atoms such as nitrogen, fluorine, or oxygen bond to hydrogen. For example, in HF (hydrogen fluoride), the structure can be described by the following "sharing" picture:

However, this structure does not tell the whole truth about the distribution of electrons in HF. Indeed, the following, "ionic" structure also respects the filled (or empty) shell rule:

In reality, HF is described by both these structures, so that the H-F bond is polar, with each atom bearing a small positive (δ+) or negative (δ-) charge. When two hydrogen fluoride molecules come close to each other, the like charges attract each other, and one gets a "molecule" of di-hydrogen fluoride as shown:

The weak "bond" between the F atom and the H is called a Hydrogen Bond, and is shown here as the dotted green line.

Hydrogen bonds can also occur between oxygen atoms and hydrogen. One of the most important types of hydrogen bonds is of this type, and is the one occurring in water. As discussed for HF, the electrons in H2O molecules are not evenly "shared": the oxygen atom has more of them than the hydrogen atoms. As a result, oxygen has a (partial) negative charge, and the hydrogens have a positive charge. When you have two water molecules close to another, a hydrogen atom on one of the molecules is attracted to the oxygen of the other molecule, to give a dimer. The structure of this dimer is shown here:

Notice how the oxygen, hydrogen, and oxygen atoms involved in the hydrogen bond are arranged more or less in a straight line. This is the preferred geometry for hydrogen bonds, and explains why only one hydrogen bond can be formed in the water dimer.

Upon going to three water molecules, it is now possible to form several hydrogen bonds. This is shown here:

How many hydrogen bonds is each water molecule involved in?

In liquid water or ice, many water molecules are close to each other, and they form dense networks of hydrogen bonds. In ice, the arrangement of the water molecules with respect to each other is regular, whereas in water, it is random. The following picture shows a typical arrangement of water molecules similar to what you might find in the liquid:

Can you see some of the hydrogen bonds? These bonds are weaker than typical covalent or ionic bonds, but nevertheless strong enough to make molecules which can hydrogen bond much more "sticky" with respect to each other than are other covalent molecules with otherwise similar properties. For example, the molecular mass of water is 18, and that of nitrogen is 28, yet nitrogen is a gas down to almost -200 degrees centigrade, whereas water is a liquid up to 100 degrees!

### Hydrogen Bonds in Biology

The cells of living things are made up of many different sorts of molecule. Two important classes of molecule are proteins and nucleic acids. In both of these molecules, parts of the (very large) molecules are involved in hydrogen bonds with other parts of the same molecules. This is very important in establishing the structure and properties of these molecules. By clicking on the link, you can view a Chime page explaining the structure of DNA, one of the most important nucleic acids, and showing the important role of hydrogen bonding.

### Conclusions

Ionic and covalent bonding are not the only kinds of bond between atoms. Some important other types of bond include metallic bonds, and hydrogen bonds. These explain the properties of metals, e.g. that they conduct electricity, and are very important in establishing the properties of water and living cells.

# Structure and Bonding in Chemistry

## Conclusions

On the previous pages, we have been able to view the structure of a number of typical chemical compounds. We have also learnt how structure is dependent upon the bonding between atoms. We have seen examples of the two most important types of chemical bond, ionic bonds and covalent ones. We have also learned that the overall properties of a compound can be related to its structure, and thus to its bonding. Finally, we have briefly examined two more types of bonding: metallic and hydrogen bonding.

This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies. If you want to look at the electronic structures of simple monatomic ions (such as Cl-, Ca2+ and Cr3+), you will find a link at the bottom of the page .

The electronic structures of atoms

Relating orbital filling to the Periodic Table

UK syllabuses for 16 - 18 year olds tend to stop at krypton when it comes to writing electronic structures, but it is possible that you could be asked for structures for elements up as far as barium. After barium you have to worry about f orbitals as well as s, p and d orbitals - and that's a problem for chemistry at a higher level. It is important that you look through past exam papers as well as your syllabus so that you can judge how hard the questions are likely to get.

This page looks in detail at the elements in the shortened version of the Periodic Table above, and then shows how you could work out the structures of some bigger atoms.

The first period

Hydrogen has its only electron in the 1s orbital - 1s1, and at helium the first level is completely full - 1s2.

The second period

Now we need to start filling the second level, and hence start the second period. Lithium's electron goes into the 2s orbital because that has a lower energy than the 2p orbitals. Lithium has an electronic structure of 1s22s1. Beryllium adds a second electron to this same level - 1s22s2.

Now the 2p levels start to fill. These levels all have the same energy, and so the electrons go in singly at first.

B  1s22s22px1

C     1s22s22px12py1

N      1s22s22px12py12pz1

Note:  The orbitals where something new is happening are shown in bold type. You wouldn't normally write them any differently from the other orbitals.

The next electrons to go in will have to pair up with those already there.

O    1s22s22px22py12pz1

F     1s22s22px22py22pz

Ne      1s22s22px22py22pz2

You can see that it is going to get progressively tedious to write the full electronic structures of atoms as the number of electrons increases. There are two ways around this, and you must be familiar with both.

Shortcut 1: All the various p electrons can be lumped together. For example, fluorine could be written as 1s22s22p5, and neon as 1s22s22p6.

This is what is normally done if the electrons are in an inner layer. If the electrons are in the bonding level (those on the outside of the atom), they are sometimes written in shorthand, sometimes in full. Don't worry about this. Be prepared to meet either version, but if you are asked for the electronic structure of something in an exam, write it out in full showing all the px, py and pz orbitals in the outer level separately.

For example, although we haven't yet met the electronic structure of chlorine, you could write it as 1s22s22p63s23px23py23pz1.

Notice that the 2p electrons are all lumped together whereas the 3p ones are shown in full. The logic is that the 3p electrons will be involved in bonding because they are on the outside of the atom, whereas the 2p electrons are buried deep in the atom and aren't really of any interest.

Shortcut 2: You can lump all the inner electrons together using, for example, the symbol [Ne]. In this context, [Ne] means the electronic structure of neon - in other words: 1s22s22px22py22pz2 You wouldn't do this with helium because it takes longer to write [He] than it does 1s2.

On this basis the structure of chlorine would be written [Ne]3s23px23py23pz1.

The third period

At neon, all the second level orbitals are full, and so after this we have to start the third period with sodium. The pattern of filling is now exactly the same as in the previous period, except that everything is now happening at the 3-level.

For example:

short version

Mg      1s22s22p63s2                               [Ne]3s2

S         1s22s22p63s23px23py13pz         [Ne]

3s23px23py13pz1

Ar       1s22s22p63s23px23py23pz2        [Ne]

3s23px23py23pz2

Note:  Check that you can do these. Cover the text and then work out these structures for yourself. Then do all the rest of this period. When you've finished, check your answers against the corresponding elements from the previous period. Your answers should be the same except a level further out.

The beginning of the fourth period

At this point the 3-level orbitals aren't all full - the 3d levels haven't been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills next.

K     1s22s22p63s23p64s1

Ca    1s22s22p63s23p64s2

There is strong evidence for this in the similarities in the chemistry of elements like sodium (1s22s22p63s1) and potassium (1s22s22p63s23p64s1)

The outer electron governs their properties and that electron is in the same sort of orbital in both of the elements. That wouldn't be true if the outer electron in potassium was 3d1.

s- and p-block elements

The elements in group 1 of the Periodic Table all have an outer electronic structure of ns1 (where n is a number between 2 and 7). All group 2 elements have an outer electronic structure of ns2. Elements in groups 1 and 2 are described as s-block elements.

Elements from group 3 across to the noble gases all have their outer electrons in p orbitals. These are then described as p-block elements.

d-block elements

Remember that the 4s orbital has a lower energy than the 3d orbitals and so fills first. Once the 3d orbitals have filled up, the next electrons go into the 4p orbitals as you would expect.

d-block elements are elements in which the last electron to be added to the atom is in a d orbital. The first series of these contains the elements from scandium to zinc, which at GCSE you probably called transition elements or transition metals. The terms "transition element" and "d-block element" don't quite have the same meaning, but it doesn't matter in the present context.

If you are interested:  A transition element is defined as one which has partially filled d orbitals either in the element or any of its compounds. Zinc (at the right-hand end of the d-block) always has a completely full 3d level (3d10) and so doesn't count as a transition element.

d electrons are almost always described as, for example, d5 or d8 - and not written as separate orbitals. Remember that there are five d orbitals, and that the electrons will inhabit them singly as far as possible. Up to 5 electrons will occupy orbitals on their own. After that they will have to pair up.

d5 means

d8 means

Notice in what follows that all the 3-level orbitals are written together, even though the 3d electrons are added to the atom after the 4s.

 Sc 1s22s22p63s23p63d14s2 Ti 1s22s22p63s23p63d24s2 V 1s22s22p63s23p63d34s2 Cr 1s22s22p63s23p63d54s1

Whoops! Chromium breaks the sequence. In chromium, the electrons in the 3d and 4s orbitals rearrange so that there is one electron in each orbital. It would be convenient if the sequence was tidy - but it's not!

 Mn 1s22s22p63s23p63d54s2 (back to being tidy again) Fe 1s22s22p63s23p63d64s2 Co 1s22s22p63s23p63d74s2 Ni 1s22s22p63s23p63d84s2 Cu 1s22s22p63s23p63d104s1 (another awkward one!) Zn 1s22s22p63s23p63d104s2

And at zinc the process of filling the d orbitals is complete.

Filling the rest of period 4

The next orbitals to be used are the 4p, and these fill in exactly the same way as the 2p or 3p. We are back now with the p-block elements from gallium to krypton. Bromine, for example, is 1s22s22p63s23p63d104s24px24py24pz1.

Useful exercise:  Work out the electronic structures of all the elements from gallium to krypton. You can check your answers by comparing them with the elements directly above them in the Periodic Table. For example, gallium will have the same sort of arrangement of its outer level electrons as boron or aluminium - except that gallium's outer electrons will be in the 4-level.

Summary

Writing the electronic structure of an element from hydrogen to krypton

• Use the Periodic Table to find the atomic number, and hence number of electrons.

• Fill up orbitals in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p - until you run out of electrons. The 3d is the awkward one - remember that specially. Fill p and d orbitals singly as far as possible before pairing electrons up.

• Remember that chromium and copper have electronic structures which break the pattern in the first row of the d-block.

Writing the electronic structure of big s- or p-block elements

Note:  We are deliberately excluding the d-block elements apart from the first row that we've already looked at in detail. The pattern of awkward structures isn't the same in the other rows. This is a problem for degree level.

First work out the number of outer electrons. This is quite likely all you will be asked to do anyway.

The number of outer electrons is the same as the group number. (The noble gases are a bit of a problem here, because they are normally called group 0 rather then group 8. Helium has 2 outer electrons; the rest have 8.) All elements in group 3, for example, have 3 electrons in their outer level. Fit these electrons into s and p orbitals as necessary. Which level orbitals? Count the periods in the Periodic Table (not forgetting the one with H and He in it).

Iodine is in group 7 and so has 7 outer electrons. It is in the fifth period and so its electrons will be in 5s and 5p orbitals. Iodine has the outer structure 5s25px25py25pz1.

What about the inner electrons if you need to work them out as well? The 1, 2 and 3 levels will all be full, and so will the 4s, 4p and 4d. The 4f levels don't fill until after anything you will be asked about at A'level. Just forget about them! That gives the full structure: 1s22s22p63s23p63d104s24p64d105s25px25py25pz1.

When you've finished, count all the electrons to make sure that they come to the same as the atomic number. Don't forget to make this check - it's easy to miss an orbital out when it gets this complicated.

Barium is in group 2 and so has 2 outer electrons. It is in the sixth period. Barium has the outer structure 6s2.

Including all the inner levels: 1s22s22p63s23p63d104s24p64d105s25p66s2.

It would be easy to include 5d10 as well by mistake, but the d level always fills after the next s level - so 5d fills after 6s just as 3d fills after 4s. As long as you counted the number of electrons you could easily spot this mistake because you would have 10 too many.

Note:  Don't worry too much about these complicated structures. You need to know how to work them out in principle.

This page explains what covalent bonding is. It starts with a simple picture of the single covalent bond, and then modifies it slightly for A'level purposes. It also takes a more sophisticated view (beyond A'level) if you are interested. You will find a link to a page on double covalent bonds at the bottom of the page.

A simple view of covalent bonding

The importance of noble gas structures

At a simple level (like GCSE) a lot of importance is attached to the electronic structures of noble gases like neon or argon which have eight electrons in their outer energy levels (or two in the case of helium). These noble gas structures are thought of as being in some way a "desirable" thing for an atom to have.

You may well have been left with the strong impression that when

other atoms react, they try to achieve noble gas structures.

As well as achieving noble gas structures by transferring electrons from one atom to another as in ionic bonding, it is also possible for atoms to reach these stable structures by sharing electrons to give covalent bonds.

Some very simple covalent molecules

Chlorine

For example, two chlorine atoms could both achieve stable structures by sharing their single unpaired electron as in the diagram.

The fact that one chlorine has been drawn with electrons marked as crosses and the other as dots is simply to show where all the electrons come from. In reality there is no difference between them.

The two chlorine atoms are said to be joined by a covalent bond. The reason that the two chlorine atoms stick together is that the shared pair of electrons is attracted to the nucleus of both chlorine atoms.

Hydrogen

Hydrogen atoms only need two electrons in their outer level to reach the noble gas structure of helium. Once again, the covalent bond holds the two atoms together because the pair of electrons is attracted to both nuclei.

Hydrogen chloride

The hydrogen has a helium structure, and the chlorine an argon structure.

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Covalent bonding at A'level

Cases where there isn't any difference from the simple view

If you stick closely to modern A'level syllabuses, there is little need to move far from the simple (GCSE) view. The only thing which must be changed is the over-reliance on the concept of noble gas structures. Most of the simple molecules you draw do in fact have all their atoms with noble gas structures.

For example:

Even with a more complicated molecule like PCl3, there's no problem. In this case, only the outer electrons are shown for simplicity. Each atom in this structure has inner layers of electrons of 2,8. Again, everything present has a noble gas structure.

Cases where the simple view throws up problems

Boron trifluoride, BF3

A boron atom only has 3 electrons in its outer level, and there is no possibility of it reaching a noble gas structure by simple sharing of electrons. Is this a problem? No. The boron has formed the maximum number of bonds that it can in the circumstances, and this is a perfectly valid structure.

Energy is released whenever a covalent bond is formed. Because energy is being lost from the system, it becomes more stable after every covalent bond is made. It follows, therefore, that an atom will tend to make as many covalent bonds as possible. In the case of boron in BF3, three bonds is the maximum possible because boron only has 3 electrons to share.

Note:  You might perhaps wonder why boron doesn't form ionic bonds with fluorine instead. Boron doesn't form ions because the total energy needed to remove three electrons to form a B3+ ion is simply too great to be recoverable when attractions are set up between the boron and fluoride ions.

Phosphorus(V) chloride, PCl5

In the case of phosphorus 5 covalent bonds are possible - as in PCl5.

Phosphorus forms two chlorides - PCl3 and PCl5. When phosphorus burns in chlorine both are formed - the majority product depending on how much chlorine is available. We've already looked at the structure of PCl3.

The diagram of PCl5 (like the previous diagram of PCl3) shows only the outer electrons.

Notice that the phosphorus now has 5 pairs of electrons in the outer level - certainly not a noble gas structure. You would have been content to draw PCl3 at GCSE, but PCl5 would have looked very worrying.

Why does phosphorus sometimes break away from a noble gas structure and form five bonds? In order to answer that question, we need to explore territory beyond the limits of current A'level syllabuses. Don't be put off by this! It isn't particularly difficult, and is extremely useful if you are going to understand the bonding in some important organic compounds.

A more sophisticated view of covalent bonding

The bonding in methane, CH4

Warning!  If you aren't happy with describing electron arrangements in s and p notation, and with the shapes of s and p orbitals, you need to read about orbitals before you go on.

What is wrong with the dots-and-crosses picture of bonding in methane?

We are starting with methane because it is the simplest case which illustrates the sort of processes involved. You will remember that the dots-and-crossed picture of methane looks like this.

There is a serious mis-match between this structure and the modern electronic structure of carbon, 1s22s22px12py1. The modern structure shows that there are only 2 unpaired electrons for hydrogens to share with, instead of the 4 which the simple view requires.

You can see this more readily using the electrons-in-boxes notation. Only the 2-level electrons are shown. The 1s2 electrons are too deep inside the atom to be involved in bonding. The only electrons directly available for sharing are the 2p electrons. Why then isn't methane CH2?

Promotion of an electron

When bonds are formed, energy is released and the system becomes more stable. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable.

There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons. The extra energy released when the bonds form more than compensates for the initial input.

Note:  People sometimes worry that the promoted electron is drawn as an up-arrow, whereas it started as a down-arrow. The reason for this is actually fairly complicated - well beyond the level we are working at. Just get in the habit of writing it like this because it makes the diagrams look tidy!

Now that we've got 4 unpaired electrons ready for bonding, another problem arises. In methane all the carbon-hydrogen bonds are identical, but our electrons are in two different kinds of orbitals. You aren't going to get four identical bonds unless you start from four identical orbitals.

Hybridisation

The electrons rearrange themselves again in a process called hybridisation. This reorganises the electrons into four identical hybrid orbitals called sp3 hybrids (because they are made from one s orbital and three p orbitals). You should read "sp3" as "s p three" - not as "s p cubed".

sp3 hybrid orbitals look a bit like half a p orbital, and they arrange themselves in space so that they are as far apart as possible. You can picture the nucleus as being at the centre of a tetrahedron (a triangularly based pyramid) with the orbitals pointing to the corners. For clarity, the nucleus is drawn far larger than it really is.

What happens when the bonds are formed?

Remember that hydrogen's electron is in a 1s orbital - a spherically symmetric region of space surrounding the nucleus where there is some fixed chance (say 95%) of finding the electron. When a covalent bond is formed, the atomic orbitals (the orbitals in the individual atoms) merge to produce a new molecular orbital which contains the electron pair which creates the bond.

Four molecular orbitals are formed, looking rather like the original sp3 hybrids, but with a hydrogen nucleus embedded in each lobe. Each orbital holds the 2 electrons that we've previously drawn as a dot and a cross.

The principles involved - promotion of electrons if necessary, then hybridisation, followed by the formation of molecular orbitals - can be applied to any covalently-bound molecule.

Note:  You will find this bit on methane repeated in the organic section of this site. That article on methane goes on to look at the formation of carbon-carbon single bonds in ethane.

The bonding in the phosphorus chlorides, PCl3 and PCl5

What's wrong with the simple view of PCl3?

This diagram only shows the outer (bonding) electrons.

Nothing is wrong with this! (Although it doesn't account for the shape of the molecule properly.) If you were going to take a more modern look at it, the argument would go like this:

Phosphorus has the electronic structure 1s22s22p63s23px13py13pz1. If we look only at the outer electrons as "electrons-in-boxes":

There are 3 unpaired electrons that can be used to form bonds with 3 chlorine atoms. The four 3-level orbitals hybridise to produce 4 equivalent sp3 hybrids just like in carbon - except that one of these hybrid orbitals contains a lone pair of electrons.

Each of the 3 chlorines then forms a covalent bond by merging the atomic orbital containing its unpaired electron with one of the

phosphorus unpaired electrons to make 3 molecular orbitals.

You might wonder whether all this is worth the bother! Probably not! It is worth it with PCl5, though.

What's wrong with the simple view of PCl5?

You will remember that the dots-and-crosses picture of PCl5 looks awkward because the phosphorus doesn't end up with a noble gas structure. This diagram also shows only the outer electrons.

In this case, a more modern view makes things look better by abandoning any pretence of worrying about noble gas structures.

If the phosphorus is going to form PCl5 it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the 3s orbital to the next available higher energy orbital.

Which higher energy orbital? It uses one of the 3d orbitals. You might have expected it to use the 4s orbital because this is the orbital that fills before the 3d when atoms are being built from scratch. Not so! Apart from when you are building the atoms in the first place, the 3d always counts as the lower energy orbital.

This leaves the phosphorus with this arrangement of its electrons:

The 3-level electrons now rearrange (hybridise) themselves to give 5 hybrid orbitals, all of equal energy. They would be called sp3d hybrids because that's what they are made from.

The electrons in each of these orbitals would then share space with electrons from five chlorines to make five new molecular orbitals - and hence five covalent bonds.

Why does phosphorus form these extra two bonds? It puts in an amount of energy to promote an electron, which is more than paid back when the new bonds form. Put simply, it is energetically profitable for the phosphorus to form the extra bonds.

The advantage of thinking of it in this way is that it completely ignores the question of whether you've got a noble gas structure, and so you don't worry about it.

A non-existent compound - NCl5

Nitrogen is in the same Group of the Periodic Table as phosphorus, and you might expect it to form a similar range of compounds. In fact, it doesn't. For example, the compound NCl3 exists, but there is no such thing as NCl5.

Nitrogen is 1s22s22px12py12pz1. The reason that NCl5 doesn't exist is that in order to form five bonds, the nitrogen would have to promote one of its 2s electrons. The problem is that there aren't any 2d orbitals to promote an electron into - and the energy gap to the next level (the 3s) is far too great.

n this case, then, the energy released when the extra bonds are made isn't enough to compensate for the energy needed to promote an electron - and so that promotion doesn't happen.

Atoms will form as many bonds as possible provided it is energetically profitable.

This page explains what co-ordinate (also called dative covalent) bonding is. You need to have a reasonable understanding of simple covalent bonding before you start.

Co-ordinate (dative covalent) bonding

A covalent bond is formed by two atoms sharing a pair of electrons. The atoms are held together because the electron pair is attracted by both of the nuclei.

In the formation of a simple covalent bond, each atom supplies one electron to the bond - but that doesn't have to be the case. A co-ordinate bond (also called a dative covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons come from the same atom.

For the rest of this page, we shall use the term co-ordinate bond - but if you prefer to call it a dative covalent bond, that's not a problem!

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The reaction between ammonia and hydrogen chloride

If these colourless gases are allowed to mix, a thick white smoke of solid ammonium chloride is formed.

Ammonium ions, NH4+, are formed by the transfer of a hydrogen ion from the hydrogen chloride to the lone pair of electrons on the ammonia molecule.

When the ammonium ion, NH4+, is formed, the fourth hydrogen is attached by a dative covalent bond, because only the hydrogen's nucleus is transferred from the chlorine to the nitrogen. The hydrogen's electron is left behind on the chlorine to form a negative chloride ion.

Once the ammonium ion has been formed it is impossible to tell any difference between the dative covalent and the ordinary covalent bonds. Although the electrons are shown differently in the diagram, there is no difference between them in reality.

Representing co-ordinate bonds

In simple diagrams, a co-ordinate bond is shown by an arrow. The arrow points from the atom donating the lone pair to the atom accepting it.

Dissolving hydrogen chloride in water to make hydrochloric acid

Something similar happens. A hydrogen ion (H+) is transferred from the chlorine to one of the lone pairs on the oxygen atom.

The H3O+ ion is variously called the hydroxonium ion, the hydronium ion or the oxonium ion.

In an introductory chemistry course (such as GCSE), whenever you have talked about hydrogen ions (for example in acids), you have actually been talking about the hydroxonium ion. A raw hydrogen ion is simply a proton, and is far too reactive to exist on its own in a test tube.

If you write the hydrogen ion as H+(aq), the "(aq)" represents the water molecule that the hydrogen ion is attached to. When it reacts with something (an alkali, for example), the hydrogen ion simply becomes detached from the water molecule again.

Note that once the co-ordinate bond has been set up, all the hydrogens attached to the oxygen are exactly equivalent. When a hydrogen ion breaks away again, it could be any of the three.

The reaction between ammonia and boron trifluoride, BF3

If you have recently read the page on covalent bonding, you may remember boron trifluoride as a compound which doesn't have a noble gas structure around the boron atom. The boron only has 3 pairs of electrons in its bonding level, whereas there would be room for 4 pairs. BF3 is described as being electron deficient.

The lone pair on the nitrogen of an ammonia molecule can be used to overcome that deficiency, and a compound is formed involving a co-ordinate bond.

Using lines to represent the bonds, this could be drawn more simply as:

The second diagram shows another way that you might find co-ordinate bonds drawn. The nitrogen end of the bond has become positive because the electron pair has moved away from the nitrogen towards the boron - which has therefore become negative. We shan't use this method again - it's more confusing than just using an arrow.

The structure of aluminium chloride

Aluminium chloride sublimes (turns straight from a solid to a gas) at about 180°C. If it simply contained ions it would have a very high melting and boiling point because of the strong attractions between the positive and negative ions. The implication is that it when it sublimes at this relatively low temperature, it must be covalent. The dots-and-crosses diagram shows only the outer electrons.

AlCl3, like BF3, is electron deficient. There is likely to be a similarity, because aluminium and boron are in the same group of the Periodic Table, as are fluorine and chlorine.

Measurements of the relative formula mass of aluminium chloride show that its formula in the vapour at the sublimation temperature is not AlCl3, but Al2Cl6. It exists as a dimer (two molecules joined together). The bonding between the two molecules is co-ordinate, using lone pairs on the chlorine atoms. Each chlorine atom has 3 lone pairs, but only the two important ones are shown in the line diagram.

Note:  The uninteresting electrons on the chlorines have been faded in colour to make the co-ordinate bonds show up better. There's nothing special about those two particular lone pairs - they just happen to be the ones pointing in the right direction.

Energy is released when the two co-ordinate bonds are formed, and so the dimer is more stable than two separate AlCl3 molecules.

Note:  Aluminium chloride is complicated because of the way it keeps changing its bonding as the temperature increases. If you are interested in exploring this in more detail, you could have a look at the page about the Period 3 chlorides. It isn't particularly relevant to the present page, though.

The bonding in hydrated metal ions

Water molecules are strongly attracted to ions in solution - the water molecules clustering around the positive or negative ions. In many cases, the attractions are so great that formal bonds are made, and this is true of almost all positive metal ions. Ions with water molecules attached are described as hydrated ions.

Although aluminium chloride is covalent, when it dissolves in water, ions are produced. Six water molecules bond to the aluminium to give an ion with the formula Al(H2O)63+. It's called the hexaaquaaluminium ion - which translates as six ("hexa") water molecules ("aqua") wrapped around an aluminium ion.

The bonding in this (and the similar ions formed by the great majority of other metals) is co-ordinate (dative covalent) using lone pairs on the water molecules.

Aluminium is 1s22s22p63s23px1. When it forms an Al3+ ion it loses the 3-level electrons to leave 1s22s22p6.

That means that all the 3-level orbitals are now empty. The aluminium re-organises (hybridises) six of these (the 3s, three 3p, and two 3d) to produce six new orbitals all with the same energy. These six hybrid orbitals accept lone pairs from six water molecules.

You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules it is possible to fit around an aluminium ion (and most other metal ions). By making the maximum number of bonds, it releases most energy and so becomes most energetically stable.

Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminium and so isn't involved in the bonding. The resulting ion looks like this:

Because of the movement of electrons towards the centre of the ion, the 3+ charge is no longer located entirely on the aluminium, but is now spread over the whole of the ion.

Note:  Dotted arrows represent lone pairs coming from water molecules behind the plane of the screen or paper. Wedge shaped arrows represent bonds from water molecules in front of the plane of the screen or paper.

Two more molecules

Carbon monoxide, CO

Carbon monoxide can be thought of as having two ordinary covalent bonds between the carbon and the oxygen plus a co-ordinate bond using a lone pair on the oxygen atom.

Nitric acid, HNO3

In this case, one of the oxygen atoms can be thought of as attaching to the nitrogen via a co-ordinate bond using the lone pair on the nitrogen atom.

In fact this structure is misleading because it suggests that the two oxygen atoms on the right-hand side of the diagram are joined to the nitrogen in different ways. Both bonds are actually identical in length and strength, and so the arrangement of the electrons must be identical. There is no way of showing this using a dots-and-crosses picture. The bonding involves delocalisation.

If you are interested:  The bonding is rather similar to the bonding in the ethanoate ion (although without the negative charge). You will find this described on a page about the acidity of organic acids.

This page explains the origin of hydrogen bonding - a relatively strong form of intermolecular attraction. If you are also interested in the weaker intermolecular forces (van der Waals dispersion forces and dipole-dipole interactions), there is a link at the bottom of the page.

Many elements form compounds with hydrogen - referred to as "hydrides". If you plot the boiling points of the hydrides of the Group 4 elements, you find that the boiling points increase as you go down the group.

The increase in boiling point happens because the molecules are getting larger with more electrons, and so van der Waals dispersion forces become greater.

If you repeat this exercise with the hydrides of elements in Groups 5, 6 and 7, something odd happens.

Although for the most part the trend is exactly the same as in group 4 (for exactly the same reasons), the boiling point of the hydride of the first element in each group is abnormally high.

In the cases of NH3, H2O and HF there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break. These relatively powerful intermolecular forces are described as hydrogen bonds.

The molecules which have this extra bonding are:

Note:  The solid line represents a bond in the plane of the screen or paper. Dotted bonds are going back into the screen or paper away from you, and wedge-shaped ones are coming out towards you.

Notice that in each of these molecules:

• The hydrogen is attached directly to one of the most electronegative elements, causing the hydrogen to acquire a significant amount of positive charge.

• Each of the elements to which the hydrogen is attached is not only significantly negative, but also has at least one "active" lone pair.

Lone pairs at the 2-level have the electrons contained in a relatively small volume of space which therefore has a high density of negative charge. Lone pairs at higher levels are more diffuse and not so attractive to positive things.

Consider two water molecules coming close together.

The + hydrogen is so strongly attracted to the lone pair that it is almost as if you were beginning to form a co-ordinate (dative covalent) bond. It doesn't go that far, but the attraction is significantly stronger than an ordinary dipole-dipole interaction.

Hydrogen bonds have about a tenth of the strength of an average covalent bond, and are being constantly broken and reformed in liquid water. If you liken the covalent bond between the oxygen and hydrogen to a stable marriage, the hydrogen bond has "just good friends" status. On the same scale, van der Waals attractions represent mere passing acquaintances!

Water as a "perfect" example of hydrogen bonding

Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of + hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding.

This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens.

In hydrogen fluoride, the problem is a shortage of hydrogens. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system.

Note:  You will find more discussion on the effect of hydrogen bonding on the properties of water in the page on molecular structures.

More complex examples of hydrogen bonding

The hydration of negative ions

When an ionic substance dissolves in water, water molecules cluster around the separated ions. This process is called hydration.

Water frequently attaches to positive ions by co-ordinate (dative covalent) bonds. It bonds to negative ions using hydrogen bonds.

Note:  If you are interested in the bonding in hydrated positive ions, you could follow this link to co-ordinate (dative covalent) bonding.

The diagram shows the potential hydrogen bonds formed to a chloride ion, Cl-. Although the lone pairs in the chloride ion are at the 3-level and wouldn't normally be active enough to form hydrogen bonds, in this case they are made more attractive by the full negative charge on the chlorine.

However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to.

Hydrogen bonding in alcohols

An alcohol is an organic molecule containing an -O-H group.

Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Such molecules will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier", and more heat is necessary to separate them.

Ethanol, CH3CH2-O-H, and methoxymethane, CH3-O-CH3, both have the same molecular formula, C2H6O.

Note:  If you haven't done any organic chemistry yet, don't worry about the names.

They have the same number of electrons, and a similar length to the molecule. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be much the same.

However, ethanol has a hydrogen atom attached directly to an oxygen - and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge.

In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur.

The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules:

 ethanol (with hydrogen bonding) 78.5°C methoxymethane (without hydrogen bonding) -24.8°C

The hydrogen bonding in the ethanol has lifted its boiling point about 100°C.

It is important to realise that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding.

Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding due to the hydrogen attached directly to the oxygen - but they aren't the same.

The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol.

Hydrogen bonding in organic molecules containing nitrogen

Hydrogen bonding also occurs in organic molecules containing N-H groups - in the same sort of way that it occurs in ammonia. Examples range from simple molecules like CH3NH2 (methylamine) to large molecules like proteins and DNA.

The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one.

Structure & Bonding

The study of organic chemistry must at some point extend to the molecular level, for the physical and chemical properties of a substance are ultimately explained in terms of the structure and bonding of molecules. This module introduces some basic facts and principles that are needed for a discussion of organic molecules.

## Electron Configurations in the Periodic Table

 1A 2A 3A 4A 5A 6A 7A 8A 1 H 1s1 2 He 1s2 3 Li 1s2 2s1 4 Be 1s2 2s2 5 B 1s2 2s22p1 6 C 1s2 2s22p2 7 N 1s2 2s22p3 8 O 1s2 2s22p4 9 F 1s2 2s22p5 10 Ne 1s2 2s22p6 11 Na [Ne] 3s1 12 Mg [Ne] 3s2 13 Al [Ne] 3s23p1 14 Si [Ne] 3s23p2 15 P [Ne] 3s23p3 16 S [Ne] 3s23p4 17 Cl [Ne] 3s23p5 18 Ar [Ne] 3s23p6

Four elements, hydrogen, carbon, oxygen and nitrogen, are the major components of most organic compounds. Consequently, our understanding of organic chemistry must have, as a foundation, an appreciation of the electronic structure and properties of these elements. The truncated periodic table shown above provides the orbital electronic structure for the first eighteen elements (hydrogen through argon). According to the Aufbau principle, the electrons of an atom occupy quantum levels or orbitals starting from the lowest energy level, and proceeding to the highest, with each orbital holding a maximum of two paired electrons (opposite spins).

Electron shell #1 has the lowest energy and its s-orbital is the first to be filled. Shell #2 has four higher energy orbitals, the 2s-orbital being lower in energy than the three 2p-orbitals. (x, y & z). As we progress from lithium (atomic number=3) to neon (atomic number=10) across the second row or period of the table, all these atoms start with a filled 1s-orbital, and the 2s-orbital is occupied with an electron pair before the 2p-orbitals are filled. In the third period of the table, the atoms all have a neon-like core of 10 electrons, and shell #3 is occupied progressively with eight electrons, starting with the 3s-orbital. The highest occupied electron shell is called the valence shell, and the electrons occupying this shell are called valence electrons.

The chemical properties of the elements reflect their electron configurations. For example, helium, neon and argon are exceptionally stable and unreactive monoatomic gases. Helium is unique since its valence shell consists of a single s-orbital. The other members of group 8 have a characteristic valence shell electron octet (ns2 + npx2 + npy2 + npz2). This group of inert (or noble) gases also includes krypton (Kr: 4s2, 4p6), xenon (Xe: 5s2, 5p6) and radon (Rn: 6s2, 6p6). In the periodic table above these elements are colored beige.

The halogens (F, Cl, Br etc.) are one electron short of a valence shell octet, and are among the most reactive of the elements (they are colored red in this periodic table). In their chemical reactions halogen atoms achieve a valence shell octet by capturing or borrowing the eighth electron from another atom or molecule. The alkali metals Li, Na, K etc. (colored violet above) are also exceptionally reactive, but for the opposite reason. These atoms have only one electron in the valence shell, and on losing this electron arrive at the lower shell valence octet. As a consequence of this electron loss, these elements are commonly encountered as cations (positively charged atoms).
The elements in groups 2 through 7 all exhibit characteristic reactivities and bonding patterns that can in large part be rationalized by their electron configurations. It should be noted that hydrogen is unique. Its location in the periodic table should not suggest a kinship to the chemistry of the alkali metals, and its role in the structure and properties of organic compounds is unlike that of any other element.

## Chemical Bonding and Valence

As noted earlier, the inert gas elements of group 8 exist as monoatomic gases, and do not in general react with other elements. In contrast, other gaseous elements exist as diatomic molecules (H2, N2, O2, F2 & Cl2), and all but nitrogen are quite reactive. Some dramatic examples of this reactivity are shown in the following equations.

 2Na + Cl2 2NaCl 2H2 + O2 2H2O C + O2 CO2 C + 2F2 CF4

Why do the atoms of many elements interact with each other and with other elements to give stable molecules? In addressing this question it is instructive to begin with a very simple model for the attraction or bonding of atoms to each other, and then progress to more sophisticated explanations.

When sodium is burned in a chlorine atmosphere, it produces the compound sodium chloride. This has a high melting point (800 ºC) and dissolves in water to to give a conducting solution. Sodium chloride is an ionic compound, and the crystalline solid has the structure shown on the right. Transfer of the lone 3s electron of a sodium atom to the half-filled 3p orbital of a chlorine atom generates a sodium cation (neon valence shell) and a chloride anion (argon valence shell). Electrostatic attraction results in these oppositely charged ions packing together in a lattice. The attractive forces holding the ions in place can be referred to as ionic bonds.

The other three reactions shown above give products that are very different from sodium chloride. Water is a liquid at room temperature; carbon dioxide and carbon tetrafluoride are gases. None of these compounds is composed of ions. A different attractive interaction between atoms, called covalent bonding, is involved here. Covalent bonding occurs by a sharing of valence electrons, rather than an outright electron transfer. Similarities in physical properties (they are all gases) suggest that the diatomic elements H2, N2, O2, F2 & Cl2 also have covalent bonds.
Examples of covalent bonding shown below include hydrogen, fluorine, carbon dioxide and carbon tetrafluoride. These illustrations use a simple
Bohr notation, with valence electrons designated by colored dots. Note that in the first case both hydrogen atoms achieve a helium-like pair of 1s-electrons by sharing. In the other examples carbon, oxygen and fluorine achieve neon-like valence octets by a similar sharing of electron pairs. Carbon dioxide is notable because it is a case in which two pairs of electrons (four in all) are shared by the same two atoms. This is an example of a double covalent bond.

These electron sharing diagrams (Lewis formulas) are a useful first step in understanding covalent bonding, but it is quicker and easier to draw Couper-Kekulé formulas in which each shared electron pair is represented by a line between the atom symbols. Non-bonding valence electrons are shown as dots. These formulas are derived from the graphic notations suggested by A. Couper and A. Kekulé, and are not identical to their original drawings. Some examples of such structural formulas are given in the following table.

Common Name

Molecular Formula

Lewis Formula

Kekulé Formula

 Methane CH4 Ammonia NH3 Ethane C2H6 Methyl Alcohol CH4O Ethylene C2H4 Formaldehyde CH2O Acetylene C2H2 Hydrogen Cyanide CHN

The sharing of two or more electron pairs, is illustrated by ethylene and formaldehyde (each has a double bond), and acetylene and hydrogen cyanide (each with a triple bond). Boron compounds such as BH3 and BF3 are exceptional in that conventional covalent bonding does not expand the valence shell occupancy of boron to an octet. Consequently, these compounds have an affinity for electrons, and they exhibit exceptional reactivity when compared with the compounds shown above.

#### Valence

The number of valence shell electrons an atom must gain or lose to achieve a valence octet is called valence. In covalent compounds the number of bonds which are characteristically formed by a given atom is equal to that atom's valence. From the formulas written above, we arrive at the following general valence assignments:

 Atom H C N O F Cl Br I Valence 1 4 3 2 1 1 1 1

The valences noted here represent the most common form these elements assume in organic compounds. Many elements, such as chlorine, bromine and iodine, are known to exist in several valence states in different inorganic compounds.

## Charge Distribution

If the electron pairs in covalent bonds were donated and shared absolutely evenly there would be no fixed local charges within a molecule. Although this is true for diatomic elements such as H2, N2 and O2, most covalent compounds show some degree of local charge separation, resulting in bond and / or molecular dipoles. A dipole exists when the centers of positive and negative charge distribution do not coincide.

#### Formal Charges

A large local charge separation usually results when a shared electron pair is donated unilaterally. The three Kekulé formulas shown here illustrate this condition.

In the formula for ozone the central oxygen atom has three bonds and a full positive charge while the right hand oxygen has a single bond and is negatively charged. The overall charge of the ozone molecule is therefore zero. Similarly, nitromethane has a positive-charged nitrogen and a negative-charged oxygen, the total molecular charge again being zero. Finally, azide anion has two negative-charged nitrogens and one positive-charged nitrogen, the total charge being minus one.
In general, for covalently bonded atoms having valence shell electron octets, if the number of covalent bonds to an atom is greater than its normal valence it will carry a positive charge. If the number of covalent bonds to an atom is less than its normal valence it will carry a negative charge. The formal charge on an atom may also be calculated by the following formula:

#### Polar Covalent Bonds

Because of their differing nuclear charges, and as a result of shielding by inner electron shells, the different atoms of the periodic table have different affinities for nearby electrons. The ability of an element to attract or hold onto electrons is called electronegativity. A rough quantitative scale of electronegativity values was established by Linus Pauling, and some of these are given in the table to the right. A larger number on this scale signifies a greater affinity for electrons. Fluorine has the greatest electronegativity of all the elements, and the heavier alkali metals such as potassium, rubidium and cesium have the lowest electronegativities. It should be noted that carbon is about in the middle of the electronegativity range, and is slightly more electronegative than hydrogen.
When two different atoms are bonded covalently, the shared electrons are attracted to the more electronegative atom of the bond, resulting in a shift of electron density toward the more electronegative atom. Such a covalent bond is polar, and will have a dipole (one end is positive and the other end negative). The degree of polarity and the magnitude of the bond dipole will be proportional to the difference in electronegativity of the bonded atoms. Thus a O–H bond is more polar than a C–H bond, with the hydrogen atom of the former being more positive than the hydrogen bonded to carbon. Likewise, C–Cl and C–Li bonds are both polar, but the carbon end is positive in the former and negative in the latter. The dipolar nature of these bonds is often indicated by a partial charge notation (δ+/–) or by an arrow pointing to the negative end of the bond.

 Electronegativity Values for Some Elements H 2.20 Li 0.98 Be 1.57 B 2.04 C 2.55 N 3.04 O 3.44 F 3.98 Na 0.90 Mg 1.31 Al 1.61 Si 1.90 P 2.19 S 2.58 Cl 3.16 K 0.82 Ca 1.00 Ga 1.81 Ge 2.01 As 2.18 Se 2.55 Br 2.96

Although there is a small electronegativity difference between carbon and hydrogen, the C–H bond is regarded as weakly polar at best, and hydrocarbons in general are considered to be non-polar compounds.

The shift of electron density in a covalent bond toward the more electronegative atom or group can be observed in several ways. For bonds to hydrogen, acidity is one criterion. If the bonding electron pair moves away from the hydrogen nucleus the proton will be more easily transfered to a base (it will be more acidic). A comparison of the acidities of methane, water and hydrofluoric acid is instructive. Methane is essentially non-acidic, since the C–H bond is nearly non-polar. As noted above, the O–H bond of water is polar, and it is at least 25 powers of ten more acidic than methane. H–F is over 12 powers of ten more acidic than water as a consequence of the greater electronegativity difference in its atoms.
Electronegativity differences may be transmitted through connecting covalent bonds by an inductive effect. Replacing one of the hydrogens of water by a more electronegative atom increases the acidity of the remaining O–H bond. Thus hydrogen peroxide, HO–O–H, is ten thousand times more acidic than water, and hypochlorous acid, Cl–O–H is one hundred million times more acidic. This inductive transfer of polarity tapers off as the number of transmitting bonds increases, and the presence of more than one highly electronegative atom has a cumulative effect. For example, trifluoro ethanol, CF3CH2–O–H is about ten thousand times more acidic than ethanol, CH3CH2–OH.

 Excellent physical evidence for the inductive effect is found in the influence of electronegative atoms on the nmr chemical shifts of nearby hydrogen atoms.

## Functional Groups

Functional groups are atoms or small groups of atoms (two to four) that exhibit a characteristic reactivity when treated with certain reagents. A particular functional group will almost always display its characteristic chemical behavior when it is present in a compound. Because of their importance in understanding organic chemistry, functional groups have characteristic names that often carry over in the naming of individual compounds incorporating specific groups. In the following table the atoms of each functional group are colored red and the characteristic IUPAC nomenclature suffix that denotes some (but not all) functional groups is also colored.

## Functional Group Tables

Group Formula

Class Name

Specific Example

IUPAC Name

Common Name

Alkene

H2C=CH2

Ethene

Ethylene

Alkyne

HC≡CH

Ethyne

Acetylene

Arene

C6H6

Benzene

Benzene

Group Formula

Class Name

Specific Example

IUPAC Name

Common Name

Halide

H3C-I

Iodomethane

Methyl iodide

Alcohol

CH3CH2OH

Ethanol

Ethyl alcohol

Ether

CH3CH2OCH2CH3

Diethyl ether

Ether

H3C-NH2

Amine

H3C-NH2

Aminomethane

Methylamine

Nitro Compound

H3C-NO2

Nitromethane

Thiol

H3C-SH

Methanethiol

Methyl mercaptan

Sulfide

H3C-S-CH3

Dimethyl sulfide

### Functional Groups with Multiple Bonds to Heteroatoms

Group Formula

Class Name

Specific Example

IUPAC Name

Common Name

Nitrile

H3C-CN

Ethanenitrile

Acetonitrile

Aldehyde

H3CCHO

Ethanal

Acetaldehyde

Ketone

H3CCOCH3

Propanone

Acetone

Carboxylic Acid

H3CCO2H

Ethanoic Acid

Acetic acid

Ester

H3CCO2CH2CH3

Ethyl ethanoate

Ethyl acetate

Acid Halide

H3CCOCl

Ethanoyl chloride

Acetyl chloride

Amide

H3CCON(CH3)2

N,N-Dimethylethanamide

N,N-Dimethylacetamide

Acid Anhydride

(H3CCO)2O

Ethanoic anhydride

Acetic anhydride

Dipole Moment - A Measure of Degree of Polarity

Molecules having two equal and opposite charges separated by certain distance are said to possess an electric dipole. In the case of such polar molecules, the centre of negative charge does not coincide with the centre of positive charge. The extent of polarity in such covalent molecules can be described by the term Dipole moment.

Dipole moment can be defined as the product of the magnitude of the charge and the distance of separation between the charges.It is represented by the Greek letter 'm'. Mathematically it is equal to

dipole moment (m) = charge (e) x distance of separation (d).It is expressed in the units of Debye and written as D

(1 Debye = 1 x 10-18e.s.u cm)

Dipole moment is a vector quantity and is represented by a small arrow with tail at the positive centre and head pointing towards a negative centre.

For example, the dipole moment of HCl molecule is 1.03 D and that of H2O is 1.84 D. The dipole of HCl may be represented as:

Dipole Moment and Molecular Structure

Diatomic molecules

A diatomic molecule has two atoms bonded to each other by a covalent bond. In such a molecule, the dipole moment of the bond gives the dipole moment of the molecule. Thus, a diatomic molecule is polar if the bond formed between the atoms is polar. Greater the electronegativity difference between the atoms, more will be the dipole moment.

The dipole moment of hydrogen halides decreases with decreasing

electronegativity of halogen atom.

Polyatomic molecules

In polyatomic molecules the dipole moment not only depends upon the individual dipole moments of the bonds but also on the spatial arrangement of the various bonds in the molecule. In such molecules the dipole moment of the molecule is the vector sum of the dipole moments of various bonds.

For example, Carbon dioxide (CO2) and water (H2O) are both triatomic molecules but the dipole moment of carbon dioxide is zero whereas that of water is 1.84 D. This is because CO2 is a linear molecule in which the two C=O (m=2.3D) bonds are oriented in opposite directions at an angle of 180°. Due to the linear geometry the dipole moment of one C = O bond cancels that of another. Therefore, the resultant dipole moment of the molecule is zero and it is a non-polar molecule.

Water molecule has a bent structure with the two OH bonds oriented at an angle of 104.5°. The dipole moment of water is 1.84D, which is the resultant of the dipole moments of two O-H bonds.

Similarly in tetra-atomic molecules such as BF3 and NH3, the dipole moment of BF3 molecule is zero while that of NH3 is 1.49 D. This suggests that BF3 has symmetrical structure in which the three B-F bonds are oriented at an angle of 120° to one another. Also the three bonds lie in one plane and the dipole moments of these bonds cancel one another giving net dipole moment equal to zero.

NH3 has a pyramidal structure. The individual dipole moments of three N-H bonds give the resultant dipole moment as 1.49 D.

Thus, the presence of polar bonds in a polyatomic molecule does not mean that the molecules are polar.

Importance of dipole moment

Dipole moment plays very important role in understanding the nature of chemical bonds.

mportance of dipole moment and problems

The measurement of dipole moment helps in distinguishing between polar and non-polar molecules. Non-polar molecules have zero dipole moment while polar molecules have some value of dipole moment.

For example

Non-polar molecules: O2, Cl2, BF3, CH4

Polar Molecules: HF (1.91 D), HCl (1.03 D), H2S (0.90 D)

Dipole moment measurement gives an idea about the degree of polarity in a diatomic molecule. The greater the dipole moment the greater is the polarity in such a molecule.

Dipole moment is used to find the shapes of molecules. This is because the dipole moment not only depends upon the individual dipole moment of the bonds but also on the arrangement of bonds.

It is possible to predict the nature of chemical bond formed depending upon the electronegativities of atoms involved in a molecule. The bond will be highly polar if the electronegativities of two atoms is large. However, when the electron is completely transferred from one atom to another, an ionic bond is formed (ionic bond is an extreme case of polar covalent bonds). The greater the difference in electronegativities of the bonded atoms, the higher is the ionic character. When the electronegativity difference between two atoms is 1.7, then the bond is 50% ionic and 50% covalent. If the electronegativitv difference is more than 1.7, then the chemical bond is largely ionic (more than 50% ionic character) and if the difference is less than 1.7, the bond formed is mainly covalent.

The percentage of ionic character can be calculated from the ratio of the observed dipole moment to the dipole moment for the complete electron transfer (100% ionic character).

In HCl molecule, the observed dipole moment is 1.03 D and its bond length is 1.275Å. Assuming 100% ionic character, the charge developed on H and Cl atoms would be 4.8 x 10-10e.s.u.

Therefore, dipole moment for 100% ionic character will be

= q x d = 4.8 x 10-10e.s.u x 1.275 x 10-8cm

=6.12x 1O-18e.s.u.cm

= 6.12 D (1D = 10-18 e.s.u. cm.)

Problem

12. Calculate the ionic character of HCl. Its measured dipole moment is 3.436 x 10-30 coulomb meter. The HCl bond length is 2.29 x 10-10 meter.

Solution

Dipole moment corresponding to 100 % ionic character of HCl

= 1.602 x 10-19 C x 1.29 x 10-10 m= 20.67 x 10-30 Cm

Actual dipole moment of HCl = 3.436 x 10-30 Cm

3. The C-Cl bond is polar but CCl4 molecule is non-polar. Explain.

1

Solution

The C-Cl bond is polar because the chlorine atom being more electronegative pulls the shared electron pair towards itself. In CCl4, there are four C-Cl bonds. Since these polar bonds are symmetrically arranged, the polarities of individual bonds cancel each other resulting in a zero dipole moment for the molecule. The net result is that CCl4 molecule is non-polar.

Two types of covalent bonds are formed depending upon the electronegativity of the combining elements.

Non-polar covalent bond

When a covalent bond is formed between two atoms of the same element, the shared electron pair will lie exactly midway between the two atoms i.e. the electrons are equally shared by the atoms. The resulting molecule will be electrically symmetrical i.e., centre of the negative charge coincides with the centre of the positive charge. This type of covalent bond is described as a non-polar covalent bond. The bonds in the molecules H2, O2, Cl2 etc., are non-polar covalent bonds.

The bond between two unlike atoms, which differ in their affinities for electrons is said to be a polar covalent bond. When a covalent bond is formed between two atoms of different elements, the bonding pair of electrons will lie more towards the atom, which has more affinity for electrons. As the said electron pair do not lie exactly midway between the two atoms, the atom with higher affinity for electrons develops a slight negative charge and the atom with lesser affinity for electrons, a slight positive charge. Such molecules are called 'polar molecules'.

In the hydrogen chloride (HCl) molecule, the bonding of hydrogen and chlorine atoms lies more towards Cl atom (because Cl is more electronegative) in the shared pair of electrons. Therefore, Cl atom acquires a slight negative charge, and H atom a slight positive charge. This causes the covalent bond between H and Cl to have an appreciable ionic character.

The compounds having polar bonds are termed polar compounds. Polar substances in their pure forms, do not conduct electricity, but give conducting solutions when dissolved in polar solvents.

Cause of polarity in bonds

A pure covalent bond is formed when the shared pair of electrons is shared equally by the two atoms. Conversely when the two combining atoms share the shared pair of electrons unequally, the bond formed is a polar covalent bond. Unequal sharing of the shared pair of electrons arises due to an unequal electron-attracting tendency of the two atoms. The electron-attracting tendency of the atoms in a molecule is described in terms of electronegativity. 'Polarity in a bond arises due to the difference in the electronegativities of the combining atoms'. Thus, the atom of an element having higher electronegativity has greater electron attracting tendency.

For example, two atoms of hydrogen combine to form a molecule of hydrogen in which H-H is a pure covalent bond. But, when the atoms of different elements (with different electronegativities) combine, the atom of the more electronegative element attracts the shared pair of electrons more towards it. This leads to the formation of a polar bond.

Hydrogen and chlorine react to give hydrogen chloride (HCl). HCl is a polar molecule because of the difference in the electronegativity values of hydrogen and chlorine.

The greater is the difference in the electronegativity values of the combining atoms, greater is the polar character in the bond so formed.

For example, in the series H - X (X=F, Cl, Br,I), the electronegativity difference between H and X atom follows the order:

H- F > H - Cl > H - Br > H - I

Therefore, the polarity in the H - X bond follows the order,

H - F > H - Cl > H - Br > H I i.e., H - F bond is the most polar and

H -I bond is the least polar in this series of compounds.

Hybridisation

Hybridisation is the phenomenon of redistribution of energies of the orbitals of slightly different energies so as to give a new set of orbitals of equivalent energies. The new orbitals are called hybrid or hybridised orbitals. The number of hybridized orbitals formed is equal to the number of atomic orbitals taking part in hybridisation.This phenomenon is more predominant in carbon containing compounds and so to understand this concept, a study of the electronic structure of carbon is essential.

Tetravalency of Carbon

Carbon forms a large number of fascinating variety of compounds. This is because of its unusual property of catenation in which one carbon unites with the other to form long chains and rings. It is this property, which is responsible for the existence of millions of compounds of carbon. This feature can be explained on the basis of tetravalency of carbon.

The electronic configuration of carbon is 1s2 2s2 2px1 2py1 2pz0. In the box notation this is represented as:

As there are two half filled orbitals in the valence shell of carbon, its bonding capacity should be two. However, in actual practice carbon exhibits a bonding capacity of four and forms molecules of the type CH4, CCl4 etc. In order to explain this tetravalency, it is proposed that one of the electrons from the '2s' filled orbital is promoted to the empty '2p' orbital (2pz), which is in a higher energy state. In this way four half filled orbitals are formed in the valence shell which can account for the bonding capacity of four bonds of carbon. This state is known as the excited state in which the electronic configuration of carbon is:

From the above configuration it is clear that all the four bonds of carbon will not be identical. This is because one bond will be formed by the overlap of '2s' orbital which will have more of 's' character. The other three bonds will be formed by the overlap of the '2p' orbitals, which will have more of 'p' character. Therefore all the four bonds will not be equivalent.

But in practice, most of the carbon compounds have all the four bonds equal. This behaviour can be explained in terms of Hybridisation.

Characterisitics of hybridisation

The hybridised orbitals are always equivalent in energy and shape.

images/SQBTN017.jpg The number of hybridised orbitals formed is equal to the number of orbitals that undergo hybridisation.

Hydridised orbitals form more stable bonds.

Hydridised orbitals orient themselves in preferred directions in space and so give a fixed geometry or shape to the molecules.

Conditions for hybridisation

Only the valence shell orbitals of the atom are hybridised.

Orbitals undergoing hybridisation should have only a small difference in their energies.

It is not necessary that only half filled orbitals participate in hybridisation. Even filled orbitals of the valence shell can take part in hybridisation.

Rearrangement by way of promotion to different orbitals is not an essential condition for hybridisation.

Also called Lewis structures, give a representation of the valence electrons surrounding an atom.

Each valence electron is represented by one dot, thus, a lone atom of hydrogen would be drawn as an H with one dot, whereas a lone atom of Helium would be drawn as an He with two dots, and so forth.

Representing two atoms joined by a covalent bond is done by drawing the atomic symbols near to each other, and drawing a single line to represent a shared pair of electrons. It is important to note: a single valence electron is represented by a dot, whereas a pair of electrons is represented by a line.

The covalent compound hydrogen fluoride, for example, would be represented by the symbol H joined to the symbol F by a single line, with three pairs (six more dots) surrounding the symbol F. The line represents the two electrons shared by both hydrogen and fluorine, whereas the six paired dots represent fluorine's remaining six valence electrons.

Dot structures are useful in illustrating simple covalent molecules, but the limitations of dot structures become obvious when diagramming even relatively simple organic molecules. The dot structures have no ability to represent the actual physical orientation of molecules, and they become overly cumbersome when more than three or four atoms are represented.

Lewis dot structures are useful for introducing the idea of covalence and bonding in small molecules, but other model types have much more capability to communicate chemistry concepts.

### Drawing electron dot structures

Some examples of electron dot structures for a few commonly encountered molecules from inorganic chemistry.

#### A note about Gilbert N. Lewis

Lewis was born in Weymouth, Massachusetts as the son of a Dartmouth-graduated lawyer/broker. He attended the University of Nebraska at age 14, then three years later transferred to Harvard. After showing an initial interest in Economics, Gilbert Newton Lewis earned first a B.A. in Chemistry, and then a Ph.D. in Chemistry in 1899.

For a few years after obtaining his doctorate, Lewis worked and studied both in the United States and abroad (including Germany and the Phillipines) and he was even a professor at M.I.T. from 1907 until 1911. He then went on to U.C. Berkeley in order to be Dean of the College of Chemistry in 1912.

In 1916 Dr. Lewis formulated the idea that a covalent bond consisted of a shared pair of electrons. His ideas on chemical bonding were expanded upon by Irving Langmuir and became the inspiration for the studies on the nature of the chemical bond by Linus Pauling.

In 1923, he formulated the electron-pair theory of acid-base reactions. In the so-called Lewis theory of acids and bases, a "Lewis acid" is an electron-pair acceptor and a "Lewis base" is an electron-pair donor.

In 1926, he coined the term "photon" for the smallest unit of radiant energy.

Lewis was also the first to produce a pure sample of deuterium oxide (heavy water) in 1933. By accelerating deuterons (deuterium nuclei) in Ernest O. Lawrence's cyclotron, he was able to study many of the properties of atomic nuclei.

During his career he published on many other subjects, and he died at age 70 of a heart attack while working in his laboratory in Berkeley. He had one daughter and two sons; both of his sons became chemistry professors themselves.

Formal :The formal charge of an atom is the charge that it would have if every bond were 100% covalent (non-polar). Formal charges are computed by using a set of rules and are useful for accounting for the electrons when writing a reaction mechanism, but they don't have any intrinsic physical meaning. They may also be used for qualitative comparisons between different resonance structures (see below) of the same molecule, and often have the same sign as the partial charge of the atom, but there are exceptions.

The formal charge of an atom is computed as the difference between the number of valence electrons that a neutral atom would have and the number of electrons that "belong" to it in the Lewis structure when one counts lone pair electrons as belonging fully to the atom, while electrons in covalent bonds are split equally between the atoms involved in the bond. The total of the formal charges on an ion should be equal to the charge on the ion, and the total of the formal charges on a neutral molecule should be equal to zero.

For example, in the hydronium ion, H3O+, the oxygen atom has 5 electrons for the purpose of computing the formal charge—2 from one lone pair, and 3 from the covalent bonds with the hydrogen atoms. The other 3 electrons in the covalent bonds are counted as belonging to the hydrogen atoms (one each). A neutral oxygen atom has 6 valence electrons (due to its position in group 16 of the periodic table); therefore the formal charge on the oxygen atom is 6 – 5 = +1. A neutral hydrogen atom has one electron. Since each of the hydrogen atoms in the hydronium atom has one electron from a covalent bond, the formal charge on the hydrogen atoms is zero. The sum of the formal charges is +1, which matches the total charge of the ion.

Formal Charge: number of valence electrons for an atom - (number of lone pair electrons + number electrons in bonds/2)

In chemistry, a formal charge (FC) on an atom in a molecule is defined as:

FC = number of valence electrons of the atom - ( number of lone pair electrons on this atom + total number of electrons participating in covalent bonds with this atom / 2).

When determining the correct Lewis structure (or predominant resonance structure) for a molecule, the structure is chosen such that the formal charge on each of the atoms is minimized.

### Examples

carbon in methane
Nitrogen in $NO_2^{-}$
double bonded oxygen in $NO_2^{-}$
single bonded oxygen in $NO_2^{-}$
 Methane (CH4): black is carbon, white is hydrogen Nitrogen dioxide (NO2): blue is nitrogen, red is oxygen

# Organic Chemistry/Spectroscopy

There are several spectroscopic techniques which can be used to identify organic molecules: infrared (IR), mass spectroscopy (MS) UV/visible spectroscopy (UV/Vis) and nuclear magnetic resonance (NMR).

IR, NMR and UV/vis spectroscopy are based on observing the frequencies of electromagnetic radiation absorbed and emitted by molecules. MS is based on measuring the mass of the molecule and any fragments of the molecule which may be produced in the MS instrument.

## UV/Visible Spectroscopy

UV/Vis Spectroscopy uses ultraviolet and/or visible light to examine the electronic properties of molecules. Irradiating a molecule with UV or Visible light of a specific wavelength can cause the electrons in a molecule to transition to an excited state. This technique is most useful for analyzing molecules with conjugated systems or carbonyl bonds.

NMR Spectroscopy

Nuclear Magnetic Resonance (NMR) Spectroscopy is one of the most useful analytical techniques for determining the structure of an organic compound. There are two main types of NMR, 1H-NMR (Proton NMR) and 13C-NMR (Carbon NMR). NMR is based on the fact that the nuclei of atoms have a quantized property called spin. When a magnetic field is applied to a 1H or 13C nucleus, the nucleus can align either with (spin +1/2) or against (spin -1/2) the applied magnetic field.

These two states have different potential energies and the energy difference depends on the strength of the magnetic field. The strength of the magnetic field about a nucleus, however, depends on the chemical environment around the nucleus. For example, the negatively charged electrons around and near the nucleus can shield the nucleus from the magnetic field, lowering the strength of the effective magnetic field felt by the nucleus. This, in turn, will lower the energy needed to transition between the +1/2 and -1/2 states. Therefore, the transition energy will be lower for nuclei attached to electron donating groups (such as alkyl groups) and higher for nuclei attached to electron withdrawing groups (such as a hydroxyl group).

In an NMR machine, the compound being analyzed is placed in a strong magnetic field and irradiated with radio waves to cause all the 1H and 13C nuclei to occupy the higher energy -1/2 state. As the nuclei relax back to the +1/2 state, they release radio waves corresponding to the energy of the difference between the two spin states. The radio waves are recorded and analyzed by computer to give an intensity versus frequency plot of the sample. This information can then be used to determine the structure of the compound.
Aromatics in H-NMR Electron Donating Groups vs. Electron Withdrawing Groups

On monosubstituted rings, electron donating groups resonate at high chemical shifts. Electron donating groups increase the electron density by releasing electrons into a reaction center, thus stabilizing the carbocation. An example of an electron donating group is methyl (-CH3).

Accordingly, electron withdrawing groups are represented at low chemical shifts. Electron withdrawing groups pull electrons away from a reacting center. This can stabilize an electron rich carbanion. Some examples of electron withdrawing groups are halogens (-Cl, -F) and carboxylic acid (-COOH).

Looking at the H NMR spectrum of ethyl benzene, we see that the methyl group is the most electron withdrawing, so it appears at the lowest chemical shift. The aromatic phenyl group is the most electron donating, so it has the highest chemical shift.

The sum of integrated intensity values for the entire aromatic region shows how many substituents are attached to the ring, so a total value of 4 indicates that the ring has 2 substituents. When a benzene ring has two substituent groups, each exerts an influence on following substitution reactions. The site at which a new substituent is introduced depends on the orientation of the existing groups and their individual directing effects. For a disubstituted benzene ring, there are three possible NMR patterns.

Note that para-substituted rings usually show two symmetric sets of peaks that look like doublets.

The order of these peaks is dependent on the nature of the two substituents. For example, the three NMR spectra of chloronitrobenzene isomers are below:

Mass Spectroscopy

A mass spectroscope measures the exact mass of ions, relative to the charge. Many times, some form of seperation is done beforehand, enabling a spectrum to be collected on a relatively pure sample. An organic sample can be introduced into a mass spectroscope and ionised. This also breaks some molecules into smaller fragments.

The resulting mass spectrum shows:

1) The heaviest ion is simply the ionised molecule itself. We can simply record its mass.

2) Other ions are fragments of the molecule and give information about its structure. Common fragments are:

 species formula mass methyl CH3+ 15 ethyl C2H5+ 29 phenyl C6H5+ 77

# Infrared spectroscopy.

Absorbing infrared radiation makes covalent bonds vibrate. Different types of bond absorb different wavelengths of infrared:

Instead of wavelength, infrared spectroscopists record the wavenumber; the number of waves that fit into 1 cm. (This is easily converted to the energy of the wave.)

For some reason the spectra are recorded backwards (from 4000 to 500 cm-1 is typical), often with a different scale below 1000 cm-1 (to see the fingerprint region more clearly) and upside-down (% radiation transmitted is recorded instead of the absorbance of radiation).

The wavenumbers of the absorbed IR radiation are characteristic of many bonds, so IR spectroscopy can determine which functional groups are contained in the sample. For example, the carbonyl (C=O) bond will absorb at 1650-1760cm-1.

## Summary of absorptions of bonds in organic molecules

Infrared Spectroscopy Correlation Table

 Bond Minimum wavenumber (cm-1) Maximum wavenumber (cm-1) Functional group (and other notes) C-O 1000 1300 Alcohols and esters N-H 1580 1650 Amine or amide C=C 1610 1680 Alkenes C=O 1650 1760 Aldehydes, ketones, acids, esters, amides O-H 2500 3300 Carboxylic acids (very broad band) C-H 2850 3000 Alkane C-H 3050 3150 Alkene (Compare intensity to alkane for rough idea of relative number of H atoms involved.) O-H 3230 3550 H-bonded in alcohols N-H 3300 3500 Amine or amide O-H 3580 3670 Free –OH in alcohols (only in samples diluted with non-polar solvent)

Absorptions listed in cm-1.

## Typical method

Typical apparatus

A beam of infra-red light is produced and split into two separate beams. One is passed through the sample, the other passed through a reference which is often the substance the sample is dissolved in. The beams are both reflected back towards a detector, however first they pass through a splitter which quickly alternates which of the two beams enters the detector. The two signals are then compared and a printout is obtained.

A reference is used for two reasons:

• This prevents fluctuations in the output of the source affecting the data

• This allows the effects of the solvent to be cancelled out (the reference is usually a pure form of the solvent the sample is in).

## Ionic and Covalent Bonds

------

Ionic bonds are formed when two ions are held together by electrostatic attraction. Shown on the left above is an ionic bond between lithium and fluorine in Li-F. These atoms are plotted with electrostatic potential surfaces. That is fancy language for saying that the overall charge is plotted in color. The plot on the left shows the surface as solid, while the plot on the right is the same except the surface is transparent so you can see the nuclei inside. The red color indicates the negative charge of the fluoride anion, the blue charge indicates the positive charge of the lithium cation. The atoms are held together because opposite charges attract each other. This is an ionic bond. Contrast this to the situation with the fuorine molecule (F-F) shown on the right in which both atoms have the same charge, so there is only green color, no red or blue. The type of bond found in the fluorine molecule is called a covalent bond and comes about because the atoms share electron density in order to each obtain a noble gas configuration.

The figure above uses three types of electron density models to compare the bonding and polarity of simple hydrides: LiH, H2, and HF. The mesh surfaces identify points where the electron density is relatively low (0.002 a.u.). These points more or less define the "edge" of the electron "cloud" in each molecule. Notice how the size of the electron cloud near H shrinks as its bonding partner changes from Li -> H -> F. Recalling the analysis of Li and Li+ given in Figure 1, you can conclude that the amount of electron density belonging to H is greatest in LiH and least in HF. In other words, Li and F do not share bonding electrons equally with H (equal sharing must occur in H2). Li donates electron density to H, but F "steals" electron density from H.

Chemists describe the ability of an atom to "steal" bonding electrons from its partner as its electronegativity. These figures demonstrate that electronegativity increases in the order: Li -- H -- F.

The colored maps show how each molecule's electrostatic potential varies on the 0.002 isodensity surfaces. The variation in potential is shown by color - RED (lowest) -> ORANGE -> YELLOW -> GREEN -> BLUE (highest) - and indicates whether a particular region is electron-rich (RED) or electron-poor (BLUE). LiH and HF are polar molecules in that the two ends of these molecules are electron-rich and electron-poor respectively. The change in potential around H is also consistent with the previous analysis based on surface size; H is electron-rich in LiH (RED), neutral in H2 (GREEN), and electron-poor in HF (BLUE).

Finally, the solid surfaces (inside the mesh surfaces) identify points where the electron density is relatively high (0.08 a.u.). Atoms that share electrons (covalently bonded) build up electron density in the region between the two nuclei. The models of H2 and HF show that these molecules contain covalent bonds - the electron density between the nuclei is 0.08 a.u. or greater. The model of LiH, on the other hand, suggests that this bond is largely ionic - although there are regions of high electron density around each nucleus, electron density is less than 0.08 between the two nuclei.

## Lewis Acids and Bases

Shown here, is the molecule BF3 represented in three different ways. The ball-and-stick model is drawn on the lower left. Boron is shown in purple and the fluorine atoms are shown in green. On the right, different views of the empty 2p-orbital belonging to boron are shown. With three bonds to fluorine (sp2 hybridization), and no lone pairs, there remains one 2p-orbital that is not hybridized and empty. Thus, it wants a lone pair of electrons to give it an octet of electrons. Also, fluorine is highly electronegative, withdrawing electron density from the boron atom. This is represented in the electrostatic potential model at the upper-left, with flourine atoms in red (partial negative charge) and boron in blue (partial positive charge). For all these reasons, the molecule BF3 is a good acceptor of electrons and therefore a good Lewis acid.

In this picture, an acid/base reaction is shown. NH3 has a lone pair of electrons. This is represented by the red area on the electrostatic potential model. The H+ is obviously positively charged as shown by its blue color. The lone pair of electrons on the NH3 molecule will donate these pairs of electrons to the H+, so here the NH3 is acting as a Lewis base as well as a Bronsted-Lowry base, and the H+ is acting a a Lewis acid and well as a Bronsted-Lowry acid. Notice that by convention, the arrow points from the electron donor to the electron acceptor. The bottom picture shows H+ bound to the NH3 molecule forming NH4+. Please note that the molecule is now sp3 hybridized and has a formal charge of +1.

In this diagram, NH3 again acts as a Lewis base. BF3 acts as a Lewis acid when it accepts the lone pair of electrons that NH3 donates. This reaction fills BF3's empty 2p-orbital, and now boron is sp3 hybridized when previously (as BF3) it was sp2 hybridized.

## Dipole Moments

For the model on the left, the white atom is hydrogen and the green atom is fluorine. The surface on the right uses color to indicate where the electrons are located in the H-F molecule. Here, the red color represents a PARTIAL negative charge (on fluorine atom), while the blue color represents PARTIAL positive charge (on hydrogen atom). There are large differences in electronegativity between hydrogen and fluorine, so that the majority of electron density in the hydrogen-fluorine bond ends up on the much more electronegative fluorine. Bonds such as this in which the electrons are not shared evenly are referred to as polar covalent bonds. Polar covalent bonds have a bond dipole moment.

For the molecular model shown on the left, the green atoms are fluorine, the light blue atom is carbon and the white atom is hydrogen. Difluoromethane (CH2F2) has two polar covalent C-F bonds as shown. For the surface on the right that indicates where their electrons are, the red color represents PARTIAL negative charge, while the blue color represents PARTIAL positive charge. As you can see, the entire molecule has a molecular dipole moment resulting from the vector sum of the two C-F bond dipole moments.

For the molecular model shown on the left, the light blue atom is carbon and the red atoms are oxygen. Carbon dioxide has two polar covalent C-O bonds. However, the bond dipole moments exactly cancel each other since they are pointing in exactly opposite directions. Thus, CO2 has no molecular dipole moment. Please note that the cancellation of bond dipole moments does not change the existence or placement of electron density. Each individual C-O bond has the normal bond dipole moment even though the molecular dipole moment is zero. By the way, you should be able to identify the carbon atom in CO2 as being sp hybridized.

For the molecular model shown on the left, the white atoms are hydrogen and the red atom is oxygen. Water has two very polar covalent bonds between oxygen and hydrogen. Because water is bent (the oxygen atom is sp3 hybridized), the two bond dipole moments add up to give water a relatively large molecular dipole moment. This molecular dipole moment combined with the individual bond dipole moments give water its unique properties that have allowed life to evolve as it has. In particular, later in the semester, we will see how these dipole moments explain water's remarkably high boiling point, as well as its ability to dissolve charged species such as the salt in the ocean.

CF4 has four polar covalent bonds. Because they are arranged in a symmetrical tetrahedral array, all of the bond dipole moment vectors exactly cancel, leaving NO MOLECULAR DIPOLE MOMENT for CF4. Hopefully, you can appreciate that deducing whether a given molecule has a molecular dipole moment is a favorite question of mine, because it forces you to synthesize everything we have learned thus far (molecular shape based on hybridization state and/or VSEPR, electronegativities) into the answer to a single question. Best of all, the prediction of molecular dipole moments will allow you to predict the properties of different molecules. You will encounter many examples of this as the semester unfolds.

### Introduction

The cleavage of a covalent heteronuclear chemical bond generally results in the formation of two oppositely charged ions, with the negative charge developing on the more electronegative atom of the bonded pair and the positive charge on the less electronegative atom. As the electronegativity difference between two bonded atoms becomes smaller and smaller, the tendency of the bond between them to break homolytically becomes larger and larger; each atom retains one of the electrons from the bond. Figure 1 presents several examples of reactions that involve both heterolytic and homolytic bond cleavage.

Figure 1

Divvy 'em up

Species in which an atom has an unpaired electron are called free radicals. The three free radicals shown in the bottom panel of Figure 1 are called chlorine atoms, hydroxyl radicals, and triphenylmethyl radicals. Chlorine atoms and hydroxyl radicals are extremely reactive materials. The triphenylmethyl radical, on the other hand, is very stable. In fact, no one has ever isolated hexaphenylethane, presumably because it decomposes spontaneously into two triphenylmethyl radicals.

According to VSEPR theory carbocations, R3C+, should be trigonal planar, while the carbanions, R3C:-, should be pyramidal. These predictions have been verified experimentally. You might expect that the geometry of a free radical, R3C., would be intermediate between that of a carbocation and a carbanion. Experimental measurements indicate, however, that carbon free radicals are essentially planar. Figure 2 compares the structures of these three species.

Figure 2

Three Forms of Trivalent Carbon

The stability of carbocations increases in the order CH3+ < CH3CH2+ < (CH3)2CH+ < (CH3)3C+. In other words, methyl < 1o < 2o < 30. The structural similarity between carbocations and carbon free radicals illustrated in Figure 2 suggests that these species should display a similar increase in stability as a function of increasing substitution at the central carbon. This expectation is borne out by experimental measurements. Presumably replacement of a hydrogen atom by an alkyl group creates the possibility for hyperconjugative stabilization of the free radical in the same way it does for the carbocation. Figure 3 demonstrates this partial obrital overlap between a C-H sigma bond of a methyl group and a p orbital on the central carbon atom of a free radical.

Figure 3

Hyperconjugation to the Rescue

When a free radical center is flanked by a pi system, a resonance interaction between the p orbital on the central carbon and the p orbitals of the pi bond(s) is possible. Figure 4 presents two views of this type of interaction for one common structure, the allylic radical. Note in this figure that the unpaired electron is originally on the carbon atom bearing the R groups. Consideration of the picture in the left hand panel reveals that overlap of the p orbital on the middle carbon with either of the flanking p orbitals is, to a first approximation, equally likely. This view is reenforced by the familiar electron pushing scheme depicted in the right hand panel. Note the use of a single barbed arrow to denote resonance delocalization of a single electron.

Figure 4

#### Polymerization

In our discussion of addition polymers we described the polymerization of alkenes in terms of an acid-catalysed process. In practice, polymerization of simple alkenes is more often initiated by a free radical than it is by a proton. Regardless of the initiator, the basic chain reaction mechanism applies. Figure 5 reviews the process which most often begins with the homolytic cleavage of the O-O bond of an organic peroxide.

Figure 5

Halogenation

At room temperature, in the absence of light, a mixture of methane and dichlorine is stable indefinitely. However, upon exposure to visible light, the components of the mixture react rapidly as shown in Equation 1. The reaction is called a free radical subsitituion.

The composition of the product depends upon the molar ratio of methane to dichlorine. The exclusive formation of chloromethane would require a very large excess of methane. However, the most interesting feature about reaction 1 is that a single photon causes the formation of thousands of molecules of chloromethane. In other words, one photon sets off a chain reaction. This and other observations led to the formulation of the 4-step mechanism shown in Figure 6.

Figure 6

Workin' on the Chain Gang

The process begins with the photochemically induced homolytic cleavage of the Cl-Cl bond to produce two chlorine atoms. This step is called initiation. In the second step, one of the chlorine atoms collides with a molecule of methane; the collision results in the transfer of a hydrogen atom, from the carbon to the chlorine. The hydrogen atom transfer results in the formation of a molecule of hydrogen chloride and a methyl radical. Since the second reaction creates a new radical, it is called a propagation reaction. A second propagation step follows as the methyl radical collides with a dichlorine molecule, an encounter that yields a molecule of chloromethane and a chlorine atom. Collision of the chlorine atom generated in the second propagation step with another molecule of methane forms a second molecule of HCl and a second methyl radical. The two propagation steps continue until the concentrations of Cl2 and/or CH4 are reduced to the point where the probability of a collision between two radicals becomes significant. Such a collision constitutes the termination step of the process.

Free radical halogenation of alkanes is a general reaction. The chlorination of 2,3-dimethylbutane provides an informative example. This simple alkane contains two types of hydrogen atoms, twelve primary and two tertiary. As shown in Equation 2, chlorination of this compound produces a mixture of 2-chloro-2,3-dimethylbutane and 1-chloro-2,3-dimethylbutane. The composition of this mixture is 82.5% 2-chloro-2,3-dimethylbutane and 17.5% 1-chloro-2,3-dimethylbutane. This product distribution is different than would be expected if the only factor governing the ratio of the two isomers were the ratio of primary and tertiary hydrogens in the starting material.

The product distribution in reaction 2 suggests that a 3o hydrogen atom is nearly six time more reactive than a 1o hydrogen atom.This increased reactivity has been attributed to the greater stability of the 3o free radical in comparison to the 1o alternative. The more stable free radical is formed more readily, i.e. faster. Figure 7 presents a reaction coordinate diagram that compares the relative energies of the reactants, intermediates, and products involved in reaction 2. The key feature of the figure is not the relative energies of the products, but rather the relative energies of the intermediates, and more importantly, the relative activation energies for their formation.

Figure 7

One Reactant, Two Products

Follow the reaction pathway highlighted in red in the figure: When a chlorine atom collides with a molecule of 2,3-dimethylbutane at one of the 3o hydrogen atoms, the C-H bond begins to break as the H-Cl bond begins to form. At the point labeled T3,1 the configuration of the C---H---Cl fragment has its maximum energy. As the H-Cl bond becomes stronger, the C-H bond weakens. As the Cl atom rebounds from the collision, it takes the H atom with it, leaving the intemediate tertiary free radical, I3, behind. Collision of this intermediate with a molecule of dichlorine produces further bonding changes that lead to the transition state labeled T3,2. From that point the path to the product is energetically downhill. A similar series of events occurs along the pathway highlighted in blue. The key difference is that E3act is less than E1act.

While we now interpret experimental outcomes such as those seen for reaction 2 in tems of the relative stabilities of the intermediates formed along the alternative reaction pathways, it is important to remember that our understanding of relative stabilities evolved from the analysis of product distribution data from many reactions similar to reaction 2.

#### Combustion

As shown in Equation 3, the chlorination of methane will produce carbon tetrachloride if there is sufficient dichlorine in the reaction mixture.

The carbon atom in this reaction undergoes a change in oxidation level from -4 to +4; it is oxidized. A similar oxidation occurs when methane is treated with dioxygen:

In this case the reaction is called combustion. Like chlorination, combustion of methane, Equation 4, involves free radical reactions.

In our introduction to MO theory we saw that dioxygen is a diradical; there is an unpaired electron on each oxygen atom. Dioxygen is a very reactive molecule. While its reactivity is essential for our existence, it can be a bane to that existence, and it has even been implicated in our demise. Dioxygen-promoted free radical reactions are definitely involved in the spoilage of food and have been implicated as contributors to the aging process.

Since many free radical reactions are chain reactions, a simple strategy for breaking the chain involves addition of a compound that will form stable, i.e. non-reactive, free radicals. For example, the free radical produced by the reaction of 2,6-di-t-butyl-4-methylphenol with dioxygen, Equation 5, has very low reactivity. Not only is the free radical center sterically hindered by the two bulky t-butyl groups, it is also stabilized by resonance interaction with the aromatic ring.

The abbreviation for 2,6-di-t-butyl-4-methylphenol is BHT, which stands for butylated hydroxytoluene. This compound is added to many packaged foods as a preservative to prevent the food from becoming stale, a process that involves oxygen- promoted free radical reactions. Another common preservative is BHA, which stands for butylated hydroxyanisole, a compound in which the CH3 group of BHT is replaced by an OCH3 group.

Free radical inhibitors are also used to prevent spoilage of food products that contain unsaturated fats and oils. The allylic C-H bonds in these structures are easily attacked by free radicals, which in turn react with dioxygen in a radical chain reaction that is called autooxidation. Figure 8 summarizes the essential features of the process using the polyunsaturated fatty acid linolenic acid as an example.

Figure 8

Autooxidation of An Unsaturated Fatty Acid

In this scheme In. represents any species that initiates the chain reaction. Note that the free radical formed in step 1 is doubly allylic. Therefore it is formed readily. Once formed, it reacts with dioxygen in step 2 to produce a peroxy radical which abstracts a hydrogen atom from another linolenic acid molecule in step 3. This produces an organic peroxide and a new free radical, thereby propagating the chain reaction. Food additives such as BHT inhibit autooxidation and retard spoilage.

Fats present in cells within our bodies are subject to similar autooxidation, which may lead to cellular damage. Vitamin E is a natural antioxidant that has received considerable attention in the poplular press recently for its "anti-aging" capabilities. Consideration of the structure of vitamin E would suggest that, to the extent that free radical reactions contribute to aging, there is some basis for these claims.

Oxidation Levels

### Introduction

Most atoms have one or two stable oxidation states. Carbon has 9!! Many of the reactions that organic molecules undergo involve changes in the oxidation level of one or more carbon atoms within the compound. For example, during the combustion of methane, which produces carbon dioxide, the oxidation level of the carbon atom changes from -4 to +4:

The procedure for calculating the oxidation level of an atom is similar to that for determining its formal charge.

#### Rule 3: Calculating Oxidation Levels

To determine the oxidation level of an atom within a molecule, separate the atom from its bonding partner(s), assigning all bonding electrons to the more electronegative of the bonded atoms. Then compare the number of electrons that "belong" to each atom to the atomic number of that atom. Figure 1uses color coding to illustrate the procedure for methane, CH4.

Figure 1

Assigning Electrons I

Since carbon is more electronegative than hydrogen, both electrons from each C-H bond are assigned to the carbon. Counting its two inner shell electrons, the carbon has 10 electrons assigned to it. Its oxidation level is the sum of its nuclear charge (atomic number) and the its electronic charge; 6+ (-10) = -4. The oxidation level of each hydrogen atom is 1 + (0) = +1. Note that the sum of the oxidation levels of all of the atoms in the molecule equals zero. This is always true of neutral molecules and provides a convenient way for you to check your calculations.

Figure 2 illustrates the assignment of electrons in carbon dioxide. No color coding is used.

Figure 2

AssigningCl Electrons II

All of the bonding electrons are assigned to the oxygen atoms. So are the lone pairs. Counting its two inner shell electrons, each oxygen has 10 electrons assigned to it. The oxidation level of each oxygen is 8 + (-10) = -2. The oxidation level of the carbon is 6 + (-2) = +4. Again, note that the sum of the oxidation levels of all the atoms in CO2 equals zero.

If two bonded atoms have the same electronegativity, the electrons they share are divided equally between the two atoms. Figure 3 shows how the electrons are assigned for acetic acid.

Figure 3

Assigning Electrons III

The oxidation level of each hydrogen atom is +1. The oxidation levels of the methyl and carboxyl carbons are -3 and +3, respectively. Right? The oxidation level of each oxygen is -2. To check: 4 x (+1) + (-3) + (+3) + 2 x (-2) = 0!

An equivalent method of calculating oxidation levels ignores the inner shell electrons; the oxidation level of an atom is the difference between its group number and the number of valence electrons assigned to it . Prove to yourself that this method works by using it to calculate the oxidation levels of all the atoms in acetic acid.

In the same way that chemists calculate the index of hydrogen deficiency of an empirical formula almost without thinking, they also perform subconscious calculations of the oxidation level of each atom within a structure. So, if you want to think like an organic chemist (which is advisable if you want to get a good grade), you should practice calculating oxidation levels until you can do it in your head.

Oxidation And Reduction Reactions in Organic Chemistry

### Introduction

As we saw in our discussion of oxidation levels, one of the unique characteristics of carbon is that it has nine stable oxidation states. It should not be surprising that organic chemists have developed reagents that allow them to alter these oxidation levels. This topic presents a survey of some of those reagents.

### Oxidizing Reagents

We have already seen several examples of such reagents in our discussion of the oxidation of alcohols. They are repeated here for the sake of completeness.

#### Chromic Acid

This reagent is prepared by mixing sodium or potassium dichromate with sulfuric acid as shown in Equation 1.

It is used to oxidize secondary alcohols to ketones:

It may also be used to oxidize primary alcohols to carboxylic acids. As Equation 3 indicates, the alcohol is initially oxidized to an aldehyde. Under the reaction conditions, a molecule of water adds to the carbonyl group to form a hydrate which is subsequently oxidized to the carboxylic acid.

#### Pyridinium Chlorochromate (PCC)

In order to prevent aldehydes from further oxidation, it is necessary to avoid the addition of water to the carbonyl group. PCC was developed as a non-aqueous alternative to chromic acid. Using this reagent, 2-phenylethanol may be oxidized to phenylacetaldehyde without subsequent oxidation to phenylacetic acid:

#### Potassium permanganate (KMnO4) and Osmium tetroxide (OsO4)

These reagents are used to convert alkenes into the corresponding 1,2-diols (glycols) by a process called syn hydroxylation. Equation 5 illustrates the process for the reaction of 1,2-dimethylcyclohexene with a dilute solution of potassium permanganate.

The reaction is thought to involve the formation of an intermediate cyclic permanganate ester which is readily hydrolysed under the reaction conditions to yield the 1,2-diol. A cyclic osmate ester is generated with OsO4.

Since aqueous KMnO4 is purple, this reaction is often used as a qualitative test for the presence of an alkene: a dilute solution of permanganate is added to a sample of the unknown compound; if the color is discharged, the test is taken as positive. The formation of a grey-black precipitate of manganese dioxide confirms the analysis.

#### Ozone

Ozone, O3, is an allotrope of oxygen. It is a highly reactive molecule that is generated by passing a stream of dioxygen over a high voltage electric discharge. (It is possible to smell ozone in the atmosphere after a lightning storm if the lightning has struck nearby.) It is not possible to draw a single structure for O3 in which each oxygen atom has a filled valence shell and is, at the same time, uncharged. Rather, resonance theory describes the structure of this compound as a hybrid of the three resonance contributors shown in Figure 1.

Figure 1

Ozone: An Allotrope of Oxygen

In a process called ozonolysis, an alkene is treated with ozone to produce intermediates called ozonides, which are reduced directly, generally with zinc metal in acetic acid, to yield aldehydes or ketones, depending on the substituents attached to the double bond of the initial alkene. Equations 6 -8 provide three specific examples.

Note that an aromatic ring is resistant to ozone.

The value of ozonolysis lies in the structural insight it affords a chemist who is trying to determine the identity of an unknown compound. Figure 2 illustrates this idea.

Figure 2

Structural Elucidation with Ozonolysis

The unknown is degraded into smaller, simpler molecules that are more readily identified. Once identified, these fragments are then mentally reconnected by joining the carbonyl carbons together to create an alkene.

Reducing Agents

In our discussion of the oxidation of alcohols, we classified this process as a 1,2-elimination of the "elements of" dihydrogen. The reverse process, the 1,2-addition of the "elements of" dihydrogen to a multiple bond, constitutes a reduction. The reagents used for the 1,2-addition of the "elements of" dihydrogen to a multiple bond depend upon the nature of the multiple bond. For homonuclear multiple bonds, i.e. alkenes and alkynes, the most common method is called catalytic hydrogenation: a solution of the alkene or alkyne is mixed with dihydrogen gas in the presence of a catalytic quantity of a transition metal. For heteronuclear multiple bonds; aldehydes, ketones, nitriles, esters, etc., addition of the "elements of" dihydrogen is genarally accomplished in two steps, addition of hydride ion, :H-, followed by addition of H+. Since the addition of hydride ion is rate determining, these reductions are called hydride ion reductions. We'll take a look at catalytic hydrogenation first.

#### Catalytic Hydrogenation

The catalyst most commonly used to reduce carbon-carbon multiple bonds consists of platinum metal dispersed over the surface of finely divided charcoal (Pt/C). Palladium and rhodium are also used (Pd/C and Rh/C). These reagents are available commercially. The catalyst is mixed with a solution of the alkene or alkyne dissolved in an inert solvent such as ethanol or diethyl ether. Since the reaction mixture is heterogeneous, it is important that the catalyst be dispersed over a large surface area in order to insure adequate contact between the reactants and the catalyst. The charcoal provides the required surface area.

The mechanistic details of catalytic hydrogenations are uncertain because of the difficulties associated with studying heterogeneous reactions.

Adsorption onto the catalytic surface brings the reactants into proximity. It also weakens the H-H and the C-C bonds, increasing the reactivity of the reactants. The details of the transfer of the hydrogen atoms from the platinum to the alkyne are uncertain. However, as the animation indicates, both hydrogen atoms add to the same side of the pi bond, leading to the formation of cis-2-butene. In other words, catalytic hydrogenation of alkenes and alkynes involves the syn addition of the "elements of" dihydrogen to the multiple bond. Equation 9 illustrates the syn addition of dihydrogen to (E)-3-chloro-2-phenyl-2-butene. Since the addition occurs to the top and bottom faces of the pi bond with equal probability, the reaction produces a racemic mixture of the two stereoisomers shown.

As reaction 10 indicates, catalytic hydrogenations are not restricted to alkenes and alkynes. Compounds containing multiple bonds between carbon and a heteroatom such as oxygen or nitrogen may also be reduced catalytically.

While the outcome depicted in reaction 10 may be desireable, it would be nice to have a method to hydrogenate heteronuclear multiple bonds while leaving carbon-carbon multiple bonds untouched. Such a method exits. It takes advantage of the fact that, unlike C-C multiple bonds, which are relatively non-polar, heteronuclear multiple bonds have a permanent bond dipole; the carbon is electron deficient while the heteroatom is electron rich. This means that the carbon atom of a heteronuclear multiple bond is inherently reactive toward negatively charged reagents, while the heteroatom is inherently reactive towards positively charged reagents. Figure 3 demonstrates this reality for negatively and positively charged hydrogen atoms.

Figure 3

Opposites Attract

The net result of the addition of :H- and H+ to the multiple bond is the 1,2-addition of the "elements of" dihydrogen. Since it is not possible to have significant concentrations of :H- and H+ in the same flask at the same time (why?), hydrogenation of heteronuclear multiple bonds is normally a 2-step process; hydride ion adds to the carbon atom in the first step, while a proton adds to the heteroatom in the second. The remainder of this topic will consider different reagents that act as a source of hydride ion.

#### Hydride Ion Reductions

Reagents that act as hydride ion donors all share one structural feature: They all contain at least one hydrogen atom that is bonded to another atom which is less electronegative than hydrogen. The greater the difference in electronegativity, the more reactive the reagent will be as a hydride donor. The most reactive source of hydride ion is lithium aluminum hydride, LiAlH4. This material is a grey solid that reacts violently with protic solvents. Most commonly it is used as a suspension in a dry, inert solvent such as diethyl ether or THF. A solution of the compound to be reduced is added to this suspension and stirred vigorously until analysis indicates that all of the starting material has reacted. At this point the mixture is acidified by the careful addition of aqueous acid. Figure 4 illustrates these two steps for the reduction of acetophenone.

Figure 4

LiAlH4 Reduction of Acetophenone

Note that all four hydrogen atoms attached to the aluminum in LiAlH4 are active; one mole of LiAlH4 will reduce four moles of the ketone.

LiAlH4 is so reactive that it will reduce almost any type of heteronuclear multiple bond. It will even reduce carboxylic acids and esters to the corresponding primary alcohols as indicated in reactions 11 and 12, and it reduces amides to amines as shown in Equation 13.

Clearly these reactions are more complicated than the mechanism shown in Figure 5 would suggest. Elimination of water as well as reduction must be involved.

Before we consider less reactive hydride donors, let's revisit the reaction of the unsaturated ketone we considered in Equation 10. Figure 5 compares the catalytic hydrogenation of pent-4-en-2-one with its reduction by LiAlH4.

Figure 5

Reduction of Homonuclear and Heteronuclear Multiple Bonds

Because simple, i.e. non-conjugated, double bonds are non-polar, they are non-reactive towards nucleophilic reagents.

While lithium aluminum hydride's high reactivity may be useful, it can also be a disadvantage if you want to selectively reduce one heteronuclear multiple bond in a compound that contains several. In that case it is desireable to have a reagent that is less reactive and more selective. One such reagent is sodium borohydride, NaBH4. This material is a white solid. It is much less reactive than LiAlH4. In fact, it is possible to reduce aldehydes, ketones, and esters with NaBH4 even in protic solvents such as ethanol. It will not reduce carboxylic acids or amides. The mechanism of the reaction is very similar to that shown in Figure 5. Figure 6 compares the results of the reduction of a compound that contains both an ester and an amide group with LiAlH4 and NaBH4.

Figure 6

LiAlH4 vs NaBH4

Now let's consider another aspect of reactivity. For compounds that belong to the carboxylic acid family, in particular carboxylic acids, esters, and amides, the oxidation level of the carboxyl carbon is +3. As you can see from the reactions in Figure 7, the oxidation level of the carboxyl carbon decreases to -1 when the carboxyl group is reduced to a primary alcohol. The question then becomes, "Can you stop the reduction at an intemediate oxidation level?" The answer is yes. Equation 14 shows how.

Here the ester is reduced to an aldehyde. The oxidation level of the carbonyl carbon decreases from +3 to +1. The trick is to use a sterically hindered reducing agent that has only one active hydrogen. In this case the reagent is called diisobutyl aluminum hydride, sometimes abbreviated DIBAL. By using 1 equivalent of DIBAL at low temperatures it is possible to reduce the ester to the corresponding aldehyde without further reduction of the aldehyde to the primary alcohol. Use of more than 1 equivalent will lead to reduction of the aldehyde.

Finally, Table 1 summarizes the reactivities of the various reducing reagents we have considered in this topic.

Table 1

 Reducing Agent Alkenes Aldehydes Ketones Esters Amides Carboxylic Acids H2, Pt/C alkanes 1o alcohols 2o alcohols NR NR NR LiAlH4 NR 1o alcohols 2o alcohols 1o alcohols amines 1o alcohols NaBH4 NR 1o alcohols 2o alcohols 1o alcohols NR NR DIBAL .... 1o alcohols 2o alcohols aldehydes aldehydes 1o alcohols

Hydride Reductions

Oxidation of Alcohols

### Introduction

The conversion of alcohols into aldehydes and ketones is one of the most common and most useful transformations available to the synthetic organic chemist. The general features of this oxidation reaction are outlined in Figure 1.

Figure 1

The Oxidation of Alcohols

A synonym for the oxidation of alcohols, dehydrogenation, suggests the structural feature that is required for this process: H-C-O-H. The OH group must be attached to a carbon atom that is bonded to at least one hydrogen atom. In other words, oxidation of alcohols involves the 1,2-elimination of "the elements of" dihydrogen, H and H.

This structural requirement means that 3o alcohols do not normally undergo oxidation. All of the alcohols shown below will not undergo oxidation under reaction conditions where 1o and 2o alcohols react readily.

### Oxidizing Agents

There is a wide variety of reagents that are used for the oxidation of alcohols. Two of the most common are chromic acid, H2Cr2O7, and pyridinium chlorochromate, PCC. Chromic acid is prepared by treatment of sodium or potassium dichromate with aquesous sulfuric acid as shown in Equation 1.

Chromic acid is most commonly used to oxidize 2o alcohols to ketones. One example is given in Equation 2.

Pyridinium chlorochromate is made by mixing chromium trioxide with pyridine and hydrochloric acid as indicated in Equation 3. The oxidizing component of PCC is the chlorochromate anion, CrClO3-.

PCC was developed especially for the oxidation of 1o alcohols to aldehydes, a transformation which is difficult to accomplish using chromic acid because aldehydes react rapidly with aqueous chromic acid to produce carboxylic acids.

Figure 2 compares the oxidation of 2-phenylethanol by chromic acid and PCC.

Figure 2

Oxidation Alternatives

Mechanism

Oxidation of alcohols is basically a two step process. The first step involves the formation of chromate esters. In our discussion of esterification, we saw that alcohols react with carboxylic acids, phosphoric acid, and sulfonic acids to produce various types of esters. The same is true for chromic acid and PCC; they react with alcohols to produce chromate esters. Once the chromate ester is formed, it undergoes an elimination reaction to generate the carbonyl group of the aldehyde or ketone. These two steps are outlined in Figure 3.

Figure 3

The Mechanism of Chromate Oxidations

### Examples

The oxidation of the secondary alcohol menthol to the ketone menthone, as outlined in Equation 4, provides a simple example of a dichromate oxidation. The product is formed in over 90% yield.

Equation 6 presents a variation on the theme discussed above. In this reaction the oxidizing agent is chromium trioxide, CrO3 dissolved in acetic acid. It converts a 2o alcohol into a ketone. This oxidation is part of a multi-step synthesis of a terpene called longifolene, which is a component of pine oil.

Synthetic chemists often pursue exotic targets. The synthesis of sirenin, Equation 6, offers a case in point. A key step in the multi-step synthesis of this material was the PCC oxidation of a 1o alcohol to an aldehyde.

Sirenin is the sperm attractant from the female gametes of a water mold. Now that's exotic.

Enolization

### Introduction-

Hydrocarbons are among the weakest acids known. Conversely, their conjugate bases are some of the strongest bases there are. Why is it so hard to remove a proton from a carbon atom? The answer is because the conjugate base is very unstable. When an excess of electron density develops on a carbon atom, the nucleus of that atom cannot offer much Coulombic attraction to stabilize it. Consider the hypothetical acid-base reactions shown in Figure 1.

Figure 1

Comparison of Four Acid-Base Reactions

In each case removal of a proton from the central atom generates an anion. The central atom of each anion has one more electron than it has protons in its nucleus. Figure 2 offers a schematic comparison of the central atom of an alkanide and an amide

Figure 2

Comparison of Charge Ratios in Alkanide and Amide Ions

The carbon has 2 electrons in its inner shell, and 5 electrons in its valence shell. The ratio of electrons to protons is 7/6 = 1.167. For the amide ion this ratio is 8/7 = 1.143. This is closer to 1/1 than in the case of carbon. Hence the amide ion is more stable than the alkanide ion. While the numerical difference in these ratios is small, the effect is huge: ammonia is 1012 more acidic than methane. Conversely, the amide ion is 1012 more stable than the methanide ion. You should extend this logic to hydroxide and fluoride ions.

Now consider the acid-base reaction of acetone shown in Equation 1.

Experiments have shown that the pKa of acetone is 19. It is 1031 times more acidic than methane! This means that the conjugate base of acetone is 1031 times more stable than the conjugate base of methane. Why? What structural feature is there in the conjugate base of acetone that affords such stabilization? Clearly the answer must be the carbonyl group. As the proton is being removed from the carbon, the geometry about the carbon begins to change. As the negative charge increases, the geometry around the a-carbon changes in order to maximize the overlap between the orbital containing the lone pair of electrons and the pi orbitals of the carbonyl group.

Interlude

#### Vocabulary

Figure 3 presents several structures with annotations that illustrate important definitions associated with this area of chemistry.

Figure 3

The Language of Enolization

Experimental Evidence

#### Isotope Exchange

When acetone is treated with base in D2O, it is slowly converted into hexadeuteroacetone as shown in Equation 2.

#### Bromination

Compounds that contain at least one hydrogen atom a to a carbonyl group quickly decolorize dilute aqueous solutions of dibromine and sodium hydroxide. Equation 3 illustrates the reaction for acetone.

The disappearance of the red-orange color indicates that the dibromine has been consumed. The products of the reaction, a-bromoacetone and sodium bromide are colorless. The reaction involves nucleophilic attack of C-a of an enolate ion on a molecule of dibromine. See Figure 5.

#### The Iodoform Reaction

Given the chemistry described in Equation 3, it shouldn't be surprising that compounds which contain at least one hydrogen atom a to a carbonyl group also decolorize aqueous solutions of diiodine and sodium hydroxide. In the case of methyl ketones, the decoloration is accompanied by the formation of a molecule of iodoform, CHI3, a yellow solid. The decoloration of the solution and the formation of the yellow solid is taken as a positive result. Equation 4 illustrates the process for 2-pentanone. Note that the starting ketone is converted into a carboxylic acid that contains one fewer carbon atoms.

In reaction 4, each of the three hydrogens on the methyl group attached to the carbonyl carbon is replaced by an iodine atom. The hydrogen atom in the iodoform comes from the acid that is added in the second step of the process.

Silylenol ethers

If a strongly basic solution of a ketone or aldehyde containing at least one a-hydrogen is "quenched" with chlorotrimethylsilane, it is possible to isolate a silyl enolether as shown in Equation 5.

Equation 5 differs from Equations 3 and 4 in two important ways. First, the equilibrium constant for the first step of reaction 5, i.e. deprotonation of the a-hydrogen, is approximately 1019, while that for reactions 3 and 4 is about 10-3. This means that there is a very high concentration of the enolate ion in reaction 5, but not in reactions 3 and 4. The importance of this fact will become obvious when we discuss aldol condensations in general, and crossed aldol condensations in particular. Second, the oxygen atom of the enolate ion acts as the nucelophile. This reflects the fact that the Si-O bond is an extremely strong bond, much stronger than a Si-C bond.

### NMR Spectroscopy

In molecules which contain hydrogen atoms that are a to two carbonyl groups, the enol tautomer may actually be the predominant form of the compound even in the absence of base. Figure 4 shows a partial 1H-NMR spectrum of 2,4-pentanedione. Integration of the signals for the methyl groups of the keto and enol forms of this compond reveals the relative amounts of each tautomer.

Figure 4

A Partial NMR Spectrum of 2,4-Pentanedione

The enol/keto ratio depends upon the solvent. In the case of 2,4-pentanedione it varies between 3/1 and 4/1.

### Mechanistic Interpretation

All of the lines of evidence described above may be rationalized by a process that involves the formation of an enolate ion. Figure 5 outlines the steps involved in the bromination of acetone.

Figure 5

Mechanistic Interpretation of the Bromination of Acetone

In the first step, the base abstracts a hydrogen atom from a methyl carbon. Electrons from the C-H bond move toward the carbonyl group, generating the enolate ion. In the second step, the negative charge on the oxygen atom moves back toward the carbon in order to regenerate the carbonyl group. As this happens, the electrons in the C-C double bond of the enolate ion form a bond to one of the bromine atoms in dibromine. This is a nucleophilic substitution reaction in which the enolate ion acts as a nucleophile and displaces a bromide ion from the dibromine.

In the isotope exchange experiment described in Equation 2, the enolate ion displaces -OD from D2O in a similar manner. The process is repeated until all of the hydrogen atoms have been replaced by deuteriums. Using D2O as the solvent insures that all of the hydrogen atoms will be exchanged.

In the iodoform reaction, the exchange of the three methyl hydrogens for iodine atoms produces a triiodomethylketone. The triiodomethyl group is then displaced by hydroxide ion as shown in Figure 6, again using 2-pentanone as an example.

Figure 6

The Last Stage of the Iodoform Reaction

Reaction of the triiodomethylketone with hydroxide ion as shown in Figure 6 is a nucleophilic acyl substitution reaction. The triiodomethanide ion is a reasonable leaving group since the negative charge on the carbon atom is stabilized by the three iodine atoms.

### Carbon vs Hydrogen

Consider the reactions shown in Equations 6 and 7.

Given that the equilibrium constant for reaction 6 is approximately 103 times greater than that for reaction 7, how is it hydroxide ion prefers to act as a base rather than as a nucleophile? The answer is that it doesn't. Reaction 6 occurs in preference to reaction 7. In fact, for every time a hydroxide ion abstracts an a-hydrogen from one acetaldehyde molecule, 1000 hydroxide ions add to the carbonyl groups of other acetaldehyde molecules. However, reaction 6 is readily reversible. Reaction 7, while reversible, may also proceed forward if the enolate ion reacts with an electrophile such as water, dibromine, diiodine, or chlorotrimethylsilane. It's sort of like the tortoise and the hare; reaction 6 gets off to a speedy start, but reaction 7 wins out in the end. Reaction 7 is the first step in an important carbon-carbon bond making process called the aldol reaction.

The Aldol Reaction

### Introduction-

Near the end of the discussion of enolization there was an analysis of the reactivity of hydroxide ion towards the carbonyl carbon compared to the hydrogen a to the carbonyl carbon in acetaldehyde. That analysis concluded that addition to the carbonyl group was approximately 10,000 times more likely than abstraction of an a-hydrogen. But because addition of hydroxide ion to the carbonyl group is readily reversible, it is a non-productive reaction.

While the abstraction of an a-hydrogen is also a reversible reaction, the enolate ion that is formed has an alternative reaction pathway available to it- addition to the carbonyl carbon of another acetaldehyde molecule. This is called the aldol reaction. Figure 1 summarizes the process.

Figure 1

The Aldol Reaction of Acetaldehyde

Note that the net effect of the aldol reaction is to add the "components of" one molecule of acetaldehyde to the carbonyl group of a second molecule of acetaldehyde, the "components of" acetaldehyde being H and HCOCH2.

The aldol reaction of two aldehydes is of limited synthetic utility. However, there are many "aldol-like" reactions which involve the essential features described in Figure 1. Figure 2 highlights these features.

Figure 2

General Features of the Aldol Reaction

The aldol reaction requires an aldehyde or ketone that contains at least one a-hydrogen. The a-carbon becomes nucleophilic when it is deprotonated by a base. The carbonyl carbon is electrophilic. Coulomb's Law brings these two oppositely charged species together to form a C-C bond.

The R groups may be H, alkyl, or aryl. When the R groups in one molecule are different than those in the other, the reaction is called a crossed-aldol reaction. The ability to join different aldehydes and ketones together is what give this process its synthetic value.

The word aldol is a common name for the product of the reaction shown in Figure 1. It is a type of compound called a b-hydroxyaldehyde. Generally the word aldol is used to refer to any b-hydroxyaldehyde or b-hydroxyketone.

Like other alcohols, b-hydroxyaldehydes and b-hydroxyketones undergo dehydration to produce alkenes. In fact, it is difficult to isolate b-hydroxyaldehydes and b-hydroxyketones because they are very prone to dehydration.

### Dehydration of b-hydroxyaldehydes and b-hydroxyketones

In order to isolate the product shown in Figure 1, the reaction conditions must be mild; the temperature must be kept low and the amount of acid used to protonate the alkoxide ion intermediate must be carefully controlled. If too much acid is added, or if the temperature is too high, the aldol will dehydrate to form a conjugated alkene as demonstrated in Figure 3.

Figure 3

Dehydration of a b-hydroxyaldehyde

The conjugated alkene shown in Figure 3 is also called an a, b-unsaturated aldehyde. This means that the double bond is between the carbon atoms a and b to the carbonyl carbon. In this position the p orbitals of the double bond may interact with those of the carbonyl group to form an extended, i.e. delocalized, pi system. The delocalization of the electrons in this pi system results in greater stabilization since the electrons experience greater nuclear attraction. Figure 4 offers two perspectives of the orbital geometry that affords this extra stabilization. The trans isomer was chosen arbitrarily.

Figure 4

Orbital Alignment in a, b-unsaturated Systems

Retrosynthetic Analysis

For synthetic organic chemists it's important to develop the ability to mentally "deconstruct" a target structure into simpler molecules from which that target may be made. The process of intellectual deconstruction is called retrosynthetic analysis. The focal point of such endeavors is inevitably the functional group(s) within the target molecule. In the case of a,b-unsaturated aldehydes or ketones the functional group of interest is the C-C double bond. Disconnecting these two carbons as animated in Figure 5 reveals the structures of the two components from which the target molecule was prepared.

Figure 5

Take One Step Back

Examples

Treatment of acetone with base results in the aldol reaction shown in Equation 1.

Experimentally reaction 1 is tricky to perform. However, if the product is separated from the reaction mixture as it is formed, it is possible to isolate the product in over 70% yield.

Acetone participates in a crossed-aldol reaction with furfural, an aldehyde produced from corn stalks, as described by Equation 2, where the carbon-carbon bond that is formed is highlighted in red.

An amazing aldol-type reaction was involved in the total synthesis of ginkolide B, one of the active components in extracts from the Ginko biloba tree. Equation 3 outlines this key step.

In this reaction the potentially nucleophilic carbon is a to the carbonyl group of an ester rather than a ketone or aldehyde. Lithium diisopropylamide (LDA) was used to deprotonate this carbon. The resulting enolate ion added to the carbonyl carbon of the complex pentacyclic ketone to form the C-C bond shown in red.

Equation 4 depicts an intramolecular crossed-aldol reaction that constituted the last step in a total synthesis of racemic progesterone. Even though the reaction conditions were very mild, the intermediate

b-hydroxyketone underwent spontaneous dehydration to produce the a, b-unsaturated ketone.

### Introduction

Resonance theory is a valuable extension of valence bond theory because it offers chemists a simple and reliable way to rationalize and/or predict the results of many reactions involving conjugated systems. In this topic we will examine a small but important group of reactions of molecules containing a carbonyl group that is conjugated to a carbon-carbon double bond. We encountered this molecular fragment during our discussion of aldol condensations. Figure 1 reviews this reaction sequence. The process begins when a base removes a proton from the a-carbon atom of the aldehyde. This generates a low concnetration of enolate ion A, which reacts with a molecule of acetaldehyde that has not been deprotonated as indicated by the arrows labeled 1, 2, and 3. Protonation of the resulting alkoxide ion B leads to the b-hydroxyaldehyde known as aldol. While it is possible to isolate this compound, it is also easy to dehydrate it; 1,2-elimination of water leads to the a,b-unsaturated aldehyde 2-butenal.

Figure 1

The Aldol Condensation

Now that you know how to prepare them, let's take a look at an important aspect of the chemistry of these molecules.

### Nucleophilic Addition Reactions of a, b-unsaturated Systems

Because of the conjugation of the double bond with the carbonyl group, a,b-unsaturated aldehydes and ketones possess two electrophilic carbons, the carbonyl carbon and the b-carbon. Figure 2 shows the familiar resonance interaction between the two functional groups in 2-butenal.

Figure 2

It is apparent that the carbonyl carbon of 2-butenal is electron deficient, simply by virtue of being bonded to a more electronegative oxygen atom, i.e. the oxygen atom draws electron density away from the carbon atom by virtue of its inductive effect (its greater electronegativity). Structure A emphasizes the fact that the resonance interaction depicted by the arrow labeled 1 reenforces the inductive withdrawl of electron density from the carbon by the oxygen. The development of a positive charge on the carbonyl carbon in A leads to the resonance interaction indicated by the arrow labeled 2. This reduces the electron density at the b-carbon as indicated by the positive charge on that atom in resonance structure B.

While the foregoing discussion should be familiar to you by now, it bears repeating because understanding resonance interactions like those shown in Figure 2 allows you to make predictions about the outcome of many chemical reactions. The prediction that is central to this topic is this: nucleophilic reagents can react with the carbonyl carbon and/or the b-carbon atom of a,b-unsaturated aldehydes and ketones. Which of these alternatives is realized in a given reaction is an issue of kinetic vs.thermodynamic control. Consider the reaction shown in Equation 1.

Of the two products, cyclohexanone is the more stable. This is primarily due to the fact that the carbon-oxygen double bond is an especially stable molecular fragment, considerably more stable than a carbon-carbon double bond. The average bond strength of a carbonyl group in aldehydes and ketones is about 178 kcal/mol, while that of the carbon-carbon double bond in alkenes is approximately 146 kcal/mol.

Whether a nucleophile adds to the carbonyl carbon or the b-carbon of an a,b-unsaturated system depends in large measure upon the reactivity of the nucleophile. With highly reactive nucleopliles, addition is kinetically controlled and the nucleophile adds to the carbonyl carbon because it is more electron deficient than the b-carbon. Less reactive nucleophiles are more selective in their choice of bonding partners and prefer to bond to the b-carbon, producing the more stable product. The change from kinetic to thermodynamic control is clearly illustrated by the product distributions obtained in the reactions of methyl lithium, methyl magnesium bromide, and lithium dimethyl cuprate with 5-methyl-2-cyclohexenone. The nucleophilic reactivity of these reagents is determined primarily by the difference in electronegativity of the methyl carbon and the metal atom to which it is bonded.

Equation 2 describes the reaction of 5-methyl-2-cyclohexenone with these three organometallic reagents in general terms. Here M stands for the metal atom. The product distribution in these three reactions is summarized in Table 1.

Table 1

 Nucleophile 1,2-Addition 1,4-Addition CH3Li >99% <1% CH3MgBr 79% 21% (CH3)2CuLi 2% 98%

Nucleophilic Addition: Kinetic vs. Thermodynamic Control

Note the almost total reversal from kinetic control to thermodynamic control as the nucleophile is changed from the highly reactive methyl lithium to the much less reactive lithium dimethyl cuprate.

Lithium dimethyl cuprate is prepared by the reaction of copper (I) iodide with methyl lithium as shown in Equation 3.

While the preparation of this complex and the determination of its structure are interesting topics in themselves, the key feature about this organometallic reagent is that the methyl groups are nucelophilic. They are less nucleophilic than a methyl group from methyl lithium or methyl magnesium bromide because the electronegativity difference between C and Cu is less than it is between C and Li or C and Mg. Now let's consider another, more familiar, way of assessing the relative reactivities of carbon nucleophiles.

### Relative Reactivity of Carbon Nucleophiles

The relative reactivity of a series of carbon nucleophiles may be assessed to a first approximation by comparing the pKa values of their conjugate acids. The weakest acids produce the strongest conjugate bases, i.e. the most reactive nucleophiles. Table 2 presents a list of carbon acids ranked in order of increasing acidity.

Table 2

 Carbon Acid Structure pKa Conjugate Base Method of Generation Alkanes ~50 Alkenes/Arenes ~44 Terminal Alkynes ~25 Aldehydes/Ketones ~19 Phosphonium Salts ~15 hydrogen cyanide ~9 commercially available NaCN or KCN b-dicarbonyl compounds ~9

pKa Values to the Rescue Again

The remainder of this topic will present a variety of reactions that involve the addition of nucleophilic reagents to the b-carbon of an a,b-unsaturated system. These reactions are also called Michael additions or conjugate additions.

A simple example of the shift from kinetic to thermodynamic control is available in the reaction of the enolate ion derived from methyl isobutyrate with cyclohexenone as shown in Scheme 1.

Scheme 1

A more dramatic example of Michael addition is illustrated in Equation 4, where conjugate addition occurs to the exclusion of 1,2-addition.

Scheme 2 outlines a transformation that constituted the first step in a multi-step synthesis of the steroid estrone.

Scheme 2

Body Builder

The nucleophilic reagent was generated by reacting cuprous iodide with the Grignard reagent derived from vinyl bromide. The divinyl cuprate added to the b-carbon of 2-methylcyclopentenone to produce an enolate ion that reacted with chlorotrimethyl silane to form the silyl enol ether in 89% yield.

A standard method for forming 6-membered rings known as the Robinson annulation (annulation means ring formation) begins with a Michael addition. Scheme 3 outlines the steps involved in this transformation.

Scheme 3

The Robinson Annulation

The sequence begins with the deprotonation of one of the a-carbons of 2-methylcyclohexanone. This is the same as the first step of an aldol condensation. However, the enolate ion, A, generated in the first step adds to the b-carbon atom of the methyl vinyl ketone that is present in the reaction mixture (arrows 4-7) to produce a new enolate ion, B, which abstracts a proton from a water molecule as shown by arrows 8-10. The resultant ketone C is then converted into its enolate D, which cyclizes as shown by arrows 12-14. This generates the b-hydroxyketone E which spontaneously dehydrates to yield the final product F.

Carbon is not the only nucleophilic species to participate in Michael additions. Scheme 4 illustrates an intramolecular Michael addition that played a key step in the first total synthesis of the antibiotic indolizomycin.

Scheme 4

Addition of the hydroperoxide anion to the b-carbon of the starting material initiated the flow of electrons indicated by arrows 1-3. Regeneration of the carbonyl group (arrow 4) of the resulting enolate ion was accompanied by an intramolecular nucleophilic substitution reaction as shown by arrows 5 and 6. The resulting epoxide was isolated in 97% yield!

Condensation Reactions of Esters

### Introduction

In our discussion of the aldol reaction we saw that competing reactions occur when aldehydes and ketones are treated with hydroxide ion. One the one hand, the hydroxide ion may add to the carbonyl carbon in a nucleophilic addition reaction, while on the other it may abstract an a-hydrogen. As the color coding in Figure 1 indicates that duality of reaction pathways stems from the fact that both the carbonyl carbon and the a-hydrogen of the aldehyde or ketone are electrophilic. For purposes of the following discussion it is important to note that addition of hydroxide ion to the carbonyl carbon generates a tetrahedral intermediate labeled T-1 in Figure 1.

Figure 1

Alternative Reaction Pathways for Aldehydes and Ketones

### Aldehydes and Ketones vs. Esters

Esters are structurally related to aldehydes and ketones: all three classes of compounds contain a carbonyl group. It shouldn't be surprising then that esters display similar reactivity patterns to aldehydes and ketones. As we shall see, they also display some interesting and significant differences. Figure 2 presents an analogous scheme to that shown in Figure 1 for the reaction of simple esters with hydroxide ion. Nucleophilic addition of hydroxide to the carbonyl carbon generates a tetrahedral intermediate T-2.

Figure 2

Alternative Reaction Pathways for Esters

One of the most favorable pathways available to these tetrahedral intermediates involves regeneration of the carbonyl group. This is favorable because the carbonyl group is an especially stable entity. Figure 3 compares the alternative ways in which the tetrahedral intermediates T-1 and T-2 might regenerate the carbonyl group.

Figure 3

Alternative Fates of Tetrahedral Intermediates

In the case of aldehydes and ketones, regeneration of the carbonyl group is a reasonable alternative only when it results in expulsion of hydroxide ion from T-1. As we have seen, regeneration of the starting material by this path allows for the less likely alternative, abstraction of an a-hydrogen and the formation of aldol condensation products which follow that event.

In the case of T-2, regeneration of the carbonyl group may be achieved by expulsion of hydroxide ion or alkoxide ion. In fact, the latter possibility is preferred. To understand why, it is necessary to consider the equilibria shown in Figure 4.

Figure 4

The equilibrium constant for expulsion of alkoxide ion from T-2 is approximately 1. Note that the product of this pathway is a carboxylic acid. Since this acid is formed in a strongly basic solution, it will be deprotonated rapidly. Given that the pKa of a carboxylic acid is about 5, the equilibrium constant for its deprotonation is approximately 1011. In other words, while the expulsion of hydroxide ion from T-2 (Figure 3) is about as likely as expulsion of alkoxide ion, the latter pathway is greatly preferred because a subsequent reaction has a much larger equilibrium constant. Consequently, treatment of an ester with aqueous sodium hydroxide results in the formation of the conjugate base of a carboxylic acid. More information is available.

#### Summary

In our discussion of the reactions of aldehydes and ketones with hydroxide ion, we saw that addition to the carbonyl carbon was more probable than abstraction of an a-hydrogen, but that the latter pathway was the one followed because the addition reaction was non-productive. In the case of structurally similar esters, i.e. esters containing at least one a-hydrogen, the more probable reaction is a productive reaction. It produces carboxylic acids. The process is called saponification.

Saponification of Esters

The formation of carboxylic acids by treatment of esters with sodium hydroxide is known as saponification. Equation 1 summarizes the net transformation for the saponification of methyl salicylate, a fragrant component in oil of wintergreen.

The general procedure involves refluxing the ester in 6M NaOH until the mixture becomes homogeneous, indicating complete formation of the water-soluble carboxylate salt, RCO2-. Acidification of the mixture during the work-up produces the carboxylic acid.

Equation 2 provides another example of saponification of an even simpler ester, ethyl acetate.

Suppose that you were to treat ethyl acetate with sodium ethoxide rather than sodium hydroxide. What would happen? The answer; the Claisen condensation.

The Claisen Condensation

### Introduction

At the end of our comparison of the reactions of aldehydes and ketones and esters with hydroxide ion, we posed the question "Suppose that you were to treat ethyl acetate with sodium ethoxide rather than sodium hydroxide. What would happen?"

Two alternatives are likely: 1. The ethoxide ion will abstract a proton from the methyl carbon a to the carbonyl group forming an enolate ion. 2. The ethoxide ion will add to the carbonyl carbon to form a tetrahedral intermediate. The latter possibility is more likely. As can be seen in Figure 1, this intemediate contains two identical ethoxy groups. Consequently, reformation of the carbonyl group has to regenerate the starting ester. As in the case of the aldol reaction, the more likely reaction is a non-productive one. And for the same reason a less likely alternative, abstraction of an a-hydrogen, becomes the productive reaction.

Figure 1

Alternative Paths for the Reaction of Ethyl Acetate with Sodium Ethoxide

Again, as in the case of the aldol condensation, abstraction of an a-hydrogen produces a low concentration of a highly reactive enolate ion. What happens when this highly reactive nucleophilic species finds itself surrounded by ethyl acetate molecules that have not been deprotonated? It reacts.

### The Claisen Condensation

The nucleophilic carbon of the enolate ion adds to the electrophilic carbon of a molecule of ethyl acetate that has not been deprotonated. A carbon-carbon bond is formed. The reaction produces yet another tetrahedral intemediate as shown in Equation 1.

Since the newly formed tetrahedral center has an electronegative atom attached to it, reformation of the carbonyl group, as shown in Equation 2, is a reasonable process.

This step results in the formation of a b-ketoester, which in this case is called ethyl acetoacetate. In the same way that b-hydroxyaldehydes and b-hydroxyketones are signature structures of the aldol reaction, b-ketoesters suggest the Claisen condensation.

A complete description of the mechanism of the Claisen condensation is, in fact, a bit more complicated than indicated in Equations 1 and 2, so, if you'd like to know more...

Retrosynthetic Analysis

To a synthetic organic chemist who is planning the synthesis of a target molecule, the presence of a b-ketoester fragment within that target should suggest the use of a Claisen condensation at some point during the synthesis. Retrosynthetic analysis of the target will then reveal the structure of the appropriate starting ester. Figure 2 demonstrates this retrosynthetic approach for a generic b-ketoester.

Figure 2

Taking Another Step Backwards

There are three points worth remembering about the retrosynthesis animated in Figure 2. First, R, R', and R'' may be the same or they may be different. Second, while R and R' may be H, R'' may not. Third, when R and R' are not the same, the condensation is called a crossed Claisen condensation.

### Examples

An intramolecular version of the Claisen condensation is known as the Dieckmann condensation. Equation 3 shows how this reaction was put to good use as part of the total synthesis of the prostaglandin PGA2.

Equation 4 offers another example of the Dieckmann condensation that was involved in the synthesis of tropinone, a degradation product that was produced during the determination of the structure of the physiologically active alkaloid atropine.

An early synthesis of cholesterol involved the "mixed Claisen reaction" shown if Equation 5.

Alkylation of Enolate Ions

### Introduction

In our discussions of the aldol condensation and the Claisen condensation we saw how deprotonation of a carbon atom a to a carbonyl group generated an enolate ion. Suppose for the moment that you wanted to react such a nucleophilic species with an alkyl halide as suggested in Figure 1 for the specific case of acetone. In other words, you want to synthesize 2-butanone from acetone.

Figure 1

A Viable Synthesis?

Is this a viable process? Let's consider the nature of this reaction. First, recall that treatment of acetone with aqueous NaOH establishes an equilibrium with an equilibrium constant of approximately 10-4. In other words, the solution contains comparatively high concentrations of acetone and NaOH and a relatively low concentration of the enolate ion. When you add CH3I to this mixture, there is a chance that it will react with the enolate ion as shown in Figure 1. However alternative pathways are also more likely. One alternative is an aldol condensation. Another is an Sn2 reaction of hydroxide ion with methyl iodide. Both of these possibilities are more likely than the desired reaction. So the answer to the question posed earlier is no, the synthesis proposed in Figure 1 is not viable. Fortunately it is possible to accomplish the transformation outlined in Figure 1 by alternative methods. We will consider two, the first a direct method, the second indirect. Then we will extend our discussion of the second method to a related system. All three approaches accomplish the same goal- alkylation of an enolate ion.

Direct Alkylation of Enolate Ions

Consider the consequences of using a stronger base than hydroxide ion to deprotonate acetone in Figure 1. First, the concentration of enolate ion at equilibrium would be higher. This would increase the likelihood of a reaction with the methyl iodide. Second, the concentration of acetone would be lower. This would reduce the probability of an aldol condensation. Both of these factors increase the likelihood of alkylation.

When lithium diisopropylamide, LDA, is used as the base to deprotonate acetone, Equation 1, the equilibrium constant for the reaction is approximately 1019. Virtually all of the acetone is deprotonated. The chances of an aldol reaction are reduced to essentially zero.

Because LDA is such a strong base, it is possible to use a stoichiometric amount of it to deprotonate all of the acetone. Thus, at equilibrium, there is no unreacted LDA to compete for any alkyl halide that might be added to the reaction mixture. Therefore the reaction of the enolate ion with added methyl iodide, Equation 2, becomes the most probable event under these conditions.

The Acetoacetic Ester Synthesis

Before the direct alkylation of lithium enolates was developed, chemists used an alternative, indirect method to achieve transformations such as that illustrated in Figure 1. Recall that the Claisen condensation of ethyl acetate produces a b-keto ester called ethyl acetoacetate. Recall, too, that the pKa of such compounds is approximately 10.

Figure 2

The Claisen Condensation (Again)

Under the reaction conditions the ethyl acetoacetate is deprotonated by the sodium ethoxide present in the mixture. The equilibrium constant for this acid-base reaction is approximately 106. In other words, the concentration of sodioacetoacetic ester is high. Addition of methyl iodide to the reaction mixture results in methylation of the enolate ion as shown in Equation 3.

The ester fragment of the product of reaction 3 is shown in blue to emphasize the idea that if we could replace that fragment with a hydrogen atom, we would have 2-butanone, the same product that was formed by direct methylation of acetone. Such a transformation is possible. It involves a 2-step, 1-pot reaction. The first step is the saponification of the ester. This results in the formation of the conjugate base of a b-ketocarboxylic acid. Acidification of the reaction mixture, followed by heating, results in the decarboxylation of the b-keto acid and the formation of the corresponding ketone. These steps are outlined in Figure 3.

Figure 3

It's a Gas

Figure 4 compares the outcomes of the direct alkylation of acetone with the indirect alkylation-saponification-decarboxylation sequence. Because the latter approach yields the desired target molecule, ethyl acetoacetate is considered to be the synthetic equivalent of acetone. Such a reagent is called a synthon. We'll see another example of a synthon when we discuss the malonic ester synthesis at the end of this topic.

Figure 4

There's More Than One Way to Skin a Cat

The decarboxylation of b-keto acids is a general phenomenon. It occurs readily because the carboxylic acid proton is transferred to the oxygen atom of the b-keto group intramolecularly. The transition state for the transfer is shown in Figure 5.

Figure 5

Decarboxylation

The immediate product of this decarboxylation is an enol which rapidly tautomerizes to the isomeric ketone, i.e. 2-butanone.

Retrosynthetic Analysis

To a synthetic organic chemist who is planning the synthesis of a target molecule, the presence of a b-ketoester fragment within that target should suggest the use of a Claisen condensation at some point during the synthesis. Retrosynthetic analysis of the target will then reveal the structure of the appropriate starting ester. Figure 6 demonstrates this retrosynthetic approach for a generic b-ketoester.

Figure 6

Taking Another Step Backwards

There are three points worth remembering about the retrosynthesis animated in Figure 6. First, R, R', and R'' may be the same or they may be different. Second, while R and R' may be H, R'' may not. Third, when R and R' are not the same, the condensation is called a crossed Claisen condensation.

The alkylation-decarboxylation sequence outlined in Figure 3 has a direct parallel in a related synthetic scheme called the malonic ester synthesis. We'll conclude this topic with a comparison of the malonic ester and the acetoacetic ester syntheses.

### The Malonic Ester Synthesis

Malonic ester, sometimes called diethyl malonate, is the diethyl ester of malonic acid. The structures of these two compounds are shown in Figure 7. Notice the structural similarities between malonic ester and acetoacetic ester. The similarity of their chemical reactivity stems from their structural relationship.

Figure 7

Meet the Malonates

Malonic ester is a common starting material for the synthesis of carboxylic acids. Figure 8 describes a typical sequence.

Figure 8

A Malonic Ester Synthesis

The Mannich Reaction

### Introduction

We have seen that attempts to alkylate simple aldehydes, ketones, and esters may be rendered ineffective by the occurrence of competing reactions, notably aldol and Claisen condensations as well as Sn2 and E2 reactions. Deprotonation of aldehydes, ketones, and esters with LDA allows for direct alkylation of these compounds, while deprotonation of dithiane derivatives of aldehydes offers an indirect method for replacing the aldehydic proton with an alkyl group. In this topic we will consider another approach to alkylation of aldehydes and ketones, namely the Mannich reaction. The development of this approach, in the early 1900s, was guided in some measure by insights into the biochemical pathways that plants use to make natural products called secondary metabolites. Secondary metabolites are compounds that an organism produces that are not essential to its survival, i.e. unlike primary metabolites, secondary metabolites are not used in the synthesis of the proteins or lipids, or nucleic acids, or energy that an organism requires for survival. However, it is now clear that these natural products are beneficial, if not essential, to the organisms that produce them.

Probably the most familiar group of secondary metabolites is the pheromones. Plants and insects alike release these compounds as a form of chemical communication, conveying simple messages such as "danger", and "this way to lunch", or "Poison! Eat at your own risk".

In plants, alkaloids constitute another group of secondary metabolites. Alkaloids are natural products that contain an amino group. The name is derived from the fact that aqueous solutions of these compounds are slightly basic, i.e. alkaline, due to the presence of the amino group. The reactions that produce alkaloids generally involve the secondary metabolism of amino acids. Figure 1 presents a highly abbreviated picture of the biosynthesis of nicotine from the amino acid ornithine.

Figure 1

An Abbreviated Biosynthesis of Nicotine

In the first step of this sequence the C-2 amino group of ornithine adds to the carbonyl group of pyridoxal phosphate to form an imine. (The pyridoxal phosphate is bound to an enzyme, which is not shown in this diagram.) Decarboxylation followed by tautomerization generates an isomeric imine which undergoes hydrolysis to produce 4-aminobutanal along with an enamine that is converted back to pyridoxal phosphate by hydrolysis. The 4-aminobutanal exists in equilibrium with the 5-membered heterocycle D1-pyrrolideine. As emphasized in the red box, an enolate ion derived from acetoacetyl coenzyme A adds to the protonated imine group of this 5-membered ring. Subsequent reactions complete the biosynthesis of nicotine. It is the addition of the enolate ion to the iminium ion that led to the development of the Mannich reaction. Synthetic methodologies that are designed in this way are referred to as biomimetic reactions. The Mannich reaction was the first biomimetic synthesis to be developed.

### The Mannich Reaction

So what is the Mannich reaction? In its simplest form, it involves the nucleophilic addition of an enol to an iminium ion formed by the reaction of formaldehyde with a secondary amine. Equation 11 provides a specific example.

The Mannich reaction involves several acid-catalysed equilibria. Like the aldol condensation, the success of the Mannich reaction depends on being able to generate both nucleophilic and electrophilic carbons in the reaction mixture at the same time. Figure 2 shows how this is done in the reaction of dimethylamine, formaldehyde, and acetone.

Figure 2

The Mannich Mechanism

When performing a Mannich reaction, it is common practice to use the hydrochloride salt of the amine as one of the starting materials. In aqueous solution the salt exists in equilibrium with the free amine. The proton that accompanies the formation of the free amine in Equilibrium 1 is available to protonate other reactants in the solution (Equilibria 2 and 3). Addition of the free amine to a protonated molecule of formaldehyde leads to the formation of the iminium ion shown at the right of Equilibria 4. The enol of acetone then adds to the carbon atom of the iminium ion in Equilibrium 5. Loss of a proton from the oxonium ion intermediate during the work-up of the reaction yields the final product.

Equations 2 and 3 provide two examples of applications of the Mannich reaction in the synthesis of natural products. Equation 4 illustrates a transformation that was achieved in 1917 and is the first example of a biomimetic synthesis.

Naturally occuring atropine is a toxic alkaloid isolated from the plant Atropa belladonna, commonly called nightshade.

Cocaine is a close structural analog of atropine, and the syntheses of these two alkaloids involve the same basic approach.

Finally, Figure 3 describes an interesting example of an intramolecular Mannich reaction that was used in the total synthesis of strychnine.

Figure 3

These Alkaloids Are Killing Me

The essential chemistry to consider in Figure 3 begins with the reaction of the secondary amine with formaldehyde. Under the conditions of the reaction, the resulting iminium ion intermediate was not isolated. Rather it underwent a thermal rearrangement, as indicated by the red arrows, to produce a new intermediate that contained an enol and an iminium ion in close proximity. As the blue arrows suggest, these electronically complementary groups reacted spontaneously to form the tricyclic structure shown at the lower right of the figure. The rest of the molecule was elaborated in a series of steps that are not relevant to this discussion. The cyclohexane ring of the final product which arose from the intramolecular Mannich reaction is highlighted in blue.

Pericyclic Reactions

### Introduction

For the synthetic organic chemist, the development of a general procedure that leads to the formation of carbon-carbon bonds is considered a laudable achievement. A general method that results in the simultaneous formation of two carbon-carbon bonds is worthy of a Nobel prize. In 1950, two chemists, Otto Diels and Kurt Alder, received that accolade for their discovery of a general method of preparing cyclohexene derivatives that is now known as the Diels-Alder reaction.

The Diels-Alder reaction is one type of a broader class of reactions that are known as pericyclic reactions. In 1965 two other Nobel laureates, Robert B.Woodward and Roald Hoffmann, published a series of short communications in which they presented a theoretical basis for these well known, but poorly understood pericyclic reactions. Their theory is called orbital symmetry theory. Subsequently other chemists published alternative interpretations of pericyclic reactions, one called frontier orbital theory, and another named aromatic transition state theory. All of these theories are based upon MO theory. In this topic we will use the Diels-Alder reaction to illustrate aspects of each of these theories.

### The Diels-Alder Reaction

The essential nature of the Diels-Alder reaction is summarized by Equation 1, where the substituents attached to the reactants have been omitted for clarity.

Figure 1 presents two examples of Diels-Alder reactions that have played vital roles in the total synthesis of biologically important molecules. The first example involves a synthesis of the steroid cortisone reported by R. B. Woodward in 1951. The second synthesis was performed by another Nobel laureate, E. J. Corey in 1969. In both cases many steps were required after the Diels-Alder reaction. The carbon atoms of the reactants are numbered to allow their identification in the final product.

Figure 1

Applying the Diels-Alder Reaction

Figure 1 demonstrates the practical side of the Diels-Alder reaction. Now let's take a look at the theoretical side, starting with orbital symmetry theory.

### Orbital Symmetry Theory

Orbital symmetry theory was a new paradigm in chemistry. As such it required new language. New words and new concepts had to be defined. The new vocabulary included the following words and terms:

• pericyclic reactions

• electrocyclic reaction

• sigmatropic rearrangement

• chelotropic reaction

• suprafacial

• antarafacial

• orbital correlation diagram

• symmetry allowed reaction

• symmetry forbidden reaction

The first term is the general reaction type described by orbital symmetry theory. The next four terms are specific types of pericyclic reactions. This discussion is limited to the first type, namely cycloaddition reactions. The term orbital correlation diagram describes the theoretical device that Woodward and Hoffmann developed to interpret pericyclic reactions. According to orbital symmetry theory the symmetry of the orbitals of the reactants must be conserved as they are transformed into the orbitals of the product. Consider the simplest example of a cycloaddition reaction, the head-to-head coupling of two ethene molecules to form cyclobutane as shown in Equation 2.

Figure 3 defines the conditions implied by the term head-to-head, namely that the reactants approach each other in parallel planes with the pi orbitals overlapping in the head-to-head fashion required for the formation of sigma bonds (shown in red in Equation 2). The sigma-bonded atoms of each ethene lie in the two parallel planes shown in black in the figure. The p orbitals on each carbon lie in the vertical plane which is shown in blue. The two planes shown in red are symmetry planes. Plane 1 bisects the C-C bond of each ethene, while plane 2 lies half way between the two planes shown in black.

Figure 3

According to the conventions of orbital symmetry theory, the reaction shown in Figure 3 is classified as a p2s + p2s cycloaddition, where p indicates that the reaction involves a p system, the number 2 is the number of electrons in the reacting p system, and the letter s stands for suprafacial: if one lobe of a p orbital is considered as the top face, while the other lobe is called the bottom face, then a suprafacial interaction is one in which the bonding occurs on the same face at both ends of the p system. The orientation of the interacting ¼ systems depicted in Figure 3 implies something about the stereochemical outcome of the reaction. Consider the p2s + p2s cycloaddition shown in Equation 3, where the two new sigma bonds are shown in red.

Because the interacting p systems approach each other in a "head-to-head" fashion, the relative stereochemistry is the same in the products as it is in the reactants, i.e. the two carboxyl groups that are "cis" to each other in the alkene are "cis" to each other in the products.

The alternative to a suprafacial interaction is an antarafacial reaction: in this case bonding occurs on the top face at one end of the p system and on the bottom face of the other. We will consider these ideas again when we discuss the Diels-Alders reaction.

 In classifying an orbital as symmetric (S) or antisymmetric (A) with respect to a symmetry plane, it is necessary to compare the phase of the lobes of the orbitals on each side of the symmetry plane: The phase may be indicated by shading or by labeling one lobe of an orbital + and the other -.

Molecular orbital theory describes the formation of the product in reaction 2 in terms of linear combinations the molecular orbitals of the reactants. Each reactant has two pi molecular orbitals of interest, y1 and y2. There are four combinations of these pi orbitals possible: y1A + y1B, y1A - y1B, y2A + y2B, and y2A - y2B, where the subscripts A and B are used to distinguish one ethene from the other. These combinations transform into the four sigma molecular orbitals in the product: s1A + s1B, s1A - s1B, s2A + s2B, and s2A - s2B. The diagram depicting the correlation of the reactant and product orbitals is shown in Figure 4. The numbers 1 2 in the diagram refer to the symmetry planes 1 and 2 in Figure 3.

Figure 4

An Orbital Correlation Diagram

The horizontal dashed line in the figure represents the energy level of an isolated p orbital. The most important feature to note about this admittedly complex diagram is that the linear combination y1A - y1B of the reactants correlates with, i.e. has the same symmetry, SA, as the s2A + s2B combination in the product. This is the central postulate of orbital symmetry theory: orbitals in the reactants must transform into product orbitals that have the same symmetry. Since the s2A + s2B orbital in the product is a high energy orbital, there must be a high activation energy for the formation of the product. Hence reaction 2 is said to be symmetry forbidden.

Since orbital correlation diagrams for more reactions involving more atoms are necessarily more complex, we will not deal further with this approach to pericyclic reactions. Rather we will examine an alternative theory known as Frontier Orbital Theory.

Frontier Orbital Theory

According to Frontier Orbital Theory it is possible to determine if a pericyclic reaction is allowed or forbidden by simply considering the symmetry relationship of the frontier orbitals of the reactants. The frontier orbitals are the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO). The interaction between these orbitals, a so-called HOMO-LUMO interaction, is a concept that is similar to Lewis acid-Lewis base chemistry which involves the interaction of a filled orbital of the base with an empty orbital of the acid.

According to Frontier Orbital Theory, a pericyclic reaction is allowed when the HOMO of one reactant has the same symmetry as the LUMO of the other.

We will now return to reaction 1, the Diels-Alder reaction, to illustrate this idea. Ethene has two pi orbitals which we will label y1E and y2E, the latter being the LUMO. 1,3-butadiene has four pi orbitals, y1B , y2B, y3B, and y4B, with y2B the HOMO. Figure 5 shows an idealized geometry for the approach of these frontier orbitals in parallel planes. The blue lines highlight the incipient overlap of the terminal lobes of the pi system of the diene with the orbitals of the alkene. The plane shown in red is a symmetry plane that bisects both molecules. Both the LUMO of ethene and the HOMO of 1,3-butadiene are antisymmetric (A) with respect to this plane. Since the HOMO-LUMO interaction shown in Figure 5 involves orbitals of the same symmetry, the reaction is allowed.

Figure 5

Frontier Orbital Interactions

Using the convention we discussed earlier, the Diels-Alder reaction is classified as a p2s + p4s cycloaddition. The alkene is a 2p electron system, while the diene is a 4p electron system. As the blue lines in Figure 5 indicate, the interaction between these two p systems occurs on the top face of each lobe of the alkene and the bottom face of each lobe of the diene. Hence, the interaction is suprafacial on both components. As we will see shortly, it is useful to regard the Diels-Alder reaction as a reference point for allowed cycloadditions.

Returning for a moment to the relationship between the relative stereochemistry of the reactants and products, Equation 4 presents a typical result for a p2s + p4s cycloaddition.

Note that, as was the case with the p2s + p2s cycloaddition summarized in Equation 3, the p2s + p4s cycloaddition results in retention of configuration.

### Electronic Complementarity

One of the values of the Diels-Alder reaction is its generality. It works well with a large variety of substituents on both the diene and the dienophile (alkene). However, the reaction works best when the two reactants are electronically complementary, which is to say that one of them is electron rich, while the other is electron poor. Fortunately we have already considered the effect of substituents on the electron density of a ¼ system in our discussion of substituent effects in electrophilic aromatic substitution reactions. Those groups that are activators in electrophilic aromatic substitution reactions exert their activating effect by increasing the electron density in the ¼ system of the aromatic ring, i.e. making it electron rich relative to benzene. Deactivators, on the other hand, reduce the electron density of the ¼ system, making it electron poor relative to benzene. In the Diels-Alder reaction our reference diene is 1,3-butadiene, while our reference dienophile is ethene (see Equation 1).

In order to use any of the three theories discussed in this topic effectively, you must be able to draw the molecular orbitals of the reactants. This is straightforward, and we will illustrate the approach with the four MOs of 1,3-butadiene. These MOs are formed by taking linear combinations of the p orbital on each of the four sigma bonded carbons. The lowest energy MO, y1= p1 + p2 + p3 + p4, has no nodes. Remember: a node is that point where the phase of a standing wave changes from positive to negative. The next orbital, y2= p1 + p2 - p3 + p4, has 1 node (the minus sign (-) indicates a node); y3 = p1 - p2 + p3 - p4, has 2 nodes and y4 = p1 - p2 - p3 - p4, has 3. The nodes in each MO are placed symmetrically. Figure 6 offers three representations of the MOs of 1,3-butadiene along with their classification as either symmetric (S) or antisymmetric (A) with respect to a symmetry plane that is perpendicular to the sigma bonded framework and bisects the C-2-C-3 bond. The red dots in MOs y2-y4 depict nodes. Notice that the symmetry of the orbitals alternates S-A-S-A as you go from one orbital to the next.

Figure 6

Assessing MOs

Aromatic Transition State Theory

How can you decide if a pericyclic reaction involves an aromatic transition state? In our discussion of aromaticity we classified structures as aromatic or anti-aromatic based upon Hückel's rule: an aromatic molecule contains a cyclic array of orbitals in which there are 4n + 2 electrons. Consider the description of the Diels-Alder reaction shown in Figure 7.

Figure 7

An Aromatic Transition State

The dashed red lines indicate bonds that are being formed in the transition state, while the dashed blue lines depict those that are breaking. Since the reaction involves the 4 pi electrons of 1,3-butadiene and the 2 pi electrons of ethene, there must be 6 electrons involved in the transition state, which, therefore, is aromatic because it conforms to Hückel's rule where n = 1. Note that this analysis assumes the geometry shown in Figure 5.

### The Fine Points

All of the theories just described involve two basic assumptions

1. The orbitals overlap suprafacially on both p systems as shown in Figures 3 and 5.

2. The reaction is thermally induced.

Given these assumptions, we can state the following:

• A thermally induced p2s + p4s cycloaddition reaction is allowed.

• A thermally induced p2s + p2s cycloaddition reaction is forbidden.

The rules are reversed when the reaction is photochemically induced:

• A photochemically induced p2s + p4s cycloaddition reaction is forbidden.

• A photochemically induced p2s + p2s reaction is allowed.

These rules are summarized in Table 1. Note that changing just one of the variables, i.e. faciality, energy source, or number of electrons (two electrons at a time), changes an allowed reaction to a forbidden one and vice versa.

Table 1

 Designation Thermal Photochemical p2s + p2s forbidden allowed p2s + p4s allowed forbidden p2s + p2a allowed forbidden p2s + p4a forbidden allowed p2a + p2a forbidden allowed p2a + p4a allowed forbidden

Pericyclic Reaction Rules

Before we move on, it is worthwhile to clarify the implications of the words allowed and forbidden. An allowed reaction is simply one with a low activation relative to some other pathway, while a forbidden reaction is a process for which there is a significant activation energy. In terms of transition states, an allowed reaction proceeds via an aromatic transition state, while a forbidden reaction does not occur because the transition state would be "anti-aromatic."

Solar Energy Storage ?

Figure 7 outlines a strategy for conversion of solar energy to chemical energy in a way that offers potential for the design of a solar energy storage device.

Figure 7

Pericyclic Reactions to the Rescue

The idea begins with a thermally allowed Diels-Alder reaction in which the 2-pi electron component is an alkyne rather than an alkene. The product, 1, is a cyclohexadiene. Irradiation of this diene should promote an allowed intramolecular 2+2 cycloaddition, leading to the formation of the highly strained tricyclic ring system 2. Since 2 cannot revert to 1 by a thermally allowed process, it is trapped in this high energy state. Recall that our definition of a forbidden reaction merely means that it is a process that has a high activation energy. This suggests that the reverse reaction, 2 ---> 1, might be enabled with an appropriate catalyst. The requirements to implement this approach then are

1. Selection of appropriate substituents, G, to insure that 1 absorbs solar radiation

2. Design of a catalyst to promote the regeneration of 1 from 2

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© M.EL-Fellah ,Chemistry Department, Benghazi University