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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.
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:
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-.
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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.
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
I n 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).
For a full discussion of
geometric isomerism follow this link. Use the
BACK button on your browser to return to this page.
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.
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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
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:
Nucleophilic Addition
I
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.
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.
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 II
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.
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
 Addition of Methanol to an Aldehyde
and a Ketone
The addition of one molecule of an alcohol to an aldehyde produces a
hemiacetal.
Note that in acetaldehyde the carbonyl carbon has two bonds to the oxygen,
one to the hydrogen, and one bond to carbon. In the hemiacetal the carbon
also has two bonds to oxygen (different oxygens), one bond to hydrogen and
one bond to carbon. The addition of one molecule of an alcohol to a ketone
produces a
hemiketal.
The prefix hemi means half: under appropriate conditions, aldehydes and
ketones may react with two molecules of an alcohol to produce
acetals
and
ketals,
respectively. Figure 4 extends the conversion of the methanol-acetaldehyde
hemiacetal to its corresponding dimethylacetal. An analogous sequence of
steps would convert the methanol hemiketal of acetone into the
dimethylketal.
Figure 4
Formation of Acetaldehyde
Dimethylacetal
Examples
Many important examples of hemiacetals, acetals, hemiketals, and ketals
may be found in sugars. The four sugars shown in Figure 5 contain examples
of hemiacetals, acetals, and ketals.
Figure 5
Hemiacetals Acetals and Ketal
Equation 7 depicts the esterification of an alcohol that was performed
during the total synthesis of taxol. Other, potentially reactive, OH groups
were protected from reaction by the formation of a cyclic ketal prior to the
esterification step.
The molecular fragment indicated in red in Equation 7 is part of a
cyclic ketal,
which, as Equation 8 indicates, may be formed by the reaction of a diol with
a ketone.
In accord with
LeChatelier's principle,
the equilibrium shown in this reaction may be shifted to the right by use of
an excess of alcohol or by removal of the water as it forms. The water is
generally removed in either of two ways. In one approach, it is physically
separated from the reaction mixture by distillation. Alternatively, a drying
agent such as calcium chloride, CaCl2,is added to the reaction
mixture. As it is formed, the water interacts with the drying agent to form
a hydrate, CaCl2(H2O)n. Since the hydrate
is not soluble in the reaction mixture, the water is not available to
participate in the equilibrium shown in Equation 8.
Conversely, the equilibrium may be shifted to the left, i.e. the diol may
be regenerated from the cyclic ketal by addition of excess aqueous acid.
Nucleophilic Addition Reactions III
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.
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.
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.
The reaction of
secondary halogenoalkanes with ammonia
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
A Free Radical
Substitution Reaction
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 + Br2 CH3Br
+ 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.
Note: If you aren't sure about the words free radical or
substitution, read the page
What is free radical substitution? Use the BACK
button on your browser to return quickly to this page.
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.
The reaction of primary
halogenoalkanes with cyanide ions
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 reaction of
tertiary halogenoalkanes with cyanide ions
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 reaction of
secondary halogenoalkanes with cyanide ions
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
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.
Back
to the Top
Electronegativity
-
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:
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.
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.
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.
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.
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 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.
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.
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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.
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Structure and Bonding in
Chemistry
I n 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:
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.
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.
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:
I n
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.
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I n 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 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
I n 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.
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.
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!
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
I onic 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.
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ELECTRONIC STRUCTURES
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 .
Important! If you
haven't already read the page on
atomic
orbitals you should follow this link before you go any further.
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.
S ummary
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.
COVALENT BONDING - SINGLE BONDS
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.
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.
Back
to the Top
CO-ORDINATE (DATIVE COVALENT) BONDING
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.
Important! If you
are uncertain about
covalent
bonding follow this link before you go on with this page.
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!
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.
INTERMOLECULAR BONDING - HYDROGEN BONDS
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.
The evidence for hydrogen bonding
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.
Back
to the Top
T he
origin of hydrogen bonding
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.
Note: If you
aren't happy about
electronegativity, you should follow this link before you go on.
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.
|
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.
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.
Back
to the Top
Ionic
Bonding
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.
Covalent
Bonding
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 |
|
Back
to the Top
Multiple bonding
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.
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.
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:
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.
|
H
2.20 |
Electronegativity Values
for Some Elements |
|
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
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
Exclusively Carbon Functional
Groups
|
Group Formula |
Class Name |
Specific Example |
IUPAC Name |
Common Name |
|
 |
Alkene |
H2C=CH2 |
Ethene |
Ethylene |
|
Alkyne |
HC≡CH |
Ethyne |
Acetylene |
|
Arene |
C6H6 |
Benzene |
Benzene |
Functional Groups with Single
Bonds to Heteroatoms
|
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 |
|
|
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
Back
to the Top
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.
Polarity in Covalent Bonds
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.
Polar covalent bond
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.
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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.
Electron Dot Structures
A lso 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)
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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).
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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
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- Nitrogen in
-
- double bonded oxygen in
-
- single bonded oxygen in
-
Methane (CH4):
black is carbon, white is hydrogen
|
Nitrogen dioxide (NO2):
blue is nitrogen, red is oxygen
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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.
Disubstituted Rings
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:
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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.
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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.
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.
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© M.EL-Fellah ,Chemistry
Department, Garyounis University
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