A So far we have decided acid and base strength by using known pKa values. If the pKa value is not known, you can use the resonance and inductive electronic effects to explain differences in acid and base strength. This program deals with the important role of resonance.
A comparison of the acid strengths of ethanol and ethanoic acid is a good example to show the concept. To do this, we need to compare the equilibria:
A good place to start is with the possible resonance forms of the species.
Draw and compare the possible resonance structures.
Q 6.3. The amide ion (H2N−) is a stronger base than hydroxide. Which of the conjugate acids, NH3 or H2O, is stronger?
Q 6.4. Ammonia has pKa 33 and propanone has pKa 20. Can the following reaction occur as drawn?
Q 6.5. Can the bicarbonate anion deprotonate methanol as shown in the fol- lowing equation?
B Possible resonance forms include:
In the case of ethanol and its ethoxide conjugate base, resonance forms are not likely.
This is because carbon is less likely than oxygen to have a negative charge. Therefore, we can say that neither ethanol nor the ethoxide conjugate base is resonance stabilized.
The resonance forms for ethanoic acid and the ethanoate anion are much more likely.
Between these two options, the resonance forms for the ethanoate anion are better.
This is because these resonance forms are equal and the resonance is symmetrical.
One of the resonance forms for ethanoic acid has charge separation, and this is a relatively high energy form. The presence of resonance stabilization in ethanoic acid (pKa 4.7) shows why it is a stronger acid than ethanol (pKa 16).
The same method can be used with base strength. Study the example of aniline (phenylamine) and its saturated counterpart cyclohexylamine.
Use possible resonance forms to show which you expect to be the stronger base?
C To compare these molecules, you should have drawn resonance forms similar to:
We can easily draw resonance forms for the aromatic aniline. However, there is no similar resonance for cyclohexylamine. The delocalization stabilizes the aniline and
E There is delocalization of the nitrogen lone pair in the case of ethanamide. Therefore ethylamine, which has no delocalization, is the stronger base.
Now we return to acidity. We can apply the above resonance method to compound classes other than the alcohol and carboxylic acid example that we studied in A.
Is propanone or ethanoic acid the stronger acid?
—Cont’d
reduces the base strength of the nitrogen lone pair. If the nitrogen in aniline is protonated, it gives a relatively acidic conjugate acid of pKa 4.6. The lone pair is used to form the bond with the proton. This means that it can no longer be delocalized into the aromatic ring.
The cyclohexylamine has a non-delocalized lone pair. This is more basic and is easily protonated to give a very weak conjugate acid of pKa 10.7.
Can you explain the use of pKa to describe the strength of bases?
D To make them easy to compare, both acids and bases are listed on the same pKa scale.
We defined base strength as “the acid strength of the conjugate acid after protonation of the base.”
A strongly acidic conjugate acid with low pKa means that the base is very weak.
A weakly acidic conjugate acid with high pKa means a stronger base. Therefore, the higher the pKa of a base, the stronger the base. This is the reverse of the situation for acids.
Work out if ethylamine or ethanamide is the stronger base.
F Here we are comparing a ketone with a carboxylic acid. We need to look at resonance forms that can make one functional group more acidic than the other. To act as an acid, the molecules must give up a proton. We need to compare the relative stability of the conjugate bases that are formed by this action.
Continued...
G As a carbon analog of a carboxylic acid, propanone has no lone pair for delocalization into the C]O. The only way to get a pair of electrons is to lose a proton. This causes the transfer of the proton to the oxygen as shown below.
These two structures are not resonance forms because the atomic arrangement has changed. Instead, they are the equilibrium between two structural isomers. This equilibrium is an example of tautomerism between the keto and enol forms of propanone.
Do you think that esters and amides show similar α-hydrogen acidity to the ketone in the reaction?
We can see the similarity between carboxylic acids and carbonyl compounds which have a hydrogen on the α-carbon. This is the same positional arrangement as the OH hydrogen in carboxylic acids. The resulting enolate anion is resonance stabilized, and this explains the relative acidity of propanone with pKa 20. However, the non-equivalent resonance forms are not as efficient as the equivalent carboxylate resonance forms.
This is because one resonance form for the enolate has the negative charge on a carbon. Both the symmetrical carboxylate resonance forms have the negative charge on oxygen. Since oxygen is more electronegative than carbon, it can better carry the negative charge. So, ethanoic acid has the greater acid strength with pKa 4.7.
For propanone, we cannot draw a resonance form similar to the one for ethanoic acid. Explain this.
—Cont’d