Acids and Bases - Organic Chemistry Concepts: An EFL Approach - page 99

H The answer is no. The resonance forms give an electronic explanation. Both esters
( pK a ≈ 24) and amides ( pK a ≈ 28) show the same resonance forms for their α -deprotonated
carbanions.

In addition, the lone pairs on the oxygen and nitrogen allow us to draw other important
resonance forms. These resonance forms do not make equal contributions to the
hybrid, but they compete directly with the delocalization of the carbanion. Because of
this, the resonance stabilization of the carbanion is not as efficient as the resonance
stabilization of the ketone enolate.

The resonance approach in this program has not been applied to other classes of
compound such as hydrocarbons and alkyl halides. This is because they have no
possibility of resonance stabilization. Therefore, they are very weak acids. Program
18 looks at the impact of the inductive effect on acidity and basicity.

Q 6.6. Arrange the following in order of decreasing acidity at the central
methylene CH 2 .

CH 3 COCH 2 COCH 3

CH 3 COCH 2 CO 2 CH 3 CH 2 (CO 2 CH 3 ) 2

Q 6.7. Draw resonance forms for the anion from the 2,4-pentandione in Q 6.6 .

PROGRAM 18 Acidity/Basicity and Inductive Effects
A In Program 17, we studied the importance of resonance on the acidity of organic
compounds. We looked at the ability of both O-H bonds (alcohols, phenols, carboxylic
acids) and the α -carbon C-H bonds (aldehydes, ketones, ester, amides) to provide a proton.
These general H-A bonds are polarized to different amounts based on the
electronegativity differences of the two atoms in the bond. This is the inductive
effect. The higher electronegativity of oxygen compared with carbon means that an
O-H bond is more dipolar than a C-H bond.