Question

$\alpha$-hydrogen of an ester is more acidic than $\alpha$-hydroden of an amide. The most suitable explanation for this observation is that :

• A. The ester enolate (1) is more stable than amide enolate (2) as a consequence of the amide enolate's carbonyl being a better acceptor of enolate's electron density than the ester enolate's carbonyl.
• B. The ester enolate (1) is more stable than the amide enolate (2) as a consequence the enolate' carbonyl being a better acceptor of the enolate's electron density than is the amide enolate's carbonyl.
• C. The ester enolate (2) is more stable than the ester enolate (1) as a consequence of the amide enolate's carbonyl being a better acceptor of enolate's electron density than is the ester enolate's carbobyl.
• D. The ester enolate (2) is more stable than ester enolate (1) as a consequence of the ester enolate's carbonyl being a better acceptor of enolate's electron density than is the amide enolate's carbonyl.

Answer 1

The correct option is B The ester enolate (1) is more stable than the amide enolate (2) as a consequence the enolate' carbonyl being a better acceptor of the enolate's electron density than is the amide enolate's carbonyl.

As we know,
-I effect of ester group is larger than that of amide group (due to high electronegativity of oxygen); also $C=O$ of amide has greater resonance contribution by $-N{H}_2$ than $C=O$ of ester by $RO$.

Hence, $C=O$ of ester is better in accommodating the negative charge by resonance than $C=O$ of amide so alhpa hydrogen of ester group is more acidic than amide group.