Aromatic Aldehydes Ketones and Acids - Result Question 9
####11. The major product formed in the following reaction is
(a)
(b)
(c)
(d)
(d)
(2019
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Answer:
Correct Answer: 11. (a)
Solution:
- In aldol condensation, generally aldehydes react at a faster rate than ketones towards base. In the given case $CH _3 CHO$ will lose $\alpha$-hydrogen faster than $\bigcirc$ reason, i.e. conjugation between benzene ring and $>C=O$ group. Along with sterically less hindered nucleophile of $CH _3 CHO$ will also add to the major product formation.
Following four products are possible in the reaction:
(i)
(Self Aldol condensation of $CH _3 CHO$ )
(ii)
$$ \begin{array}{cl} \stackrel{H}{\mid} & \begin{array}{l} \text { Cross Aldol condensation of } \ CH _3 CHO \text { and } CH _3 COC _6 H _5 \end{array} \ CH _3-C-CH _2-C-C _6 H _5 & \begin{array}{l} \text { in which }>C=O \text { group of } \ CH _3 CHO \text { is carbanion } \end{array} \ OH & \begin{array}{l} \text { acceptor }) . \end{array} \end{array} $$
(iii)
(Cross Aldol condensation of $CH _3 CHO$ and $CH _3 COC _6 H _5$ in which $>C=O$ group of $C _6 H _5 COCH _3$ is carbanion acceptor).
(iv)
(Self Aldol condensation product of $C _6 H _5 COCH _3$ )
$O H$ of base will prefer to attack on $-CH _3$ group of $CH _3 CHO$ for the formation of carbanion and as among the $>C=O$ groups available, the $>C=O$ group of $CH _3 CHO$ is the best carbanion acceptor. Hence, self condensation product of $CH _3 CHO$ will be the major product.
