Aromatic Aldehydes Ketones and Acids - Result Question 22

####26. The compound $Y$ is

(2018 Adv.)

(a)

(c)

(b)

(d)

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Answer:

Correct Answer: 26. (a)

Solution:

  1. Given,

(Major product) For this question we require only reaction 1 to 4 written above. Let us explore them one by one.

Reaction 1 It is called formylation or Gatterman Koch reaction. $A-CHO$ group is introduced to benzene ring through this reaction as

The attacking electrophile is $H-\stackrel{+}{C}=O$ which is generated as

(i) $CO+HCl \rightleftharpoons H-\underset{O}{C}-Cl$

(ii) $H-\underset{O}{C}-Cl+AlCl _3 \rightleftharpoons H-\underset{O _0^{C}}{\stackrel{+}{O}}+AlCl _4^{-}$

Reaction 2 It is Perkin condensation which results in $\alpha, \beta$ unsaturated acid as

Note Besides $CH _3 COO^{-} Na^{+}$, quinoline, pyridine, $Na _2 CO _3$, triethylamine can also be used as bases in this reaction.

Reaction 3 It is simple addition of bromine to unsaturated acid formed through reaction 2 .

Cinnamic acid

$Na _2 CO _3$ works as a base in the reaction to trap $H^{+}$to be released

by $-COOH$ of the reactant. in the reaction as the minor product.

Reaction 4 It is decarboxylation and dehydrohalogenation of product produced by reaction 3 as

Hence, $Y$ is



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