Aromatic Aldehydes Ketones and Acids - Result Question 22
####26. The compound $Y$ is
(2018 Adv.)
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Answer:
Correct Answer: 26. (c)
Solution:
- Given,
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 .
$Na _2 CO _3$ 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
