Aromatic Aldehydes Ketones and Acids - Result Question 22
Passage
Treatment of benzene with $\mathrm{CO} / \mathrm{HCl}$ in the presence of anhydrous $\mathrm{AlCl}_3 / \mathrm{CuCl}$ followed by reaction with $\mathrm{Ac}_2 \mathrm{O} / \mathrm{NaOAc}$ gives compound $X$ as the major product. Compound $X$ upon reaction with $\mathrm{Br}_2 / \mathrm{Na}_2 \mathrm{CO}_3$ followed by heating at $473$ K with moist KOH furnishes $Y$ as the major product. Reaction of $X$ with $\mathrm{H}_2 / \mathrm{Pd}-\mathrm{C}$, followed by $\mathrm{H}_3 \mathrm{PO}_4$ treatment gives $Z$ as the major product.
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
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
in the reaction as the minor product.
Reaction 4 It is decarboxylation and dehydrohalogenation of product produced by reaction 3 as
BETA
