Aromatic Aldehydes Ketones And Acids Result Question 31
Passage
In the following reactions sequence, the compound $J$ is an intermediate.
$J\left(\mathrm{C}_9 \mathrm{H}_8 \mathrm{O}_2\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_3$ and positive Baeyer’s test.
31. The compound I, is
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Answer:
Correct Answer: 31. ( a )
Solution:
- The first step of reaction is Perkin’s condensation.
$J$ being a carboxylic acid gives effervescence with $\mathrm{NaHCO} _3$. Also, $J$ has olefinic bond, it will decolourise Baeyer’s reagent.
In the second step, $J$ on treatment with $\mathrm{H} _2 / \mathrm{Pd} / \mathrm{C}$ undergo hydrogenation at olefinic bond only as :
$ J+\mathrm{H} _2 / \mathrm{Pd} \longrightarrow \mathrm{C} _6 \mathrm{H} _5-\mathrm{CH} _2-\mathrm{CH} _2-\mathrm{COOH} $
The hydrogenated acid, on treatment with $\mathrm{SOCl} _2$ gives acid chloride.
$ \mathrm{C} _6 \mathrm{H} _5-\mathrm{CH} _2-\mathrm{CH} _2-\mathrm{COOH}+\mathrm{SOCl} _2 \longrightarrow \mathrm{C} _6 \mathrm{H} _5-\mathrm{CH} _2-\mathrm{CH} _2-\mathrm{COCl}+\mathrm{HCl}+\mathrm{SO} _2 $
In the final step, acid chloride formed above undergo intramolecular Friedel-Craft acylation as:
BETA
