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The reduction of $p - O_{2} {NC}_{6} H_{4} {CH}_{2} COOH$ with ${LiAlH}_{4} g$ gives $p - O_{2} {NC}_{6} H_{5} {CH}_{2} {CH}_{2} OH .$

The reaction of ${Ph}_{2} C = {CHCH}_{3}$ with ${BH}_{3}$ in tetrahydrofuran followed by $H_{2} O_{2} / {OH}^{-}$ gives ${Ph}_{2} C (OH) {CH}_{2} {CH}_{3}$

Cyclopentylmethylcarbinol does not give iodoform test.

An alcohol containing $- CH (OH) {CH}_{3}$ gives iodoform test.

answer is D.

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Detailed Solution

(a) Both ${NO}_{2}$ and $COOH$ groups are reduced. The product is $p - H_{2} {NC}_{6} H_{4} {CH}_{2} {CH}_{2} OH .$

(b) The hydration involves anti-Markovanikov addition.

(c) An alcohol containing a methyl carbinol with at least one H on the carbinol C gives iodoform test. Cyclopentylmethylcarbinol shows iodoform test and the product is cyclopentane-carboxylic acid.

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