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(D) The starting material is $2	ext{-hydroxy-2-methylpropanenitrile}$.$1$. Reaction with $LiAlH_4$ followed by $H_3O^+$: $LiAlH_4$ is a strong reduci

Vedclass · Accepted Answer

$A = 	ext{2-hydroxy-2-methylpropan-1-amine}, B = 	ext{2-methylprop-2-enoic acid}$

Vedclass · Answer

(D) The starting material is $2	ext{-hydroxy-2-methylpropanenitrile}$.$1$. Reaction with $LiAlH_4$ followed by $H_3O^+$: $LiAlH_4$ is a strong reducing agent that reduces the nitrile group $(-CN)$ to a primary amine $(-CH_2NH_2)$. The hydroxyl group $(-OH)$ remains unaffected. Thus,product $A$ is $2	ext{-hydroxy-2-methylpropan-1-amine}$.$2$. Reaction with $H_3O^+$ followed by $H_2SO_4$: Acidic hydrolysis of the nitrile group $(-CN)$ yields a carboxylic acid $(-COOH)$. The intermediate is $2	ext{-hydroxy-2-methylpropanoic acid}$. Subsequent treatment with $H_2SO_4$ (dehydrating agent) causes the elimination of water from the tertiary alcohol to form an alkene. Thus,product $B$ is $2	ext{-methylprop-2-enoic acid}$ (methacrylic acid).

The major products $A$ and $B$ in the following set of reactions are:

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