Animated Solution for Chemistry - Organic Chemistry: The major products A and B in the following set of reactions are

Welcome to a fascinating exploration of organic reactivity! In this problem, we are presented with a single starting molecule—a cyanohydrin specifically known as $2$-hydroxy-$2$-methylpropanenitrile. This molecule is a fantastic playground for chemists because it features two distinct functional groups: a tertiary hydroxyl group ($-OH$) and a nitrile group ($-C\equiv N$).

Our mission is to predict how this molecule behaves when subjected to two completely different chemical environments. Let's break down the pathways step-by-step.

The Power of Lithium Aluminium Hydride

In the first reaction, our molecule is treated with Lithium Aluminium Hydride ($LiAl{H}_4$), followed by an acidic workup ($H_3{O}^{+}$). $LiAl{H}_4$ is renowned in organic chemistry as a heavy-duty, powerful reducing agent. It is a fantastic source of nucleophilic hydride ions ($H^{-}$).

When $LiAl{H}_4$ encounters our molecule, it evaluates the functional groups. The hydroxyl group is already fully reduced; oxygen is holding onto its electrons tightly, so the hydride ions ignore it. However, the nitrile group is a prime target. The carbon in the $-C\equiv N$ bond is highly electrophilic.

The hydride ions attack this carbon, breaking the triple bond. Through a sequence of additions, the nitrile is completely reduced. Two hydrogen atoms are added to the carbon, and two are added to the nitrogen, transforming the $-C\equiv N$ group into a primary amine ($-C{H}_{2}N{H}_2$). Thus, Product A is $1$-amino-$2$-methylpropan-$2$-ol.

Acidic Hydrolysis of Nitriles

Now, let's shift gears and look at the second reaction. Here, the starting molecule is thrown into a bath of aqueous sulfuric acid ($H_3{O}^{+}/H_{2}S{O}_4$). These are the classic, textbook conditions for the acidic hydrolysis of nitriles.

In this acidic, watery environment, the nitrile group undergoes a profound transformation. The electrophilic carbon is attacked by water molecules. The reaction proceeds through a primary amide intermediate ($-CON{H}_2$), which is rapidly hydrolyzed further under these strong acidic conditions. Ultimately, the nitrogen atom is expelled as an ammonium ion ($N{H}_4^{+}$), leaving behind a carboxylic acid group ($-COOH$).

Once again, our sturdy tertiary alcohol remains untouched (assuming standard temperatures where dehydration doesn't dominate). Therefore, Product B is $2$-hydroxy-$2$-methylpropanoic acid.

If we look at the correct option (c), you might notice something slightly unsettling about the skeletal drawing for Product A. The line extends from the central carbon and ends directly with an $-N{H}_2$ label. In strict IUPAC skeletal notation, a line ending in a heteroatom implies a direct bond to that atom, which would mean the carbon from the original nitrile group magically vanished!

However, chemically, we know the carbon is still there, forming a $-C{H}_{2}N{H}_2$ group. This is a very common typographical shortcut found in competitive exams. The line itself is often loosely intended to represent the $C{H}_2$ group. The golden rule here is: Always trust your chemical mechanism over a slightly ambiguous drawing!

By relying on our solid understanding of reagents, we confidently arrive at the correct answer.