JEE Chemistry 2025 Question with Solution

Given: The starting compound is propyne $\text{CH}_3-\text{C}\equiv\text{CH}$. It is treated successively with $Hg^{2+}, H_2SO_4$, then HCN, and finally $H_2/Ni$.

Find: The structure of product (A).

From the given solution:

1. Hydration of propyne in the presence of $Hg^{2+}, H_2SO_4$ gives the ketone acetone.

$$\text{CH}_3-\text{C}\equiv\text{CH} \longrightarrow \text{CH}_3-\text{CO}-\text{CH}_3$$

2. Reaction of acetone with HCN forms the corresponding cyanohydrin.

$$\text{CH}_3-\text{CO}-\text{CH}_3 \longrightarrow \text{CH}_3-\text{C}(\text{OH})(\text{CN})-\text{CH}_3$$

3. Hydrogenation with $H_2/Ni$ reduces the nitrile group $(-\text{CN})$ to $(-\text{CH}_2\text{NH}_2)$.

$$\text{CH}_3-\text{C}(\text{OH})(\text{CN})-\text{CH}_3 \longrightarrow \text{CH}_3-\text{C}(\text{OH})(\text{CH}_2\text{NH}_2)-\text{CH}_3$$

Therefore, the final product is 2-amino-2-methylpropanol, corresponding to option B.

Given: A three-step conversion of terminal alkyne is shown.

Find: Which option matches the final product.

The first reagent set $Hg^{2+}, H_2SO_4$ carries out Markovnikov hydration of the terminal alkyne, giving a methyl ketone. Thus propyne forms acetone.

The carbonyl group of acetone then undergoes nucleophilic addition with HCN to give a cyanohydrin. In this product, the same carbon bears both OH and CN groups.

In the last step, catalytic hydrogenation with $H_2/Ni$ reduces the nitrile group to a primary amine side chain $(-\text{CH}_2\text{NH}_2)$, while the OH group remains on the same carbon.

So the final arrangement is:

$$\text{CH}_3-\text{C}(\text{OH})(\text{CH}_3)-\text{CH}_2\text{NH}_2$$

This matches option B.