JEE Chemistry 2023 Question with Solution

Given: An unsaturated alcohol undergoes reaction with $\text{H}_3\text{O}^+$ to form major product B.

Find: The correct product formed in the acid-catalyzed cyclization.

The reaction proceeds through acid-catalyzed intramolecular cyclization of the alcohol.

1. Protonation of the alcohol: The $-\text{OH}$ group is protonated by $\text{H}_3\text{O}^+$, converting it into a better leaving group.
2. Carbocation formation: Water leaves and a carbocation is generated.
3. Cyclization: The double bond attacks intramolecularly, leading to ring closure. The preferred pathway is the one that forms the more stable tertiary carbocation and gives a six-membered ring arrangement.
4. Rearrangement and deprotonation: After the favorable carbocation forms, loss of a proton gives the final cyclic product.

From the mechanism shown, the major product corresponds to option (4).

Therefore, the correct option is D.

Given: The substrate contains both an alcohol group and a double bond, so intramolecular attack is possible under acidic conditions.

Find: Why the major product is option (4) and not the others.

Under acidic conditions, the reaction does not stop at a simple substitution step. Instead, it proceeds through a carbocation intermediate. Such reactions are governed mainly by:

• carbocation stability, and
• favorable ring formation.

The intramolecular attack occurs in a way that produces the more stable tertiary carbocation. The pathway shown in the mechanism also includes a methyl shift, which further stabilizes the intermediate before the final product is formed.

Because of this stabilization sequence, the product corresponding to option (4) is favored over the structures in options (1), (2), and (3).

Hence, the major product B is option (4), that is, D.