Animated Solution for Chemistry - Organic Chemistry: The product of the reaction given below is

The reaction conditions are given as NBS (N-bromosuccinimide) in the presence of light ($h u$), followed by aqueous potassium carbonate (${\text{H}}_2\text{O}/{\text{K}}_2{\text{CO}}_3$).

This is a classic two-step sequence. The first step is an allylic bromination, and the second step is a nucleophilic substitution (hydrolysis).

Allylic Bromination NBS is a highly specific reagent. Instead of adding bromine across the double bond, it provides a low, steady concentration of bromine radicals (${\text{Br}}^{\bullet }$). These radicals abstract an allylic hydrogen atom to form a resonance-stabilized allylic radical.

In 1-methylcyclohexene, we have two distinct allylic positions on the ring: Carbon-3 and Carbon-6. If a hydrogen is abstracted from Carbon-3, we get a radical at C3. If it's abstracted from Carbon-6, we get a radical at C6.

To determine which radical is formed preferentially, we must look at their resonance structures.

The radical at Carbon-3 is adjacent to the double bond. Through resonance, the unpaired electron can delocalize to Carbon-1. This means the intermediate is a hybrid of a secondary radical (at C3) and a tertiary radical (at C1).

On the other hand, the radical at Carbon-6 delocalizes to Carbon-2. This intermediate is a hybrid of two secondary radicals.

Because the C3 radical has a tertiary contributor, it is significantly more stable. Therefore, the bromination occurs preferentially at Carbon-3, yielding 3-bromo-1-methylcyclohexene as the major intermediate.

Hydrolysis In the second step, we treat the allylic bromide with water and potassium carbonate. The ${\text{K}}_2{\text{CO}}_3$ acts as a mild base to neutralize the hydrobromic acid ($\text{HBr}$) formed during the reaction, driving the hydrolysis forward.

Water acts as a nucleophile, replacing the bromide leaving group with a hydroxyl ($-\text{OH}$) group. This substitution gives us our final major product: 3-methylcyclohex-2-en-1-ol.

Looking at our options, this perfectly matches option (a). The beauty of this problem lies in understanding how resonance stabilization dictates the regioselectivity of the initial radical reaction!