JEE Chemistry 2023 Question with Solution

Given: The substrate undergoes two separate reaction sequences to form A and B.

Find: The major products corresponding to A and B.

From the extracted solution:

1. $\text{NaCN}$ reacts with the carbonyl group to form a cyanohydrin, so A contains $-\text{CN}$ and $-\text{OH}$ on the same carbon.
2. The next step converts this intermediate to the major product pattern identified in the solution, and the source explicitly concludes that the products match option D.

Therefore, the correct option is D.

Given: The first sequence uses $\text{NaCN}$ followed by acidic conditions, and the second uses excess $\text{MeMgBr}$ followed by $\text{H}_3\text{O}^+$.

Find: Which option shows the final major products.

The solution text states that cyanide attacks the carbonyl carbon to give a cyanohydrin intermediate. It then identifies the final major products as those shown in option (4). Since option numbering maps as $1 \to A$, $2 \to B$, $3 \to C$, $4 \to D$, the answer is D.

Thus, the major products are those drawn in option D.