JEE Chemistry 2023 Question with Solution

Given: The substrate is an $\alpha,\beta$-unsaturated ketone treated with $(1)\ \text{MeMgBr/CuI}$ and then $(2)\ \text{nPrI}$.

Find: The major product formed in the reaction.

The given reaction involves two steps.

1. Reaction with $\text{MeMgBr/CuI}$: In the presence of $\text{CuI}$, the Grignard reagent behaves as a softer nucleophile and gives $1,4$-conjugate addition to the enone system rather than direct $1,2$-addition to the carbonyl group.

2. This places a methyl group at the $\beta$-carbon and generates an enolate intermediate.

3. Reaction with $\text{nPrI}$: The enolate is then alkylated at the $\alpha$-carbon by $\text{nPrI}$.

So, the final product is the cyclohexanone retaining the carbonyl group, with propyl substitution at the $\alpha$-position and gem-dimethyl substitution at the $\beta$-position.

Therefore, the correct option is C.

Given: An enone reacts with $\text{MeMgBr/CuI}$ followed by $\text{nPrI}$.

Find: Why the product corresponds to option C.

For an $\alpha,\beta$-unsaturated ketone, a normal Grignard reagent often favors $1,2$-addition. However, in the presence of $\text{CuI}$, an organocuprate-like species is formed, and such reagents preferentially undergo $1,4$-addition.

After $1,4$-addition of $\text{Me}$, the double bond is saturated and an enolate is formed:

This enolate then reacts with $\text{nPrI}$ by $\alpha$-alkylation, giving the ketone with a new propyl group next to the carbonyl.

Hence, the product must:

• retain the carbonyl group,
• have an additional methyl delivered to the $\beta$-position, and
• have a propyl group at the $\alpha$-position.

Among the given structures, this matches option C.