JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Chapter_pending

Here the copper(I) iodide is the decisive clue. A simple Grignard reagent adds to the carbonyl in a 1,2 fashion, but in the presence of $\mathrm{CuI}$ the methyl nucleophile is delivered in a conjugate 1,4 fashion to the $\beta$-carbon of the enone. Our starting enone is $3\text{-methylcyclohex-2-en-1-one}$, so the methyl adds to the already methyl-bearing $\beta$-carbon $(\mathrm{C}3)$ and creates a quaternary $gem$-dimethyl centre there. What results from this addition is not a neutral ketone yet but a magnesium enolate, with the negative charge delocalised onto the $\alpha$-carbon $(\mathrm{C}2)$. That nucleophilic $\alpha$-carbon is then trapped by $n$-propyl iodide in an $S_{N}2$ alkylation, installing the $n$-propyl group right next to the new $gem$-dimethyl carbon. The carbonyl is regenerated on workup, so the ring stays a ketone; the alcohol and the fully saturated hydrocarbon are therefore ruled out. The major product is $3,3\text{-dimethyl-}2\text{-propylcyclohexan-1-one}$.