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Question

(D) Step $1$: Benzyl iodide $(C_6H_5CH_2I)$ reacts with $KCN$ to form benzyl cyanide $(C_6H_5CH_2CN)$.Step $2$: Benzyl cyanide reacts with methylmagne

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$C_9 H_{10} O$

Vedclass · Answer

(D) Step $1$: Benzyl iodide $(C_6H_5CH_2I)$ reacts with $KCN$ to form benzyl cyanide $(C_6H_5CH_2CN)$.Step $2$: Benzyl cyanide reacts with methylmagnesium bromide $(CH_3MgBr)$ to form an imine intermediate $(C_6H_5CH_2C(CH_3)=NMgBr)$.Step $3$: Acidic hydrolysis $(H_3O^+)$ of the imine intermediate yields phenylacetone $(C_6H_5CH_2COCH_3)$.The molecular formula of phenylacetone is $C_{10}H_{12}O$. However,reviewing the options provided,let us re-evaluate the structure. The product is $C_6H_5-CH_2-CO-CH_3$. Counting the atoms: $6$ (ring) $+ 1$ (methylene) $+ 1$ (carbonyl) $+ 1$ (methyl) $= 9$ carbons. Wait,$C_6H_5-CH_2-CO-CH_3$ has $6+1+1+1 = 9$ carbons. Let us re-count: $C_6H_5$ $(6C, 5H)$ $+ CH_2$ $(1C, 2H)$ $+ CO$ $(1C)$ $+ CH_3$ $(1C, 3H)$. Total: $C_9H_{10}O$. Thus,the correct option is $D$.

The molecular formula of the final product of the following synthetic scheme is:

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