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(A) The starting material is $1$-indanone. Step $(1)$: Reaction with $HCN$ adds a cyano group to the carbonyl carbon,forming a cyanohydrin. Step $(2)$

Vedclass · Accepted Answer

$1-$tetralone

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(A) The starting material is $1$-indanone. Step $(1)$: Reaction with $HCN$ adds a cyano group to the carbonyl carbon,forming a cyanohydrin. Step $(2)$: Reduction with $LiAlH_4$ reduces the cyano group to a primary amine $(-CH_2NH_2)$ and the hydroxyl group (from the cyanohydrin) is removed via deoxygenation or reduction processes in these conditions,leading to an expansion or rearrangement. However,looking at the sequence $HCN$,$LiAlH_4$,and $NaNO_2/H^+$,this is a classic Tiffeneau-Demjanov rearrangement. The cyanohydrin is reduced to a $\beta$-amino alcohol,which upon treatment with $NaNO_2/H^+$ undergoes diazotization followed by ring expansion. Starting with a $5$-membered ring ketone ($1$-indanone),the ring expansion leads to a $6$-membered ring ketone,which is $1$-tetralone.

The major product $(A)$ is:

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