Product of the reaction is

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(C) The reaction involves the reduction of an imide ($N$-phenylglutarimide) using an excess of $LiAlH_4$ (Lithium Aluminium Hydride). $LiAlH_4$ is a s

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$N$-phenylpiperidine

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(C) The reaction involves the reduction of an imide ($N$-phenylglutarimide) using an excess of $LiAlH_4$ (Lithium Aluminium Hydride). $LiAlH_4$ is a strong reducing agent that reduces imides to amines. Specifically,the two carbonyl groups $(C=O)$ of the imide are reduced to methylene groups $(-CH_2-)$. Therefore,the product formed is $N$-phenylpiperidine.

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