Question

In the given reactions, 'X' and 'Y' respectively are: \[ \ce{C6H5CH2NH2 ⇔[X] C6H5CONH2 ⇒[Y] C6H5NH2} \]

• A. \ce{LiAlH4, H2O} ; \ce{Br2 / OH^-}
• B. \ce{Br2 / OH^-} ; \ce{LiAlH4, H2O}
• C. \ce{Br2 / H^+} ; \ce{NaBH4}
• D. \ce{NaBH4} ; \ce{Br2 / H^+}

Hint:

To convert an amide to a primary amine with one carbon less, use Hofmann rearrangement (\ce{Br2 / OH^-}). To reduce an amide back to an amine with same carbon count, use LiAlH\textsubscript{4}.

Answer 1

The Correct Option is A

Let us break the overall reaction sequence into two steps: \begin{itemize} \item The forward reaction converts \ce{C6H5CH2NH2} (benzylamine) to \ce{C6H5CONH2} (benzamide). This is an oxidation process, converting a primary amine to an amide. \item However, since the direction of the reaction is indicated as \ce{C6H5CH2NH2 ⇔C6H5CONH2}, it implies that X is a reagent that helps in converting \ce{C6H5CH2NH2} to \ce{C6H5CONH2}, i.e., it involves oxidation, and the reverse of reduction. \item But in the given context, more clearly: X is LiAlH\textsubscript{4, H\textsubscript{2}O} used to reduce benzamide (\ce{C6H5CONH2}) back to benzylamine (\ce{C6H5CH2NH2}). So the left-to-right direction is amide to amine, which is a reduction by LiAlH\textsubscript{4}. \item For the second part, converting \ce{C6H5CONH2} to \ce{C6H5NH2} (aniline) removes the -CO part and replaces the amide with a direct amine. This is the classic Hofmann bromamide reaction (also called Hofmann rearrangement). \item Hofmann bromamide reaction is done using Br\textsubscript{2}/OH\textsuperscript{–} to convert an amide to a primary amine with one carbon less. \end{itemize} So, the correct reagent sequence is: \begin{align*} X &: \ce{LiAlH4, H2O} \quad \text{(reduction of amide to amine)}
Y &: \ce{Br2 / OH^-} \quad \text{(Hofmann rearrangement of amide to amine)} \end{align*} \[ \boxed{\text{Correct Answer: (1)}} \]