Question

Consider the following reaction: \[ \mathrm{Ph{-}CH_2NH_2 \xrightarrow[\text{Py}]{PhCOCl} X \xrightarrow[]{LiAlH_4} Y} \] The correct structure of \(Y\) is:

• A. \( \mathrm{PhCH_2NHCOPh} \)
• B. \( \mathrm{Ph{-}CH_2NHCH_2Ph} \)
• C. \( \mathrm{PhNH_2CH_2Ph} \)
• D. \( \mathrm{PhCH_3} \)

Hint:

Remember:
Acid chloride + amine \(\rightarrow\) amide
\( \mathrm{LiAlH_4} \) reduces amides to amines

Answer 1

The Correct Option is B

Concept:
Acid chlorides react with amines to form amides.
\( \mathrm{LiAlH_4} \) reduces amides to amines by converting the \(\mathrm{CO}\) group into a \(\mathrm{CH_2}\) group.
Step 1: Identify the starting compound. The given compound is benzylamine: \[ \mathrm{Ph{-}CH_2NH_2} \]
Step 2: Reaction with benzoyl chloride (\( \mathrm{PhCOCl} \)) in presence of pyridine. This is an acylation reaction, forming an amide: \[ \mathrm{Ph{-}CH_2NH_2 \xrightarrow[]{PhCOCl/Py} Ph{-}CH_2NHCOPh} \] Thus, \[ X = \mathrm{Ph{-}CH_2NHCOPh} \]
Step 3: Reduction with \( \mathrm{LiAlH_4} \). Lithium aluminium hydride reduces amides as: \[ \mathrm{RCONHR' \xrightarrow[]{LiAlH_4} RCH_2NHR'} \] Applying this: \[ \mathrm{Ph{-}CH_2NHCOPh \xrightarrow[]{LiAlH_4} Ph{-}CH_2NHCH_2Ph} \]
Step 4: Final product. \[ \boxed{Y = \mathrm{Ph{-}CH_2NHCH_2Ph}} \]