What are X and Y respectively in the following reactions?
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The Hofmann Bromamide reaction is a key reaction in organic chemistry, allowing the conversion of an amide (\( CONH_2 \)) to an amine (\( NH_2 \)) with one fewer carbon atom.
Step 1: Understanding the Reaction Sequence 1. Hofmann Bromamide Reaction (Formation of X) - The reaction of amide with bromine (\( Br_2 \)) and sodium hydroxide (\( NaOH \)) is known as the Hofmann Bromamide Reaction. - This reaction leads to converting an amide into an amine with one fewer carbon atoms. \[ \text{CONH}_2 \xrightarrow{NaOH + Br_2} \text{NH}_2 \] - Here, compound Y (amide) undergoes Hofmann bromamide reaction to form compound X (amine). 2. Reduction of Amide to Amine (Formation of Y) - The amide (\( CONH_2 \)) can be formed from the corresponding acid or acid derivative. - The amide reduction using lithium aluminum hydride (\( LiAlH_4 \)) gives an amine. \[ \text{COOH} \xrightarrow{LiAlH_4 + H_2O} \text{CONH}_2 \] - This reduction step forms Y (amide).
Step 2: Evaluating the Given Options - Option (1): Incorrect, as it misplaces the Hofmann bromamide reaction step. - Option (2): Correct, as Hofmann bromamide reaction forms X and LiAlH\(_4\) reduces to Y. - Option (3): Incorrect, as NaBH\(_4\) is not strong enough to reduce amides. - Option (4): Incorrect, as the order of reactions is incorrect. Thus, the correct answer is
Option (2).
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