Question:
Write the structures of A and B in the following reaction:
\[
\text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{NH}_3, \Delta} A \xrightarrow{\text{Br}_2/\text{NaOH}} B
\]
Write the structures of A and B in the following reaction:
\[
\text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{NH}_3, \Delta} A \xrightarrow{\text{Br}_2/\text{NaOH}} B
\]
Show Hint
- The Hoffmann Bromamide Reaction is used for amide to amine conversion by removing the carbonyl group.
- This reaction reduces carbon count by 1, making it useful for amine synthesis.
Updated On: Feb 25, 2025
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Solution and Explanation
- \( A \) = Benzamide (\( C_6H_5CONH_2 \))
- \( B \) = Aniline (\( C_6H_5NH_2 \))
Step 1: Conversion of Benzoic Acid to Benzamide When benzoic acid (\( C_6H_5COOH \)) is heated with ammonia, it undergoes amide formation, forming benzamide (\( C_6H_5CONH_2 \)). \[ \text{C}_6\text{H}_5\text{COOH} + NH_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5CONH_2 \] Step 2: Hoffmann Bromamide Degradation When benzamide (\( C_6H_5CONH_2 \)) reacts with Br\(_2\)/NaOH, the carbonyl (-CO) group is removed, leading to the formation of aniline (\( C_6H_5NH_2 \)). \[ \text{C}_6\text{H}_5CONH_2 + Br_2 + NaOH \rightarrow \text{C}_6\text{H}_5NH_2 + Na_2CO_3 \]
- \( B \) = Aniline (\( C_6H_5NH_2 \))
Step 1: Conversion of Benzoic Acid to Benzamide When benzoic acid (\( C_6H_5COOH \)) is heated with ammonia, it undergoes amide formation, forming benzamide (\( C_6H_5CONH_2 \)). \[ \text{C}_6\text{H}_5\text{COOH} + NH_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5CONH_2 \] Step 2: Hoffmann Bromamide Degradation When benzamide (\( C_6H_5CONH_2 \)) reacts with Br\(_2\)/NaOH, the carbonyl (-CO) group is removed, leading to the formation of aniline (\( C_6H_5NH_2 \)). \[ \text{C}_6\text{H}_5CONH_2 + Br_2 + NaOH \rightarrow \text{C}_6\text{H}_5NH_2 + Na_2CO_3 \]
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