Remember the reactions of carbonyl compounds, amides, and amines, including haloform reaction, Hoffmann bromamide degradation, and carbylamine reaction.
1. Reaction of acetophenone with Br\(_2\)/OH\(^{-}\) followed by H\(^{+}\):
This is a haloform reaction.
\( \text{C}_6\text{H}_5\text{COCH}_3 + 3\text{Br}_2 + 4\text{OH}^{-} \rightarrow \text{C}_6\text{H}_5\text{COO}^{-} + 3\text{HBr} + \text{CHBr}_3 \)
\( \text{C}_6\text{H}_5\text{COO}^{-} + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH} \)
A is benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \)).
2. Reaction of benzoic acid with NH\(_3\) and heat:
\( \text{C}_6\text{H}_5\text{COOH} + \text{NH}_3 \rightarrow \text{C}_6\text{H}_5\text{COONH}_4 \)
\( \text{C}_6\text{H}_5\text{COONH}_4 \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CONH}_2 + \text{H}_2\text{O} \)
B is benzamide (\( \text{C}_6\text{H}_5\text{CONH}_2 \)).
3. Reaction of benzamide with Br\(_2\)/NaOH:
This is Hoffmann bromamide degradation.
\( \text{C}_6\text{H}_5\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O} \)
C is aniline (\( \text{C}_6\text{H}_5\text{NH}_2 \)).
4. Reaction of aniline with CHCl\(_3\)/alc. KOH:
This is the carbylamine reaction.
\( \text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \rightarrow \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
D is benzene isocyanide (\( \text{C}_6\text{H}_5\text{NC} \)).
Therefore, D is benzene isocyanide.
Final Answer:
Benzene isocyanide.
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