Question:
Aniline on treatment with NaNO\(_2\) and HCl at 278K gives compound (Y). When the compound (Y) is treated with Cu/HBr, compound 'Z' is obtained. The compound 'Z' is
Aniline on treatment with NaNO\(_2\) and HCl at 278K gives compound (Y). When the compound (Y) is treated with Cu/HBr, compound 'Z' is obtained. The compound 'Z' is
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Sandmeyer reaction: ArN\(_2^+\)X\(^-\) + CuX → ArX + N\(_2\). CuCl gives chlorobenzene, CuBr gives bromobenzene, CuCN gives benzonitrile.
Updated On: Apr 24, 2026
- Benzene
- Benzene diazonium bromide
- Phenol
- Bromobenzene
- Chlorobenzene
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
This is a sequence of reactions: diazotization followed by a Sandmeyer reaction.
Step 2: Detailed Explanation:
1. Aniline (C\(_6\)H\(_5\)NH\(_2\)) reacts with NaNO\(_2\) and HCl at 273-278 K (0-5\(^\circ\)C) to form benzene diazonium chloride, which is compound (Y). The reaction is diazotization. 2. When benzene diazonium chloride is treated with Cu/HBr (a Sandmeyer reaction), the diazonium group (-N\(_2^+\)) is replaced by a bromine atom (-Br), forming bromobenzene (C\(_6\)H\(_5\)Br), which is compound 'Z'.
Step 3: Final Answer:
Compound 'Z' is bromobenzene.
This is a sequence of reactions: diazotization followed by a Sandmeyer reaction.
Step 2: Detailed Explanation:
1. Aniline (C\(_6\)H\(_5\)NH\(_2\)) reacts with NaNO\(_2\) and HCl at 273-278 K (0-5\(^\circ\)C) to form benzene diazonium chloride, which is compound (Y). The reaction is diazotization. 2. When benzene diazonium chloride is treated with Cu/HBr (a Sandmeyer reaction), the diazonium group (-N\(_2^+\)) is replaced by a bromine atom (-Br), forming bromobenzene (C\(_6\)H\(_5\)Br), which is compound 'Z'.
Step 3: Final Answer:
Compound 'Z' is bromobenzene.
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