Question:
Product C of the following reaction sequence will be [assuming sequence: Aniline $\xrightarrow{Br_2/H_2O}$ A $\xrightarrow{NaNO_2/HCl}$ B $\xrightarrow{H_3PO_2}$ C]
Product C of the following reaction sequence will be [assuming sequence: Aniline $\xrightarrow{Br_2/H_2O}$ A $\xrightarrow{NaNO_2/HCl}$ B $\xrightarrow{H_3PO_2}$ C]
Updated On: Apr 10, 2026
- 1-Bromo-4-nitrobenzene
- 1, 3, 5-Tribromo-2-nitrobenzene
- 4-Bromo-1-nitrobenzene
- 1, 3, 5-Tribromobenzene
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept:
Aniline is highly reactive towards electrophilic substitution. Bromine water causes poly-substitution. Subsequent diazotization and deamination allow for the removal of the amino group, leaving the substituted halogens on the ring.
Step 2: Key Formula or Approach:
1. Step 1: Bromination of Aniline. 2. Step 2: Diazotization ($-NH_2 \to -N_2^+Cl^-$). 3. Step 3: Reduction (Deamination) using $H_3PO_2$ or Ethanol.
Step 4: Detailed Explanation:
1. Aniline + $Br_2/H_2O$: The $-NH_2$ group is strongly activating. It directs bromine to both ortho and para positions, yielding 2,4,6-Tribromoaniline (Product A). 2. 2,4,6-Tribromoaniline + $NaNO_2/HCl$: The amino group is converted into a diazonium salt, yielding 2,4,6-Tribromobenzenediazonium chloride (Product B). 3. Product B + $H_3PO_2$: Hypophosphorous acid reduces the diazonium group to a hydrogen atom. 4. The final product is 1,3,5-Tribromobenzene (Product C).
Step 5: Final Answer:
Product C is 1, 3, 5-Tribromobenzene.
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