Question

Give structures of A, B and C: Aniline $\xrightarrow{Br_2/H_2O}$ A $\xrightarrow{NaNO_2+HCl, 0-5^\circ C}$ B $\xrightarrow{H_3PO_2+H_2O}$ C

Hint:

$H_3PO_2$ replaces diazonium group with $-$H. This is used to prepare compounds not directly accessible by electrophilic substitution.

Answer 1

Step 1: Step 1 — Bromination (A).
Aniline + $Br_2/H_2O$: the $-NH_2$ group strongly activates ortho/para positions. All three available positions (2, 4, 6) are brominated simultaneously. Compound A: 2,4,6-tribromoaniline ($2,4,6$-$Br_3C_6H_2NH_2$).
Step 2: Step 2 — Diazotization (B).
A + $NaNO_2 + HCl$ at $0-5^\circ C$: the primary $-NH_2$ is converted to a diazonium group $(-N_2^+Cl^-)$. Compound B: 2,4,6-tribromobenzenediazonium chloride.
Step 3: Step 3 — Replacement with $H_3PO_2$ (C).
B + $H_3PO_2/H_2O$: hypophosphorous acid reduces the diazonium group, replacing $-N_2^+$ with $-H$. The three Br substituents remain. Compound C: 1,3,5-tribromobenzene.
Step 4: Summary.
A: 2,4,6-tribromoaniline; B: 2,4,6-tribromobenzenediazonium chloride; C: 1,3,5-tribromobenzene.