The compound \(B\) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Bromobenzene
Aniline reacts with nitrous acid (\( \ce{NaNO2} \) + HCl) at 273–278 K to give benzenediazonium chloride; this is A. This diazotisation is carried out cold because the diazonium salt decomposes on warming.
\( \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273-278 K}} \ce{C6H5N2+Cl- + NaCl + 2H2O} \)
With \( \ce{Cu2Br2} \)/HBr the –\( \ce{N2^{+}} \) group is replaced by bromine and nitrogen gas is given off (Sandmeyer reaction). B is bromobenzene:
\( \ce{C6H5N2+Cl-} \xrightarrow{\ce{Cu2Br2/HBr}} \ce{C6H5Br + N2} \)
The diazonium group is a very good leaving group, so the same salt gives many products, each with its own reagent:
Sandmeyer reaction (Cu(I) salts): \( \ce{C6H5N2+Cl-} \xrightarrow{\ce{Cu2Cl2/HCl}} \ce{C6H5Cl + N2} \); with \( \ce{CuCN/KCN} \) it gives \( \ce{C6H5CN} \).
Gattermann reaction (copper powder and the halogen acid): \( \ce{C6H5N2+Cl-} \xrightarrow{\ce{Cu/HBr}} \ce{C6H5Br + N2 + CuCl} \)
Iodide ion: \( \ce{C6H5N2+Cl- + KI -> C6H5I + KCl + N2} \)
Water, warm (above 283 K): \( \ce{C6H5N2+Cl- + H2O -> C6H5OH + N2 + HCl} \)
Why the other options are incorrect
Option (1) Benzene — Incorrect. Benzene forms when the diazonium salt is reduced by a mild reducing agent such as hypophosphorous acid (or ethanol), not with \( \ce{Cu2Br2} \)/HBr:
\( \ce{C6H5N2+Cl- + H3PO2 + H2O -> C6H6 + N2 + H3PO3 + HCl} \)
Option (2) Nitrobenzene — Incorrect. The –\( \ce{NO2} \) group is introduced by heating the diazonium fluoroborate with aqueous \( \ce{NaNO2} \) in the presence of copper; no nitrite is present here:
\( \ce{C6H5N2+Cl- + HBF4 -> C6H5N2+BF4- + HCl} \)
\( \ce{C6H5N2+BF4-} \xrightarrow[\text{Cu, }\Delta]{\ce{NaNO2(aq)}} \ce{C6H5NO2 + NaBF4 + N2} \)
Option (4) Phenylcopper (C6H5–Cu) — Incorrect. Cu(I) only helps the replacement of the –\( \ce{N2^{+}} \) group by the halide ion; copper does not become bonded to the ring in the product.
NCERT (2026-27 edition):
Reactions Involving Displacement of Nitrogen – “Diazonium group being a very good leaving group, is substituted by other groups such as Cl–, Br–, I–, CN– and OH– which displace nitrogen from the aromatic ring. The nitrogen formed escapes from the reaction mixture as a gas.
1. Replacement by halide or cyanide ion: The Cl–, Br– and CN– nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction.”