JEE Advanced 2025 Paper 1 · Q15 · Named Reactions & Reagents

Question

The major products obtained from the reactions in List-II (above) are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

List-I (named reactions):
(P) Stephen reaction
(Q) Sandmeyer reaction
(R) Hoffmann bromamide degradation reaction
(S) Cannizzaro reaction

List-II (reactant-producing schemes):
(1) Toluene $\xrightarrow{	ext{(i) CrO}_2	ext{Cl}_2/	ext{CS}_2;\,(ii)\,	ext{H}_3	ext{O}^+}$ (→ benzaldehyde, by Étard reaction)
(2) Benzoic acid $\xrightarrow{	ext{(i) PCl}_5;\,(ii)\,	ext{NH}_3;\,(iii)\,	ext{P}_4	ext{O}_{10},\Delta}$ (→ benzonitrile)
(3) Nitrobenzene $\xrightarrow{	ext{(i) Fe, HCl;\,(ii)\,HCl, NaNO}_2,\,273	ext{-}278\,	ext{K, H}_2	ext{O}}$ (→ benzene-diazonium chloride)
(4) Toluene $\xrightarrow{	ext{(i) Cl}_2/h
u, 	ext{H}_2	ext{O};\,(ii)\,	ext{Tollen's;\,(iii)\,SO}_2	ext{Cl}_2;\,(iv)\,	ext{NH}_3}$ (→ benzamide)
(5) Aniline $\xrightarrow{	ext{(i) (CH}_3	ext{CO)}_2	ext{O, Py;\,(ii)\,HNO}_3,\,	ext{H}_2	ext{SO}_4;\,(iii)\,	ext{aq. NaOH}}$ (→ p-nitroaniline)

SolveKar AI · Accepted Answer

Identify the reactant each named reaction needs, then map.

(P) Stephen reaction: nitrile + SnCl$_2$/HCl → aldehyde. Reactant must be a nitrile. Scheme (2) gives benzonitrile. P→2.

(Q) Sandmeyer reaction: aryl diazonium salt + CuX → aryl halide. Reactant must be an aryldiazonium. Scheme (3) gives benzene-diazonium chloride. Q→3.

(R) Hoffmann bromamide degradation: primary amide + Br$_2$/NaOH → amine with one less C. Reactant is a 1° amide. Scheme (4) ends at benzamide. R→4.

(S) Cannizzaro reaction: aldehyde without α-H + conc. base → alcohol + carboxylate. Reactant must be an aldehyde with no α-H, e.g. benzaldehyde. Scheme (1) gives benzaldehyde via Étard reaction. S→1.

Match: P→2, Q→3, R→4, S→1 → option (B).

JEE Advanced 2025 Paper 1 · Q15 · Named Reactions & Reagents

Solution