JEE Advanced 2025 · Paper 1 · Chemistry Q15

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ChemistryList match+4 / -1Question 47 of 48

The major products obtained from the reactions in List-II 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

P
Stephen reaction
Q
Sandmeyer reaction
R
Hoffmann bromamide degradation reaction
S
Cannizzaro reaction

List-II

1
Toluene $\xrightarrow{\text{(i) } \ce{CrO2Cl2/CS2};\ \text{(ii) } \ce{H3O+}}$
2
Benzoic acid $\xrightarrow{\text{(i) } \ce{PCl5};\ \text{(ii) } \ce{NH3};\ \text{(iii) } \ce{P4O10}, \Delta}$
3
Nitrobenzene $\xrightarrow{\text{(i) Fe, HCl};\ \text{(ii) HCl, } \ce{NaNO2}\ (273\text{-}278\ \mathrm{K}), \ce{H2O}}$
4
Toluene $\xrightarrow{\text{(i) } \ce{Cl2}/h\nu, \ce{H2O};\ \text{(ii) Tollen's reagent};\ \text{(iii) } \ce{SO2Cl2};\ \text{(iv) } \ce{NH3}}$
5
Aniline $\xrightarrow{\text{(i) } \ce{(CH3CO)2O}\text{, pyridine};\ \text{(ii) } \ce{HNO3}, \ce{H2SO4}, 288\ \mathrm{K};\ \text{(iii) aq. NaOH}}$

P → 2; Q → 4; R → 1; S → 3

P → 2; Q → 3; R → 4; S → 1

P → 5; Q → 3; R → 4; S → 2

P → 5; Q → 4; R → 2; S → 1

Answer and worked solution

Official answer: B

Products of List-II.

Toluene with $\ce{CrO2Cl2}$ (Étard reaction) gives benzaldehyde.
Benzoic acid → benzoyl chloride → benzamide, then $\ce{P4O10}$ dehydrates it to benzonitrile.
Nitrobenzene → aniline, then diazotisation gives benzenediazonium chloride.
Toluene → benzaldehyde → benzoic acid → benzoyl chloride → benzamide.
Aniline → acetanilide → p-nitroacetanilide → p-nitroaniline.

Matching to the named reactions.

(P) Stephen reaction needs a nitrile → (2).
(Q) Sandmeyer reaction needs a diazonium salt → (3).
(R) Hoffmann bromamide degradation needs an amide → (4).
(S) Cannizzaro reaction needs an aldehyde without α-H (benzaldehyde) → (1).

The match is P→2, Q→3, R→4, S→1, which is option (B).

Topics:Named reactions Diazonium salts Acid derivatives

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JEE Advanced 2025 · Paper 1, Chemistry question 15