JEE Challenger
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ChemMedAdv 2025 • Paper 1

Matching Reaction Products to Named Organic Reactions

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-IList-II(P) Stephen reaction(1) Toluene→(ii) H3O+(i) CrO2Cl2/CS2(Q) Sandmeyer reaction(2) Benzoic acid→(iii) P4O10, Δ(i) PCl5(ii) NH3(R) Hoffmann bromamide degradation reaction(3) Nitrobenzene→(273–278 K), H2O(i) Fe, HCl(ii) HCl, NaNO2(S) Cannizzaro reaction(4) Toluene→(iii) SO2Cl2(iv) NH3(i) Cl2/hν,H2O(ii) Tollen’s reagent(5) Aniline→(iii) aq. NaOH(i) (CH3CO)2O, Pyridine(ii) HNO3,H2SO4,288 K\begin{array}{ll} \textbf{List-I} & \textbf{List-II} \\[6pt] \text{(P) Stephen reaction} & \text{(1) } \text{Toluene} \xrightarrow[\text{(ii) } \text{H}_3\text{O}^+]{\text{(i) } \text{CrO}_2\text{Cl}_2/\text{CS}_2} \\[12pt] \text{(Q) Sandmeyer reaction} & \text{(2) } \text{Benzoic acid} \xrightarrow[\text{(iii) } \text{P}_4\text{O}_{10},\, \Delta]{\substack{\text{(i) } \text{PCl}_5 \\ \text{(ii) } \text{NH}_3}} \\[18pt] \text{(R) Hoffmann bromamide degradation reaction} & \text{(3) } \text{Nitrobenzene} \xrightarrow[(273\text{--}278\text{ K}),\, \text{H}_2\text{O}]{\substack{\text{(i) } \text{Fe, HCl} \\ \text{(ii) } \text{HCl, NaNO}_2}} \\[18pt] \text{(S) Cannizzaro reaction} & \text{(4) } \text{Toluene} \xrightarrow[\substack{\text{(iii) } \text{SO}_2\text{Cl}_2 \\ \text{(iv) } \text{NH}_3}]{\substack{\text{(i) } \text{Cl}_2/\text{h}\nu,\, \text{H}_2\text{O} \\ \text{(ii) Tollen's reagent}}} \\[24pt] & \text{(5) } \text{Aniline} \xrightarrow[\text{(iii) aq. } \text{NaOH}]{\substack{\text{(i) } (\text{CH}_3\text{CO})_2\text{O, Pyridine} \\ \text{(ii) } \text{HNO}_3,\, \text{H}_2\text{SO}_4,\, 288\text{ K}}} \end{array}

Options

A

P→2; Q→4; R→1; S→3\text{P} \rightarrow 2;\, \text{Q} \rightarrow 4;\, \text{R} \rightarrow 1;\, \text{S} \rightarrow 3

B

P→2; Q→3; R→4; S→1\text{P} \rightarrow 2;\, \text{Q} \rightarrow 3;\, \text{R} \rightarrow 4;\, \text{S} \rightarrow 1

Correct
C

P→5; Q→3; R→4; S→2\text{P} \rightarrow 5;\, \text{Q} \rightarrow 3;\, \text{R} \rightarrow 4;\, \text{S} \rightarrow 2

D

P→5; Q→4; R→2; S→1\text{P} \rightarrow 5;\, \text{Q} \rightarrow 4;\, \text{R} \rightarrow 2;\, \text{S} \rightarrow 1

Topics & Concepts

Aldehydes, Ketones and Carboxylic AcidsAldehydes and KetonesCarboxylic Acids and DerivativesAminesOxidation and Reduction (Organic)

Step-by-Step Solution

To determine the correct matching, we first find the major organic product formed in each sequence of reactions given in List-II:

  1. Reaction (1): Toluene (C6H5CH3)→(ii) H3O+(i) CrO2Cl2/CS2Benzaldehyde (C6H5CHO)\text{Toluene } (\text{C}_6\text{H}_5\text{CH}_3) \xrightarrow[\text{(ii) } \text{H}_3\text{O}^+]{\text{(i) } \text{CrO}_2\text{Cl}_2/\text{CS}_2} \text{Benzaldehyde } (\text{C}_6\text{H}_5\text{CHO}) Benzaldehyde has no α\alpha-hydrogen atoms and undergoes the Cannizzaro reaction upon treatment with a concentrated base.   ⟹  S→1\implies \mathbf{S \rightarrow 1}

  2. Reaction (2): Benzoic acid (C6H5COOH)→(i) PCl5C6H5COCl→(ii) NH3C6H5CONH2→(iii) P4O10, ΔBenzonitrile (C6H5CN)\text{Benzoic acid } (\text{C}_6\text{H}_5\text{COOH}) \xrightarrow{\text{(i) } \text{PCl}_5} \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{(ii) } \text{NH}_3} \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{(iii) } \text{P}_4\text{O}_{10},\, \Delta} \text{Benzonitrile } (\text{C}_6\text{H}_5\text{CN}) Benzonitrile (C6H5CN\text{C}_6\text{H}_5\text{CN}) acts as a substrate in the Stephen reaction (reduction with SnCl2/HCl\text{SnCl}_2/\text{HCl} followed by hydrolysis to yield benzaldehyde).   ⟹  P→2\implies \mathbf{P \rightarrow 2}

  3. Reaction (3): Nitrobenzene (C6H5NO2)→(i) Fe, HClAniline (C6H5NH2)→273–278 K(ii) NaNO2+HClBenzenediazonium chloride (C6H5N2+Cl−)\text{Nitrobenzene } (\text{C}_6\text{H}_5\text{NO}_2) \xrightarrow{\text{(i) } \text{Fe, HCl}} \text{Aniline } (\text{C}_6\text{H}_5\text{NH}_2) \xrightarrow[\text{273--278 K}]{\text{(ii) } \text{NaNO}_2 + \text{HCl}} \text{Benzenediazonium chloride } (\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-) Benzenediazonium chloride is the primary starting material for the Sandmeyer reaction (reaction with CuCl/HCl\text{CuCl}/\text{HCl}, CuBr/HBr\text{CuBr}/\text{HBr}, or CuCN/KCN\text{CuCN}/\text{KCN}).   ⟹  Q→3\implies \mathbf{Q \rightarrow 3}

  4. Reaction (4): Toluene (C6H5CH3)→(i) Cl2/hν, H2OBenzaldehyde (C6H5CHO)→(ii) Tollens’ reagentC6H5COOH→(iii) SO2Cl2C6H5COCl→(iv) NH3Benzamide (C6H5CONH2)\text{Toluene } (\text{C}_6\text{H}_5\text{CH}_3) \xrightarrow{\text{(i) } \text{Cl}_2/h\nu,\, \text{H}_2\text{O}} \text{Benzaldehyde } (\text{C}_6\text{H}_5\text{CHO}) \xrightarrow{\text{(ii) Tollens' reagent}} \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{(iii) } \text{SO}_2\text{Cl}_2} \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{(iv) } \text{NH}_3} \text{Benzamide } (\text{C}_6\text{H}_5\text{CONH}_2) Benzamide is a primary carboxamide and undergoes the Hoffmann bromamide degradation reaction when treated with Br2\text{Br}_2 and aqueous NaOH\text{NaOH} to give aniline.   ⟹  R→4\implies \mathbf{R \rightarrow 4}

Combining these matching pairs gives: P→2;Q→3;R→4;S→1\text{P} \rightarrow 2;\quad \text{Q} \rightarrow 3;\quad \text{R} \rightarrow 4;\quad \text{S} \rightarrow 1

Therefore, the correct option is B.

Matching Reaction Products to Named Organic Reactions | Chemistry PYQ Solution - JEE Challenger