1. Interpretation of the question

We need to identify the major product formed in each entry of List-II, and then match that product to the reactant required for the named reaction in List-I.

So we first find the final product of each sequence in List-II.

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2. Analyze List-II

(1)

$\text{Toluene} \xrightarrow[\text{(ii) } H_3O^+]{\text{(i) } CrO_2Cl_2/CS_2}$

This is Etard reaction on toluene.

Toluene gives benzaldehyde: $C_6H_5CH_3 \rightarrow C_6H_5CHO$

So, (1) gives benzaldehyde.

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(2)

$\text{Benzoic acid} \xrightarrow[\text{(ii) } NH_3]{\text{(i) } PCl_5} \xrightarrow[\Delta]{\text{(iii) } P_4O_{10}}$

Stepwise:

• Benzoic acid $\xrightarrow{PCl_5}$ benzoyl chloride
• Benzoyl chloride $\xrightarrow{NH_3}$ benzamide
• Benzamide $\xrightarrow{P_4O_{10},\Delta}$ benzonitrile (dehydration)

Thus (2) gives benzonitrile: $C_6H_5CONH_2 \rightarrow C_6H_5CN$

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(3)

$\text{Nitrobenzene} \xrightarrow{Fe/HCl} \xrightarrow[273\text{-}278K, H_2O]{HCl, NaNO_2}$

Stepwise:

• Nitrobenzene $\xrightarrow{Fe/HCl}$ aniline
• Aniline $\xrightarrow{NaNO_2/HCl}$ benzene diazonium chloride
• Diazonium salt $\xrightarrow{H_2O}$ phenol

So (3) gives phenol.

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(4)

$\text{Toluene} \xrightarrow{Cl_2/h\nu, H_2O} \xrightarrow{\text{Tollen's reagent}} \xrightarrow{SO_2Cl_2} \xrightarrow{NH_3}$

Stepwise:

• Side-chain chlorination/hydrolysis of toluene gives benzaldehyde (via benzal chloride / hydrolysis type conversion)
• Benzaldehyde with Tollen's reagent gives benzoic acid
• Benzoic acid with $SO_2Cl_2$ effectively forms acid chloride-related derivative, then with $NH_3$ gives benzamide

Hence (4) gives benzamide.

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(5)

$\text{Aniline} \xrightarrow{(CH_3CO)_2O/\text{pyridine}} \xrightarrow{HNO_3/H_2SO_4, 288K} \xrightarrow{aq.NaOH}$

Stepwise:

• Aniline is acetylated to acetanilide
• Nitration gives mainly p-nitroacetanilide
• Hydrolysis with aqueous NaOH removes acetyl group

Final product = p-nitroaniline.

So (5) gives p-nitroaniline.

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3. Now match with List-I reactions

(P) Stephen reaction

Stephen reaction converts a nitrile into an aldehyde.

Required reactant: benzonitrile

This is obtained in (2).

So, $P \to 2$

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(Q) Sandmeyer reaction

Sandmeyer reaction uses an aryl diazonium salt as reactant.

From sequence (3), nitrobenzene is converted to aniline, then diazotized; this directly involves diazonium salt as the key reactant for Sandmeyer reaction.

Hence, $Q \to 3$

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(R) Hoffmann bromamide degradation reaction

Hoffmann bromamide degradation starts from an amide.

From (4), the product is benzamide.

Therefore, $R \to 4$

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(S) Cannizzaro reaction

Cannizzaro reaction requires an aldehyde without $\alpha$-hydrogen.

From (1), the product is benzaldehyde, which has no $\alpha$-hydrogen.

So, $S \to 1$

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4. Final matching

$P \to 2,\quad Q \to 3,\quad R \to 4,\quad S \to 1$

This corresponds to Option B.

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5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So the derived answer agrees with the stored answer.