Question

Identify the final product formed when benzenamine reacts with the given reagents in the sequential order as: (i) $\mathrm{(CH_3CO)_2O}$/Pyridine, (ii) Conc. $\mathrm{HNO_3 + H_2SO_4}$ followed by $\mathrm{H_3O^+}$, (iii) $\mathrm{NaNO_2/HCl}$ (273 K) followed by $\mathrm{H_3PO_2(aq)}$.

• A. 2-Chloro-4-Nitrophenol
• B. Nitrobenzene
• C. 4-Nitrophenol
• D. 4-Chloro-2-Nitroaniline

Hint:

Important sequence to remember: \[ \mathrm{Aniline \rightarrow Diazonium\ Salt \xrightarrow{H_3PO_2} Benzene} \] Hypophosphorous acid removes the diazonium group and replaces it by hydrogen. This reaction is called deamination.

Answer 1

The Correct Option is B

Concept: This question involves a multistep conversion starting from aniline. The sequence includes:
• Protection of amino group
• Electrophilic nitration
• Deprotection
• Diazotization
• Reduction of diazonium salt Such sequences are extremely important in aromatic chemistry because amino groups are highly activating and often need protection before substitution reactions.

Step 1: Protection of Amino Group Benzenamine (aniline): \[ \mathrm{C_6H_5NH_2} \] reacts with acetic anhydride: \[ \mathrm{(CH_3CO)_2O} \] in presence of pyridine. The amino group gets acetylated forming acetanilide: \[ \mathrm{C_6H_5NHCOCH_3} \] Reaction: \[ \mathrm{C_6H_5NH_2 \rightarrow C_6H_5NHCOCH_3} \] This step is necessary because the free amino group is highly activating and may undergo oxidation during nitration. Acetylation decreases its activating effect.

Step 2: Nitration of Acetanilide Now acetanilide is treated with nitrating mixture: \[ \mathrm{Conc.\ HNO_3 + H_2SO_4} \] The group: \[ \mathrm{-NHCOCH_3} \] is ortho-para directing. Because of steric hindrance at ortho position, the para product dominates. Thus: \[ \mathrm{p\text{-}nitroacetanilide} \] is formed as the major product.

Step 3: Hydrolysis Acidic hydrolysis with: \[ \mathrm{H_3O^+} \] removes the acetyl protecting group. Therefore: \[ \mathrm{p\text{-}nitroacetanilide} \rightarrow \mathrm{p\text{-}nitroaniline} \]

Step 4: Diazotization Now p-nitroaniline reacts with: \[ \mathrm{NaNO_2/HCl} \] at 273 K. This converts the amino group into diazonium salt: \[ \mathrm{p\text{-}nitrobenzene\ diazonium\ chloride} \]

Step 5: Reduction of Diazonium Salt Hypophosphorous acid: \[ \mathrm{H_3PO_2} \] reduces diazonium group and replaces it by hydrogen. Thus: \[ \mathrm{ArN_2^+Cl^- \rightarrow ArH} \] Hence: \[ \mathrm{p\text{-}nitrobenzene\ diazonium\ chloride} \rightarrow \mathrm{Nitrobenzene} \] Therefore, the final product formed is: \[ \boxed{\mathrm{Nitrobenzene}} \] Hence, the correct answer is: \[ \boxed{(B)} \]