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

Understanding the 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: \[ \boxed{(B)} \]