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JEE Chemistry 2026 Question with Solution

Given below are two statements:

Statement I: Aniline can be synthesized from propylbenzene using simpler reagents in the order i) Acidic KMnO4KMnO_4, ii) Ammonia, iii) Bromine and alkali

Reaction scheme image associated with Statement I showing the organic structure or conversion sequence needed for synthesis from propylbenzene.

Statement II: Aniline can be converted into 1,3,51,3,5-tribromobenzene using reagents in the order i) Bromine-H2OH_2O ii) NaNO2/HClNaNO_2/HCl (0−5 C0 - 5 \, \text{C}) (iii) H3PO2H_3PO_2.

Reaction scheme image associated with Statement II showing bromination, diazotization, and reduction sequence leading to tribromobenzene.

In the light of the above statements, choose the correct answer from the options given below

  • A

    Statement I is false but Statement II is true

  • B

    Both Statement I and Statement II are false

  • C

    Statement I is true but Statement II is false

  • D

    Both Statement I and Statement II are true

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Two synthetic sequences involving aniline, propylbenzene, and conversion to 1,3,51,3,5-tribromobenzene are to be checked.

Find: Which statement is correct.

For Statement I:

Propylbenzene undergoes side-chain oxidation with acidic KMnO4KMnO_4 to give benzoic acid. Benzoic acid on reaction with ammonia and heat gives benzamide. Hoffmann bromamide degradation using bromine and alkali converts benzamide into aniline.

Propylbenzene→acidic KMnO4Benzoic acid→NH3,heatBenzamide→Br2/NaOHAniline\text{Propylbenzene} \xrightarrow{\text{acidic } KMnO_4} \text{Benzoic acid} \xrightarrow{NH_3, \text{heat}} \text{Benzamide} \xrightarrow{Br_2/NaOH} \text{Aniline}

So, Statement I is true.

For Statement II:

Aniline reacts with bromine water to form 2,4,62,4,6-tribromoaniline. Diazotization with NaNO2/HClNaNO_2/HCl at 0−5 C0 - 5 \, \text{C} converts the −NH2-NH_2 group into a diazonium salt. Reduction with H3PO2H_3PO_2 replaces the diazonium group by hydrogen, giving 1,3,51,3,5-tribromobenzene.

Aniline→Br2/H2O2,4,6-tribromoaniline→NaNO2/HCl, 0−5 ∘Cdiazonium salt→H3PO21,3,5-tribromobenzene\text{Aniline} \xrightarrow{Br_2/H_2O} 2,4,6\text{-tribromoaniline} \xrightarrow{NaNO_2/HCl,\, 0-5\,^\circ C} \text{diazonium salt} \xrightarrow{H_3PO_2} 1,3,5\text{-tribromobenzene}

So, Statement II is true.

Therefore, both statements are correct and the correct option is D.

Stepwise Reaction Logic

Given: The truth of two reaction sequences must be evaluated.

Find: Whether each statement is true or false.

Statement I logic:

  1. Any alkyl benzene having at least one benzylic hydrogen gives benzoic acid on strong oxidation with acidic KMnO4KMnO_4.
  2. Benzoic acid with ammonia on heating forms benzamide.
  3. Hoffmann bromamide degradation converts an amide to an amine having one carbon less in the carbonyl chain; thus benzamide gives aniline.

Hence the sequence is chemically valid.

Statement II logic:

  1. Aniline is a strongly activating group, so bromine water gives 2,4,62,4,6-tribromoaniline.
  2. On treatment with NaNO2/HClNaNO_2/HCl at low temperature, the amino group is converted into the benzenediazonium group.
  3. H3PO2H_3PO_2 reduces the diazonium group to hydrogen.
  4. Therefore, the amino substituent is removed and the product becomes 1,3,51,3,5-tribromobenzene.

Thus Statement II is also chemically valid.

Therefore, the correct option is D.

Common mistakes

  • Assuming side-chain oxidation of propylbenzene preserves the full propyl chain is incorrect. Acidic KMnO4KMnO_4 oxidizes the benzylic side chain completely to benzoic acid. Use benzylic oxidation rules for alkyl benzenes.

  • Thinking that ammonia converts benzoic acid directly into aniline is wrong. The intermediate formed on heating is benzamide, and only then Hoffmann bromamide degradation with bromine and alkali gives aniline.

  • Confusing the product of bromination of aniline with monobromination is a conceptual error. Because −NH2-NH_2 strongly activates the ring, bromine water gives 2,4,62,4,6-tribromoaniline. Apply directing effects before predicting the product.

  • Assuming H3PO2H_3PO_2 forms a substituted diazonium product instead of removing the diazonium group is incorrect. H3PO2H_3PO_2 reduces Ar−N2+Ar-N_2^+ to Ar−HAr-H. Use diazonium reduction chemistry here.

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