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ChemistrySingle correct+4 / -1Question 68 of 75

JEE Main 2026 · 5 Apr · Shift 1, Chemistry question 68

Identify the incorrect statements.

A. Benzylamine is a stronger base than aniline.

B. 4-Methoxyaniline can be synthesized by Gabriel phthalimide synthesis.

C. 2-Phenylacetamide →BrX2, NaOH\xrightarrow{\ce{Br2},\ \ce{NaOH}} primary aromatic amine

D. 4-Nitroaniline →(i) NaNOX2, HCl, 0∘C; (ii) Δ\xrightarrow{\text{(i) }\ce{NaNO2},\ \ce{HCl},\ 0^\circ\mathrm C;\ \text{(ii) }\Delta} product will dissolve in NaOH

Choose the correct answer from the options given below:

Structures in the statements: A benzylamine and aniline; B 4-methoxyaniline; C 2-phenylacetamide; D 4-nitroaniline.
Choose one option
Answer and worked solution

Official answer: C

Each statement.

  • A: In benzylamine the lone pair is not delocalised into the ring, so it is more basic than aniline. Correct.
  • B: Gabriel synthesis cannot make aryl amines, because aryl halides do not undergo SN2S_N2 with the phthalimide anion. Incorrect.
  • C: Hofmann bromamide degradation of CX6HX5CHX2CONHX2\ce{C6H5CH2CONH2} gives benzylamine CX6HX5CHX2NHX2\ce{C6H5CH2NH2}. That is not an aromatic amine (the N is not on the ring). Incorrect.
  • D: Diazotisation followed by warming gives 4-nitrophenol, which is acidic and dissolves in NaOH. Correct.

Incorrect: B and C.

(C)

Topics:AminesDiazonium salts
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