JEE Advanced 2016 · Paper 1 · Chemistry Q30 · Dronacharya Exams

My attemptsTake a paper

← JEE Advanced 2016 · Paper 1

ChemistryMultiple correct+4 / -2Question 30 of 54

The product(s) of the following reaction sequence is(are)

Aniline → i) Acetic anhydride/pyridine; ii) $\ce{KBrO3/HBr}$ ; iii) $\ce{H3O+}$ , heat; iv) $\ce{NaNO2/HCl}$ , 273-278 K; v) $\ce{Cu/HBr}$

Aniline (benzene ring with NH₂) over an arrow labelled with five steps: i) acetic anhydride/pyridine, ii) KBrO₃/HBr, iii) H₃O⁺, heat, iv) NaNO₂/HCl, 273-278 K, v) Cu/HBr.

A

1,3-Dibromobenzene (meta-dibromobenzene): a benzene ring with Br at positions 1 and 3.

B

1,4-Dibromobenzene (para-dibromobenzene): a benzene ring with Br at positions 1 and 4.

C

1,2,3-Tribromobenzene: a benzene ring with three adjacent Br atoms.

D

1,2,3,5-Tetrabromobenzene: a benzene ring with Br at positions 1, 2, 3 and 5.

Answer and worked solution

Official answer: B

Acetic anhydride/pyridine protects $\ce{NH2}$ as acetanilide. This moderates its activation, so bromination stops at one substitution.
$\ce{KBrO3/HBr}$ generates $\ce{Br2}$ , which brominates para, giving p-bromoacetanilide.
$\ce{H3O+}$ , heat hydrolyses the amide to p-bromoaniline.
$\ce{NaNO2/HCl}$ , 0-5 °C forms the diazonium salt.
$\ce{Cu/HBr}$ (Gattermann/Sandmeyer) replaces $\ce{N2+}$ with Br.

Product: 1,4-dibromobenzene.

(B)

Topics:Diazonium salts Amines Aromatic hydrocarbons and electrophilic substitution

← Previous question Next question →