JEE Advanced 2018 · Paper 2 · Chemistry Q3

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ChemistryMultiple correct+4 / -2Question 21 of 54

Aniline reacts with mixed acid (conc. $\mathrm{HNO_3}$ and conc. $\mathrm{H_2SO_4}$ ) at 288 K to give P (51 %), Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are)

R → [1) Ac $_2$ O, pyridine; 2) Br $_2$ , CH $_3$ CO $_2$ H; 3) H $_3$ O $^+$ ; 4) NaNO $_2$ , HCl/273-278 K; 5) EtOH, Δ] → S → [1) Sn/HCl; 2) Br $_2$ /H $_2$ O (excess); 3) NaNO $_2$ , HCl/273-278 K; 4) H $_3$ PO $_2$ ] → major product(s)

Two-stage scheme: R is converted to S with 1) Ac₂O, pyridine; 2) Br₂, CH₃CO₂H; 3) H₃O⁺; 4) NaNO₂, HCl/273-278 K; 5) EtOH, Δ. S is then converted to the major product(s) with 1) Sn/HCl; 2) Br₂/H₂O (excess); 3) NaNO₂, HCl/273-278 K; 4) H₃PO₂.

Answer and worked solution

Official answer: D

Nitration. In mixed acid aniline is mostly protonated, which makes it meta-directing. That gives p- (51%), m- (47%) and o-nitroaniline (2%). So R is o-nitroaniline.

R → S.

$\ce{Ac2O}$ protects the amine as NHAc.
$\ce{Br2}$ goes para to NHAc, giving 4-bromo-2-nitroacetanilide.
$\ce{H3O+}$ removes the acetyl group.
Diazotisation.
EtOH replaces $\ce{N2+}$ by H.

S is 1-bromo-3-nitrobenzene (Br and $\ce{NO2}$ end up meta).

S → product.

Sn/HCl gives 3-bromoaniline.
Excess bromine water brominates all free positions ortho/para to $\ce{NH2}$ (2, 4, 6), giving 2,3,4,6-tetrabromoaniline.
Diazotisation.
$\ce{H3PO2}$ removes the amino group.

The product is 1,2,3,5-tetrabromobenzene.

(D)

Topics:Diazonium salts Aromatic hydrocarbons and electrophilic substitution Amines

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