JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q60

← JEE Main 2014 · 19 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 60 of 90

In a set of reactions p - nitrotoluene yielded a product E
$\text{p-nitrotoluene}\ce{->[Br2][FeBr3] B ->[Sn/HCl] C ->[NaNO2][HCl] D ->[CuBr][HBr] E}$
The product E would be :

Scheme: p-nitrotoluene (CH₃ and NO₂ para on a benzene ring) → Br₂/FeBr₃ → B → Sn/HCl → C → NaNO₂/HCl → D → CuBr/HBr → E.

2,4,6-Tribromotoluene

Toluene with Br at both positions ortho to CH₃ and at the para position.

2,4-Dibromotoluene

2,3-Dibromotoluene

Toluene with Br at the two adjacent positions 2 and 3.

4-Bromobenzyl bromide

Benzene ring with CH₂Br and, para to it, Br.

Answer and worked solution

Official answer: B

$\ce{Br2/FeBr3}$ brominates ortho to $\ce{CH3}$ (meta to $\ce{NO2}$ ): B is 2-bromo-4-nitrotoluene.
Sn/HCl reduces $\ce{NO2}$ to $\ce{NH2}$ (C).
Diazotisation gives the diazonium salt (D).
CuBr/HBr (Sandmeyer) replaces it with Br.

E is 2,4-dibromotoluene, structure (2).

Topics:Diazonium salts Aromatic hydrocarbons and electrophilic substitution

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JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 60