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JEE Advanced 2019 · Paper 1, Chemistry question 18

Schemes 1 and 2 describe the conversion of P to Q and R to S, respectively. Scheme 3 describes the synthesis of T from Q and S. The total number of Br atoms in a molecule of T is ______

Scheme 1: P (aniline) with (i) Br₂ (excess), H₂O, (ii) NaNO₂, HCl, 273 K, (iii) CuCN/KCN, (iv) H₃O⁺, Δ, (v) SOCl₂, pyridine gives Q (major). Scheme 2: R (benzene) with (i) oleum, (ii) NaOH, Δ, (iii) H⁺, (iv) Br₂, CS₂, 273 K gives S (major). Scheme 3: S with (i) NaOH, (ii) Q gives T (major).
Answer and worked solution

Official answer: 4

Q (Scheme 1). Aniline with bromine water gives 2,4,6-tribromoaniline. Diazotisation and CuCN give the nitrile, hydrolysis gives 2,4,6-tribromobenzoic acid, and SOClX2\ce{SOCl2} gives its acid chloride. Q has 3 Br.

S (Scheme 2). Benzene is sulfonated to benzenesulfonic acid, and NaOH fusion then HX+\ce{H+} gives phenol. BrX2\ce{Br2} in CSX2\ce{CS2} at 273 K monobrominates it, giving 4-bromophenol. S has 1 Br.

T (Scheme 3). The phenoxide attacks the acid chloride, giving the ester 4-bromophenyl 2,4,6-tribromobenzoate. That is 3+1=43+1=4 Br.

Answer: 4

Topics:Diazonium saltsPhenolsAcid derivatives
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