Consider the following reactions sequence [figure] When the product (P) is subjected to Carius analysis using AgNO3, 1.0 g of the product (P) will produce g of the precipitate of AgBr. (Nearest Integer) (Given : molar mass in g mol-1 C : 12, H : 1, O : 16, N : 14, Br : 80, Ag : 108)
Answer
1
Solution
(i) Sn/HCl, then OH : –NO2 → –NH2 (p-toluidine)
(ii) (CH3CO)2O: –NH2 → –NHCOCH3 (activation is moderated, monosubstitution)
(iii) Br2/AlBr3: para position is blocked by –CH3; –NHCOCH3 is the stronger o/p director ⇒ Br enters ortho to –NHCOCH3
(iv) H3O : amide hydrolysed back to –NH2
⇒ P = 2-bromo-4-methylaniline [figure] , C7H8BrN
M(P) = 84 + 8 + 80 + 14 = 186 g mol-1; M(AgBr) = 188 g mol-1
1 mol P (1 Br) → 1 mol AgBr ⇒ mass of AgBr = 1.0186 188 = 1.01 g ≈ 1 g
Answer: 1