JEE Advanced 2021 Paper 2 · Q02 · Amines & Diazonium
Correct option(s) for the following sequence of reactions is(are):
PhCH$_3$ $\xrightarrow{\text{Br}_2,\ \text{light}}$ P $\xrightarrow{\text{1. Q; 2. H}_2,\ \text{Pd/C}}$ R $\xrightarrow{\text{CHCl}_3,\ \text{KOH}}$ S (foul smelling)
Also: PhCH$_3$ $\xrightarrow{\text{i. KMnO}_4,\ \text{KOH, heat; ii. H}_3\text{O}^+}$ T $\xrightarrow{\text{1. NH}_3;\ \text{2. heat}}$ U $\xrightarrow{\text{1. NH}_3;\ \text{2. heat}}$ and R $\xrightarrow{\text{W}}$ V
(Refer to the boxed scheme: P = PhCH$_2$Br, R = PhCH$_2$NH$_2$, S = PhCH$_2$NC (foul smelling, carbylamine), T = PhCOOH, U = PhCONH$_2$, V (from R via W) and Q, W are reagents to be identified.)
Reveal answer + step-by-step solution
Correct answer:C, D
Solution
PhCH$_3$ $\xrightarrow{\text{Br}_2,\ \text{light}}$ PhCH$_2$Br (P, benzyl bromide).
Step P $\rightarrow$ R: '1. Q; 2. H$_2$, Pd/C'. With Q = AgNO$_2$, benzyl bromide gives the nitro compound PhCH$_2$NO$_2$ (silver nitrite, S$_N$2 favours the C-nitro product), which on hydrogenation over Pd/C gives benzylamine R = PhCH$_2$NH$_2$ (phenylmethanamine).
R $\xrightarrow{\text{CHCl}_3,\ \text{KOH}}$ S: this is the carbylamine reaction; primary amine + CHCl$_3$ + alc. KOH gives benzyl isocyanide PhCH$_2$NC, which is foul smelling. Confirms S.
Side branch: PhCH$_3$ $\xrightarrow{\text{KMnO}_4/\text{KOH, heat; then H}_3\text{O}^+}$ PhCOOH (T) $\xrightarrow{\text{1. NH}_3;\ \text{2. heat}}$ PhCONH$_2$ (U, benzamide) — consistent.
R $\xrightarrow{\text{W}}$ V: To convert PhCH$_2$NH$_2$ into V the only sensible reagent set giving an isolable product is W = AgCN, which converts the amine via prior $\beta$-step to V = PhCH$_2$NC. Matching the scheme, the consistent identifications are R = phenylmethanamine and Q = AgNO$_2$ (option C) and W = LiAlH$_4$, V = AgCN — note: the official FIITJEE key marks (C) and (D); option (C) fixes Q and R; option (D) fixes W and V via the si … [truncated]
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