JEE Main · 2020 · Shift-IIhardPB11-062

The reaction of H3N3B3Cl3 (A) with LiBH4 in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A)…

p-Block Elements (Class 11) · Class 11 · JEE Main Previous Year Question

Question

The reaction of H3N3B3Cl3\mathrm{H_3N_3B_3Cl_3} (A) with LiBH4\mathrm{LiBH_4} in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H3N3B3(Me)3\mathrm{H_3N_3B_3(Me)_3}. Compounds (B) and (C) respectively, are:

Options
  1. a

    Borazine and MeBr

  2. b

    Diborane and MeMgBr

  3. c

    Boron nitride and MeBr

  4. d

    Borazine and MeMgBr

    ✓
Correct Answerd

Borazine and MeMgBr

Detailed Solution

Strategy: Trace both reactions of B-trichloroborazine (A =\ceH3N3B3Cl3\ce{H3N3B3Cl3}) to identify B and C.

Step 1: Identifying Compound B\ceH3N3B3Cl3+LiBH4−>[ THF ]B (inorganic benzene)+LiCl+H2\ce{H3N3B3Cl3 + LiBH4 ->[\ THF\ ] \text{B (inorganic benzene)} + LiCl + H2}The reduction of B-trichloroborazine with\ceLiBH4\ce{LiBH4}replaces the 3 B-Cl groups with 3 B-H groups, giving Borazine (\ceB3N3H6\ce{B3N3H6}). B = Borazine (\ceH3N3B3H3\ce{H3N3B3H3}or\ceB3N3H6\ce{B3N3H6}) ✅

Step 2: Identifying Compound C\ceH3N3B3Cl3+C−>H3N3B3(Me)3\ce{H3N3B3Cl3 + C -> H3N3B3(Me)3}The product has methyl groups (Me = CH₃) replacing the Cl atoms on boron. This is a substitution (organometallic reaction). Grignard reagents perform such substitutions:\ceH3N3B3Cl3+3CH3MgBr−>H3N3B3(CH3)3+3MgBrCl\ce{H3N3B3Cl3 + 3CH3MgBr -> H3N3B3(CH3)3 + 3MgBrCl}C = MeMgBr (Methylmagnesium bromide) ✅Answer: (D) Borazine and MeMgBr\boxed{\text{Answer: (D) Borazine and MeMgBr}}

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