Q47.Hoffmann bromamide degradation of benzamide gives product A, which upon heating with CHCl3 and NaOH gives product B. The structures of A and B are: JEE Main 2021 (17 Mar Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)
What This Question Tests
This question requires identifying the products of a two-step reaction sequence: Hoffmann bromamide degradation of benzamide to form aniline, followed by the carbylamine reaction of aniline to form phenyl isocyanide.
Concepts Tested
📚 NCERT Sections This Tests
9.9 — Give The Structures Of A, B And C In The Following Reactions:
Chemistry Class 12 · Chapter 9
9.9 Give the structures of A, B and C in the following reactions: B NaOH Br 2 C (i) CH 3 CH 2 I NaCN A Partial OHhydrolysis CuCN H 2 O/H NH 3 (ii) C 6 H 5 N 2 Cl A B C 4 B HNO 2 C (iii) CH 3 CH 2 Br KCN A LiAlH 0 C 2 2 O/H HCl B H C (iv) C6 H 5 NO 2 Fe/HCl A NaNO273 K 3 A NaOBr B NaNO 2 /HCl C (v) CH 3 COOH NH 2 B C 6 H 5 OH C (vi) C6 H 5 NO 2 Fe/HCl A HNO273K 279 Amines Reprint 2025-26
9.7 — Write Short Notes On The Following:
Chemistry Class 12 · Chapter 9
9.7 Write short notes on the following: (i) Carbylamine reaction (ii) Diazotisation (iii) Hofmann’s bromamide reaction (iv) Coupling reaction (v) Ammonolysis (vi) Acetylation (vii) Gabriel phthalimide synthesis.
8.17 — Complete Each Synthesis By Giving Missing Starting Material, Reagent Or Products
Chemistry Class 12 · Chapter 8
8.17 Complete each synthesis by giving missing starting material, reagent or products
📋 Question Details
- Chapter
- Nitrogen Compounds
- Topic
- Hoffmann bromamide degradation and Carbylamine reaction
- Year
- 2021
- Shift
- 17 Mar Shift 1
- Q Number
- Q47
- Type
- MCQ
- NCERT Ref
- Class 12 Chemistry Ch 13: Amines
More from this Chapter
Q81.An organic compound having molecular mass 60 is found to contain C = 20%, H = 6.67% and N = 46.67% while rest is oxygen. On heating it gives NH3 alongwith a solid residue. The solid residue give violet colour with alkaline copper sulphate solution. The compound is (1) CH3NCO (2) CH3CONH2 (3) (NH2)2CO (4) CH3CH2CONH2
Q57.Hinsberg’s reagent is: (1) C6H5SO2Cl (2) SOCl2 (3) (COCl)2 (4) C6H5COCl
Q80.In the chemical reaction, CH3CH2NH2 −CHCl3 + 3KOH ⟶(A) + (B) + 3H2O, the compound (A) and (B) are respectively (1) C2H5CN and 3KCl (2) CH3CH2CONH2 and 3KCl (3) C2H5NC and K2CO3 (4) C2H5NC and 3KCl
Q81.Which one of the following is the strongest base in aqueous solution? (1) Trimethylamine (2) Aniline (3) Dimethylamine (4) Methylamine