Q59.A compound 𝑥 with molar mass 108𝑔 mol−1 undergoes acetylation to give product with molar mass 192𝑔 mol−1. The number of amino groups in the compound 𝑥 is ________.
What This Question Tests
This question requires calculating the number of amino groups in a compound by analyzing the change in molar mass upon acetylation, as each amino group reacts with an acetyl group.
Concepts Tested
Formulas Used
Change in molar mass = n * (Molar mass of acetyl group - Molar mass of H)
📚 NCERT Sections This Tests
9.5 — How Will You Convert:
Chemistry Class 12 · Chapter 9
9.5 How will you convert: (i) Ethanoic acid into methanamine (ii) Hexanenitrile into 1-aminopentane (iii) Methanol to ethanoic acid (iv) Ethanamine into methanamine (v) Ethanoic acid into propanoic acid (vi) Methanamine into ethanamine (vii) Nitromethane into dimethylamine (viii) Propanoic acid into ethanoic acid? 9.6 Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.
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.
1.28 — Calculate The Mass Percentage Of Aspirin (C9H8O4) In Acetonitrile (Ch3Cn) When
Chemistry Class 11 · Chapter 1
1.28 Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN.
📋 Question Details
- Chapter
- Nitrogen Compounds
- Topic
- Acetylation of amino groups
- Year
- 2024
- Shift
- 31 Jan Shift 2
- Q Number
- Q59
- Type
- Numerical
- NCERT Ref
- Class 12 Chemistry Ch 13: Nitrogen Compounds
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