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Amines
Chemistry · Class 12NCERT pp. 259–2772 sets × 25 questions
Key facts and formulas for Amines, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Structure of amine N | Three bonds + one lone pair, sp³, pyramidal; basic because the lone pair accepts H⁺ |
| Classification | 1°, 2°, 3° by number of groups on N (not on carbon); tert-butylamine is primary |
| Basic strength | Availability of N lone pair; aliphatic amines > NH₃ > aniline > amides; in water ethyl series: 2° > 3° > 1° > NH₃ |
| pKa + pKb = 14 | Kb = 10^(−pKb); for a weak base [OH⁻] = √(Kb × C), then pOH, pH = 14 − pOH |
| Preparation | Reduction of nitro compounds and nitriles; ammonolysis gives a mixture; Gabriel gives pure aliphatic 1° amines only |
| Hoffmann bromamide | RCONH₂ → RNH₂ with one carbon fewer, 1 : 1 moles, configuration of R retained |
| Tests with reagents | Carbylamine (CHCl₃/KOH) for 1° amines; Hinsberg: 1° soluble in KOH, 2° insoluble, 3° no reaction |
| Nitrous acid | Aliphatic 1° → alcohol + N₂; aromatic 1° at 273–278 K → diazonium salt; 2° → N-nitrosoamine; 3° aromatic → ring p-nitroso |
| Diazonium reactions | Sandmeyer CuCl/CuBr/CuCN; KI for iodide; H₃PO₂ removes N₂⁺; coupling with phenol or amine gives azo dyes |
| Aniline reactions | Br₂ water → 2,4,6-tribromoaniline; direct nitration gives some meta via anilinium ion; Kjeldahl: mmol N = mmol NH₃ |
Three sample questions
Q1. The nitrogen atom in an amine is
- sp² hybridised with a planar shape
- sp³ hybridised with a pyramidal shape
- sp hybridised with a linear shape
- sp³ hybridised with a planar shape
Hint and solution
Hint: Count the bond pairs and lone pairs on nitrogen, then ask what that many electron domains need.
- N: 3 bond pairs + 1 lone pair = 4 domains — Count the electron domains around nitrogen in an amine: three sigma bonds and one lone pair.
- 4 domains → sp³ hybridisation — Four domains need four equivalent orbitals, so nitrogen is sp³ hybridised.
- Lone pair repels bonds → pyramidal — The lone pair takes more space than a bond pair, so the three bonds are squeezed into a pyramid, not a flat triangle.
- Answer: sp³, pyramidal — So nitrogen is sp³ hybridised with a pyramidal shape. Option B.
Answer: B. Three bond pairs and one lone pair use four sp³ orbitals; the lone pair makes the shape pyramidal.
Q2. The number of structural isomers of C₃H₉N is
- 2
- 3
- 5
- 4
Hint and solution
Hint: Sort the isomers by class: primary, secondary and tertiary. Primary amines have a three carbon chain in two ways.
- Primary: CH₃CH₂CH₂NH₂ and (CH₃)₂CHNH₂ — Put the NH₂ group on the end or the middle carbon of a three carbon chain. That gives two primary amines.
- Secondary: CH₃NHC₂H₅ — Split the three carbons as one and two on the nitrogen. That gives N-methylethanamine.
- Tertiary: (CH₃)₃N — Three methyl groups on nitrogen give N,N-dimethylmethanamine.
- Total = 2 + 1 + 1 = 4 — Adding the classes gives four isomers. Option D.
Answer: D. Propan-1-amine, propan-2-amine, N-methylethanamine and N,N-dimethylmethanamine.
Q3. The IUPAC name of (C₂H₅)₃N is
- N,N-diethylethanamine
- Triethylamine-N
- N-ethyldiethanamine
- Triethanamine
Hint and solution
Hint: Choose one ethyl group as the parent chain. The other two groups sit on nitrogen and need the locant N.
- Parent = longest chain on N = ethanamine — Pick one ethyl chain as the parent and call it ethanamine.
- Other two ethyls on N → N,N-diethyl — The remaining two ethyl groups are on nitrogen, so they get the locant N twice.
- Name = N,N-diethylethanamine — Join the substituents and the parent in alphabetical form.
- Answer: N,N-diethylethanamine — This matches the first choice. Option A.
Answer: A. The ethanamine chain is the parent; the two other ethyl groups on N are written N,N-diethyl.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.