Is it correct to say: Thinking the Hoffmann bromamide product keeps the same number of carbons.?

The amine has one carbon fewer: CH₃CONH₂ → CH₃NH₂.

Is it correct to say: Proposing Gabriel synthesis to make aniline.?

Gabriel synthesis gives only 1° aliphatic amines; aryl halides do not react with the phthalimide anion.

Is it correct to say: Writing (CH₃)₃N as the strongest base in water because it has the most alkyl groups.?

In water: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. The order 3° > 2° > 1° holds only in the gas phase.

Is it correct to say: Saying 2° amines give the carbylamine test.?

Only 1° amines, aliphatic or aromatic, give isocyanides.

Is it correct to say: Carrying out diazotisation at room temperature.?

Use 273–278 K. When warmed, the diazonium salt reacts with water to give phenol and N₂.

Is it correct to say: Expecting only p-nitroaniline from direct nitration of aniline.?

The anilinium ion formed in acid is meta-directing, so a lot of m-nitroaniline forms, along with tarry oxidation products. Protect as acetanilide first.

📄 Formula sheet (PDF)

Amines

Preparation of amines (Gabriel, Hoffmann bromamide), basicity and pKb, carbylamine and Hinsberg tests, electrophilic substitution in aniline, diazonium salts and their reactions — NCERT Class 12 Chemistry Ch 9

📐 19 formulas✏️ 4 examples🎯 6 practice⚖️ 6 marks🏫 CBSE📚 Class 12✓ 2025–26 syllabus

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Board Exam Tips

→Basicity orders (aqueous and gas phase) and the reasons behind them (+I effect, solvation, steric hindrance) are standard questions. Learn the methyl and ethyl series separately.
→To identify 1°, 2° and 3° amines, use the Hinsberg test (solubility of the sulphonamide in alkali) and the carbylamine test (1° only). Write the equations, not just the observations.
→Track the carbon count: Hoffmann bromamide removes one carbon; nitrile reduction adds one compared with the alkyl halide; Gabriel synthesis and amide reduction keep it the same.
→Diazonium salts are the hub of aromatic conversions: –NH₂ can become –Cl, –Br, –CN, –I, –F, –OH or –H, or the N₂ can be kept in a coupling reaction to make an azo dye.
→To control the very reactive –NH₂ group in bromination or nitration, protect it by acetylation, carry out the reaction, then hydrolyse.

📐 Formulas(19)

Reduction of Nitro Compounds

C_{6} H_{5} NO_{2} Fe + HCl (or Sn + HCl, or H_{2} / Pd) C_{6} H_{5} NH_{2}

Ammonolysis of Alkyl Halides

R-X + NH_{3} \to R-NH_{3}^{+} X^{-} NaOH R-NH_{2} + NaX + H_{2} O

Reduction of Nitriles and Amides

R-C \equiv N H_{2} / Ni or LiAlH_{4} R-CH_{2} -NH_{2}; R-CONH_{2} (i) LiAlH_{4} (ii) H_{2} O R-CH_{2} -NH_{2}

Gabriel Phthalimide Synthesis★ Board fav

C_{6} H_{4} (CO)_{2} NH KOH (ethanolic) C_{6} H_{4} (CO)_{2} N^{-} K^{+} R-X C_{6} H_{4} (CO)_{2} N-R NaOH(aq) R-NH_{2}

Hoffmann Bromamide Degradation★ Board fav

R-CONH_{2} + Br_{2} + 4 NaOH \to R-NH_{2} + Na_{2} CO_{3} + 2 NaBr + 2 H_{2} O

Base Dissociation Constant

K_{b} = \frac{[ R-NH _{3}^{+} ] [ OH ^{-} ]}{[ R-NH _{2} ]}, p K_{b} = - lo g K_{b}

Symbol	Meaning
$K_{b}$	Base dissociation constant of the amine in water
$p K_{b}$	Negative logarithm of K_b (smaller value = stronger base)

Basicity of Aliphatic Amines in Water★ Board fav

(C_{2} H_{5})_{2} NH > (C_{2} H_{5})_{3} N > C_{2} H_{5} NH_{2} > NH_{3}; (CH_{3})_{2} NH > CH_{3} NH_{2} > (CH_{3})_{3} N > NH_{3}

Aromatic Amines are Weaker Bases

Basicity: CH_{3} NH_{2} > NH_{3} > C_{6} H_{5} NH_{2} (p K_{b} = 3.38, 4.75, 9.38)

Carbylamine (Isocyanide) Test★ Board fav

R-NH_{2} + CHCl_{3} + 3 KOH Δ R-NC + 3 KCl + 3 H_{2} O

Nitrous Acid with 1° Aliphatic Amines

R-NH_{2} + HNO_{2} NaNO_{2} + HCl [R-N_{2}^{+} Cl^{-}] H_{2} O R-OH + N_{2} ↑ + HCl

Acylation of Amines

C_{2} H_{5} NH_{2} + CH_{3} COCl pyridine C_{2} H_{5} NHCOCH_{3} + HCl

Hinsberg Test (benzenesulphonyl chloride)★ Board fav

C_{6} H_{5} SO_{2} Cl + H-NHR \to C_{6} H_{5} SO_{2} NHR + HCl

Bromination of Aniline

C_{6} H_{5} NH_{2} + 3 Br_{2} H_{2} O 2,4,6-Br_{3} C_{6} H_{2} NH_{2} ↓ + 3 HBr

Controlled Nitration through Acetanilide

C_{6} H_{5} NH_{2} (CH_{3} CO)_{2} O C_{6} H_{5} NHCOCH_{3} HNO_{3} / H_{2} SO_{4} p-O_{2} NC_{6} H_{4} NHCOCH_{3} H^{+} or OH^{-} p-O_{2} NC_{6} H_{4} NH_{2}

Diazotisation★ Board fav

C_{6} H_{5} NH_{2} + NaNO_{2} + 2 HCl 273 - 278 K C_{6} H_{5} N_{2}^{+} Cl^{-} + NaCl + 2 H_{2} O

Sandmeyer and Gattermann Reactions

ArN_{2}^{+} X^{-} CuCl/HCl, CuBr/HBr or CuCN/KCN ArCl, ArBr or ArCN + N_{2}

Replacement by Iodide and Fluoride

ArN_{2}^{+} Cl^{-} + KI \to ArI + KCl + N_{2}; ArN_{2}^{+} Cl^{-} HBF_{4} ArN_{2}^{+} BF_{4}^{-} Δ ArF + BF_{3} + N_{2}

Replacement by H and by OH

ArN_{2}^{+} Cl^{-} + H_{3} PO_{2} + H_{2} O \to ArH + N_{2} + H_{3} PO_{3} + HCl; ArN_{2}^{+} Cl^{-} + H_{2} O \to ArOH + N_{2} + HCl

Coupling Reactions (azo dyes)

C_{6} H_{5} N_{2}^{+} Cl^{-} + C_{6} H_{5} OH OH^{-} C_{6} H_{5} -N=N-C_{6} H_{4} -OH (p) + Cl^{-} + H_{2} O

✏️ Solved Examples

1Solved Exampleeasy3 steps

Convert benzene into aniline.

Nitrate benzene with the nitrating mixture

C_{6} H_{6} conc. HNO_{3} + conc. H_{2} SO_{4} C_{6} H_{5} NO_{2}

2Solved Exampleboard3 steps

Convert aniline into p-bromoaniline.

Bromine water would give the tribromo product, so first protect –NH₂ by acetylation

C_{6} H_{5} NH_{2} + (CH_{3} CO)_{2} O pyridine C_{6} H_{5} NHCOCH_{3} + CH_{3} COOH

3Solved Exampleboard3 steps

Convert benzoic acid into aniline.

Form the ammonium salt with ammonia

C_{6} H_{5} COOH + NH_{3} \to C_{6} H_{5} COO^{-} NH_{4}^{+}

4Solved ExampleHOTS3 steps

Convert aniline into 1,3,5-tribromobenzene.

Use the strong activating power of –NH₂ to put Br at 2, 4 and 6

C_{6} H_{5} NH_{2} + 3 Br_{2} H_{2} O 2,4,6-Br_{3} C_{6} H_{2} NH_{2} + 3 HBr

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Thinking the Hoffmann bromamide product keeps the same number of carbons.
✓The amine has one carbon fewer: CH₃CONH₂ → CH₃NH₂.
2
Proposing Gabriel synthesis to make aniline.
✓Gabriel synthesis gives only 1° aliphatic amines; aryl halides do not react with the phthalimide anion.
3
Writing (CH₃)₃N as the strongest base in water because it has the most alkyl groups.
✓In water: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃. The order 3° > 2° > 1° holds only in the gas phase.
4
Saying 2° amines give the carbylamine test.
✓Only 1° amines, aliphatic or aromatic, give isocyanides.
5
Carrying out diazotisation at room temperature.
✓Use 273–278 K. When warmed, the diazonium salt reacts with water to give phenol and N₂.
6
Expecting only p-nitroaniline from direct nitration of aniline.
✓The anilinium ion formed in acid is meta-directing, so a lot of m-nitroaniline forms, along with tarry oxidation products. Protect as acetanilide first.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Give a chemical test to distinguish ethanamine from N-ethylethanamine.
Q2
Arrange in increasing order of basic strength in water: NH₃, C₂H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N.
Q3
Name the amine formed when propanamide is treated with Br₂ and NaOH.
Q4
The pK_b of aniline is 9.38. Calculate its K_b.
$K_{b} = 1 0^{- p K_{b}}$
Q5
Why can aniline not be prepared by Gabriel phthalimide synthesis?
Q6
Why does aniline not undergo Friedel–Crafts reaction?

📝 Notes

Amines

Amines are derivatives of ammonia in which one, two or three H atoms are replaced by alkyl or aryl groups, giving 1°, 2° and 3° amines. The lone pair on nitrogen explains almost everything: basicity, nucleophilic reactions (acylation, Hinsberg, carbylamine) and the strong activation of the ring in aniline.

Preparation — watch the carbon count

Same carbons: reduction of nitro compounds, reduction of amides, Gabriel synthesis (1° aliphatic only), ammonolysis (gives a mixture).
One carbon more than R–X: R–X → R–CN → R–CH₂NH₂.
One carbon fewer: Hoffmann bromamide degradation, RCONH₂ → RNH₂.

Basicity — three competing effects

The +I effect of alkyl groups increases electron density on N. In water, the protonated amine is also stabilised by H-bonding (solvation), and bulky groups add steric hindrance. The result: the 2° amine is the strongest base in both the methyl and ethyl series, while the 1°/3° order depends on the alkyl group (methyl: 1° > 3°; ethyl: 3° > 1°). Aniline ( $\text{p}K_b$ = 9.38) is much weaker than NH₃ (4.75) because its lone pair is delocalised into the ring.

Diazonium salts — the conversion hub

Benzenediazonium chloride, made from aniline with NaNO₂ + HCl at 273–278 K, is a key intermediate. N₂ is an excellent leaving group, so it can be replaced by Cl, Br, CN (Sandmeyer), I (KI), F (HBF₄, heat), OH (water) or H (H₃PO₂). Alternatively, the N=N can be kept by coupling with phenol or aniline to make azo dyes. A common trick: use –NH₂ to direct substitution, then remove it through the diazonium salt.