Amines & Diazonium Salts
Complete Organic Chemistry Revision Guide for CBSE
1. Introduction & Structure
Amines constitute an important class of organic compounds derived by replacing one or more hydrogen atoms of ammonia molecule ($NH_3$) by alkyl or aryl groups. They occur widely in nature (proteins, vitamins, alkaloids, and hormones).
- Structure: Like ammonia, the nitrogen atom of amines is $sp^3$ hybridized. The three $sp^3$ hybridized orbitals are involved in bond formation, while the fourth contains an unshared (lone) pair of electrons. Due to the presence of this unshared pair, the geometry of amines is pyramidal (bond angle < 109.5°).
- Classification:
- Primary (1°): One H replaced by R/Ar group. ($R-NH_2$)
- Secondary (2°): Two H's replaced. ($R_2NH$)
- Tertiary (3°): All three H's replaced. ($R_3N$)
2. Master Cheat Sheet: Top Name Reactions
A. Golden Preparation Methods
- 1. Gabriel Phthalimide Synthesis: Used ONLY for the preparation of pure primary aliphatic amines.
$$\text{Phthalimide} \xrightarrow{KOH} \text{Potassium Phthalimide} \xrightarrow{R-X} \text{N-Alkylphthalimide} \xrightarrow{NaOH(aq)} 1^\circ \text{ Amine } (R-NH_2)$$Exam Alert: Aromatic primary amines (like aniline) cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution ($S_N2$) with the anion formed by phthalimide under normal conditions.
- 2. Hoffmann Bromamide Degradation: A method to convert an amide into a primary amine with one carbon less than the parent amide. This is a crucial step-down reaction.
$$R-CO-NH_2 + Br_2 + 4NaOH \rightarrow R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$$
B. Identifying & Distinguishing Tests
- 3. Carbylamine Reaction (Isocyanide Test): Only primary amines (both aliphatic and aromatic) on heating with chloroform ($CHCl_3$) and ethanolic KOH form foul-smelling isocyanides (carbylamines).
$$R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC \text{ (Foul smell)} + 3KCl + 3H_2O$$Secondary and tertiary amines do not give this test.
- 4. Hinsberg's Test: Used to distinguish 1°, 2°, and 3° amines using Benzenesulphonyl chloride ($C_6H_5SO_2Cl$).
- 1° Amine: Forms N-alkylbenzenesulphonamide, which has an acidic hydrogen attached to N. It is soluble in alkali.
- 2° Amine: Forms N,N-dialkylbenzenesulphonamide. It has no acidic hydrogen on N. It is insoluble in alkali.
- 3° Amine: Does not react with Hinsberg's reagent.
3. Exhaustive Theoretical Concepts
3.1 Basicity of Amines (The Most Important Topic)
Amines act as Lewis bases due to the presence of a lone pair of electrons on the nitrogen atom. The basic strength is expressed in terms of $K_b$ or $pK_b$. Larger $K_b$ (or smaller $pK_b$) means a stronger base.
Basicity in Gas Phase (Non-aqueous medium)
In the gaseous phase, there is no solvation effect (no water molecules to stabilize the ion). The basicity depends purely on the +I effect (electron-donating effect) of the alkyl groups. Alkyl groups push electron density towards nitrogen, making the lone pair more readily available for donation.
Order: 3° Amine > 2° Amine > 1° Amine > Ammonia ($NH_3$)
Basicity in Aqueous Phase (Highly Tested!)
In aqueous solution, basic strength depends on three competing factors:
- +I Effect of alkyl groups (Favors 3° > 2° > 1°)
- Solvation (Hydration) Effect: The protonated amine is stabilized by hydrogen bonding with water. More H-bonds = more stability. (Favors 1° > 2° > 3°)
- Steric Hindrance: Bulky alkyl groups block the attack of water molecules. (Favors 1° > 2° > 3°)
The delicate balance of these three forces results in unique orders depending on the size of the alkyl group:
- For Methyl group ($-CH_3$): 2° > 1° > 3° > $NH_3$
$(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ - For Ethyl group ($-C_2H_5$): 2° > 3° > 1° > $NH_3$
$(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$
Why is Aniline less basic than Aliphatic Amines?
In aniline ($C_6H_5NH_2$), the lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$ system of the benzene ring. It gets delocalized over the ring through resonance. As a result, the lone pair is less available for protonation. Furthermore, the anilinium ion formed after accepting a proton has only two resonance structures, compared to five for aniline, making protonation energetically unfavorable.
3.2 Electrophilic Substitution of Aniline
The $-NH_2$ group is highly activating and ortho/para directing. Because it activates the ring so strongly, electrophilic substitution reactions are difficult to control.
- Bromination: Direct reaction with bromine water yields a white precipitate of 2,4,6-tribromoaniline instantly.
To get monobrominated product (e.g., p-bromoaniline): The activating power of $-NH_2$ must be controlled. This is done by protecting the group by acetylation with acetic anhydride to form acetanilide. The lone pair of N is now in resonance with the carbonyl group, reducing its availability for the ring. After bromination, the amide is hydrolyzed back to the amine. - Nitration: Direct nitration of aniline with conc. $HNO_3$ / conc. $H_2SO_4$ yields a complex mixture, including a surprisingly large amount of m-nitroaniline (47%).
Why? In the strongly acidic medium, aniline is protonated to form the anilinium ion ($-NH_3^+$). This group is a strong electron-withdrawing, meta-directing group.
To get p-nitroaniline: Again, protect the amino group via acetylation, nitrate it, and then hydrolyze.
3.3 Diazonium Salts
General formula: $R-N_2^+ X^-$. Aromatic diazonium salts are highly stable at low temperatures (0 - 5°C) due to resonance stabilization of the $C_6H_5-N_2^+$ ion.
Preparation (Diazotization): Treat primary aromatic amine (aniline) with nitrous acid ($NaNO_2 + HCl$) at 273-278 K.
Key Reactions:
- Sandmeyer Reaction: $Ar-N_2^+Cl^- \xrightarrow{CuCl/HCl} Ar-Cl$ (or use $CuBr/HBr$, $CuCN/KCN$).
- Gattermann Reaction: Uses $Cu$ powder instead of cuprous salts. $Ar-N_2^+Cl^- \xrightarrow{Cu/HCl} Ar-Cl$.
- Replacement by H: Treat with mild reducing agents like Hypophosphorous acid ($H_3PO_2$) or Ethanol ($C_2H_5OH$) to yield Benzene.
- Azo Coupling Reaction: Diazonium salts act as weak electrophiles. They react with highly activated rings like phenol or aniline to form brightly colored azo dyes ($-N=N-$ linkage).
With Phenol (in basic medium, pH 9-10) $\rightarrow$ p-Hydroxyazobenzene (Orange dye).
With Aniline (in mildly acidic medium, pH 4-5) $\rightarrow$ p-Aminoazobenzene (Yellow dye).
4. Most Important CBSE Board Questions
Curated from latest CBSE sample papers, emphasizing reasoning, basicity orders, and distinguishing tests.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. The correct order of basic strength of amines in aqueous solution is: 1 Mark
- $(CH_3)_3N > (CH_3)_2NH > CH_3NH_2 > NH_3$
- $(CH_3)_2NH > (CH_3)_3N > CH_3NH_2 > NH_3$
- $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$
- $CH_3NH_2 > (CH_3)_2NH > (CH_3)_3N > NH_3$
Explanation: In aqueous solution, basicity of methylamines is governed by a combination of +I effect, hydration enthalpy, and steric hindrance. The 2° amine strikes the best balance, making it the most basic. Hydration overcomes the +I effect of the 3rd methyl group, pushing 3° below 1°.
Q2. Which of the following will give a primary amine on reduction with $LiAlH_4$? 1 Mark
- Methyl isocyanide
- Acetamide
- Nitrobenzene
- N-Methylacetamide
Explanation: Acetamide ($CH_3CONH_2$) reduces to ethylamine ($CH_3CH_2NH_2$), a 1° amine. Methyl isocyanide ($CH_3NC$) reduces to dimethylamine (a 2° amine). N-Methylacetamide reduces to a 2° amine. Nitrobenzene reduces to aniline (a 1° aromatic amine, but typically $Sn/HCl$ is preferred as $LiAlH_4$ can form azo compounds).
Assertion-Reasoning
Q3. Assertion (A): Gabriel phthalimide synthesis is not used for the preparation of aniline.
Reason (R): Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. 1 Mark
Explanation: In aryl halides (like chlorobenzene), the C-X bond has partial double bond character due to resonance. Thus, the bulky phthalimide anion cannot displace the halide ion to form N-phenylphthalimide.
Section B: 2 Mark Questions (Very Short Answer)
Q4. Give reasons for the following:
(i) Aniline does not undergo Friedel-Crafts reaction.
(ii) Amines are less acidic than alcohols of comparable molecular masses. 2 Marks
(ii) Loss of a proton from an amine gives an amide ion ($R-NH^-$), while an alcohol gives an alkoxide ion ($R-O^-$). Oxygen is more electronegative than nitrogen, so it can accommodate the negative charge much more easily, making the alkoxide ion more stable than the amide ion. Hence, alcohols release protons more readily.
Q5. Complete the following reactions:
(i) $CH_3CH_2NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta}$
(ii) $C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow$ 2 Marks
$$CH_3CH_2NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3CH_2NC (\text{Ethyl isocyanide}) + 3KCl + 3H_2O$$
(ii) Reduction of diazonium salt:
$$C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow C_6H_6 (\text{Benzene}) + N_2 + H_3PO_3 + HCl$$
Section C: 3 Mark Questions (Short Answer)
Q6. Give a simple chemical test to distinguish between the following pairs:
(i) Aniline and Ethylamine
(ii) Methylamine and Dimethylamine
(iii) Aniline and N-Methylaniline 3 Marks
(ii) Methylamine (1°) and Dimethylamine (2°): Carbylamine test. Warm with chloroform and alcoholic KOH. Methylamine will emit a highly offensive (foul) smell of methyl isocyanide. Dimethylamine gives no such smell.
(iii) Aniline (1°) and N-Methylaniline (2°): Carbylamine test. Aniline will give the foul smell of isocyanide. N-methylaniline will not. (Alternatively, use Hinsberg's test. Aniline forms a product soluble in NaOH; N-methylaniline forms an insoluble product).
Q7. How will you bring about the following conversions?
(i) Benzene to Aniline
(ii) Aniline to p-bromoaniline
(iii) Benzamide to Toluene 3 Marks
Nitrate benzene to nitrobenzene, then reduce.
$C_6H_6 \xrightarrow{Conc. HNO_3 + Conc. H_2SO_4} C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2$
(ii) Aniline to p-bromoaniline:
Protect the amine, brominate, then deprotect.
$C_6H_5NH_2 \xrightarrow{(CH_3CO)_2O} C_6H_5NHCOCH_3 \xrightarrow{Br_2/CH_3COOH} \text{p-Br-}C_6H_4NHCOCH_3 \xrightarrow{H^+/H_2O} \text{p-Br-}C_6H_4NH_2$
(iii) Benzamide to Toluene:
Hoffmann degradation $\rightarrow$ Diazotization $\rightarrow$ Reduction $\rightarrow$ Friedel Crafts.
$C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} C_6H_5NH_2 \xrightarrow{NaNO_2/HCl, 0^\circ C} C_6H_5N_2Cl \xrightarrow{H_3PO_2, H_2O} C_6H_6$
$C_6H_6 \xrightarrow{CH_3Cl, \text{ Anhy. } AlCl_3} C_6H_5CH_3 \text{ (Toluene)}$
Section D: 4 Mark Case-Based Questions
Q8. Read the passage given below and answer the following questions:
(i) Why are tertiary amines incapable of undergoing acylation reactions? (1 Mark)
(ii) Write the structure of Hinsberg's reagent. (1 Mark)
(iii) Why is the sulphonamide formed by a primary amine soluble in aqueous NaOH, whereas that formed by a secondary amine is insoluble? (2 Marks)
(ii) Benzenesulphonyl chloride. Structure: $C_6H_5SO_2Cl$.
(iii) The sulphonamide formed by a primary amine ($C_6H_5SO_2NHR$) has a hydrogen atom attached to the nitrogen. This hydrogen is strongly acidic due to the powerful electron-withdrawing nature of the sulphonyl group. Hence, it easily dissolves in a strong base like aqueous NaOH to form a soluble salt. The sulphonamide formed by a secondary amine ($C_6H_5SO_2NR_2$) has no such acidic hydrogen attached to nitrogen, making it completely insoluble in alkali.
Section E: 5 Mark Long Answer Questions
Q9. An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with $Br_2$ and $KOH$ forms a compound 'C' of molecular formula $C_6H_7N$. Compound 'C' on treatment with $NaNO_2/HCl$ at 0-5°C forms compound 'D'. Compound 'D' on treatment with phenol in mild alkaline medium forms an orange dye 'E'.
(i) Identify A, B, C, D, and E.
(ii) Write all the chemical equations involved. 5 Marks
- "Compound 'C' has formula $C_6H_7N$" $\rightarrow$ This is a classic formula for Aniline ($C_6H_5NH_2$).
- "'B' heated with $Br_2/KOH$ forms 'C'" $\rightarrow$ This is the Hoffmann bromamide degradation. It means 'B' is an amide with one more carbon than aniline. Thus, 'B' must be Benzamide ($C_6H_5CONH_2$).
- "'A' + aqueous $NH_3$ + heat forms Benzamide" $\rightarrow$ Compound 'A' must be a carboxylic acid. Thus, 'A' is Benzoic Acid ($C_6H_5COOH$).
- "Aniline 'C' + $NaNO_2/HCl$ at 0-5°C forms 'D'" $\rightarrow$ This is diazotization. 'D' is Benzenediazonium chloride ($C_6H_5N_2^+Cl^-$).
- "'D' + Phenol forms orange dye 'E'" $\rightarrow$ This is a coupling reaction. 'E' is p-Hydroxyazobenzene.
1. Formation of B (Amide synthesis):
$$C_6H_5COOH \text{ (A)} + NH_3 \rightarrow C_6H_5COO^-NH_4^+ \xrightarrow{\Delta} C_6H_5CONH_2 \text{ (B)} + H_2O$$
2. Formation of C (Hoffmann bromamide):
$$C_6H_5CONH_2 \text{ (B)} + Br_2 + 4KOH \rightarrow C_6H_5NH_2 \text{ (C)} + K_2CO_3 + 2KBr + 2H_2O$$
3. Formation of D (Diazotization):
$$C_6H_5NH_2 \text{ (C)} + NaNO_2 + 2HCl \xrightarrow{0-5^\circ C} C_6H_5N_2^+Cl^- \text{ (D)} + NaCl + 2H_2O$$
4. Formation of E (Azo Coupling):
$$C_6H_5N_2^+Cl^- \text{ (D)} + C_6H_5OH \xrightarrow{pH 9-10} C_6H_5-N=N-C_6H_4OH \text{ (E, p-Hydroxyazobenzene)} + HCl$$
Mastering 'Amines'
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The Basicity Trap: When asked to arrange amines in order of basic strength, read the question twice to see if it says "gas phase" or "aqueous solution". If aqueous, look carefully if it is a methyl or ethyl series. It changes everything!
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Hoffmann = Step Down: Remember that Hoffmann Bromamide reaction is the ONLY reaction in your syllabus designed specifically to reduce the length of a carbon chain by exactly one carbon. If a roadmap problem loses a carbon, look for this reaction.
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Why Protect Aniline?: The reasoning question "Why is aniline acetylated before nitration or bromination?" appears almost every alternate year. The answer has two parts: 1) To control the extreme reactivity to prevent poly-substitution, and 2) To prevent the oxidation of the amine group by strong acids like nitric acid.
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Carbylamine = Primary only: If you need to distinguish between ANY primary amine (aliphatic or aromatic) and a secondary or tertiary amine, the Carbylamine (isocyanide) test is your fastest and safest answer.
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