CBSE Class 12 Chemistry Amines Important Questions Set 07

Very Short Answer Questions (1 Mark)

Question 1. Aniline gets coloured on standing in air for a long time. Why ?
or Why does aniline turn blackish brown in open air. (MP Board 2013)
Answer:
Due to strong electron-donating effect (+R-effect) of the NH₂ group, the electron density on the benzene ring increases. As a result, aniline is easily oxidised on standing in air for a long time to form coloured blackish brown products.

Teacher's Note:
1. Freshly distilled aniline is colourless; the colour that develops is due to atmospheric oxidation.

Question 2. Give an example of a zwitterion.
Answer:
H₃N⁺–C₆H₄–SO₃⁻ (p-isomer), sulphanilic acid.

Teacher's Note:
1. A zwitterion carries both a positive and a negative charge in the same molecule; this ionic character gives sulphanilic acid its high melting point.

Question 3. Mention two important uses of sulphanilic acid.
Answer:
Sulphanilic acid is used in the manufacture of (i) dyes (ii) sulpha drugs.

Teacher's Note:
1. Sulphanilic acid is diazotised and coupled to make azo dyes such as methyl orange.

Question 4. Name one reagent used for the separation of primary, secondary and tertiary amines.
Answer:
Hinsberg's reagent (C₆H₅SO₂Cl)

Teacher's Note:
1. Hinsberg's reagent is benzenesulphonyl chloride: the 1° amine product dissolves in alkali, the 2° product does not, and the 3° amine does not react.

Question 5. A (C₃H₉N) reacts with benzenesulphonyl chloride to give a solid insoluble in alkali. Give the structure of A.
Answer:
Since A (C₃H₉N) reacts with benzenesulphonyl chloride to give a solid insoluble in alkali, (A) must be a 2° amine ; i.e., CH₃–NH–CH₂CH₃ (ethylmethylamine).

Teacher's Note:
1. The sulphonamide of a 2° amine has no H on N, so it cannot form a salt with alkali; this is the basis of Hinsberg's test.

Question 6. Mention the chief use of quaternary ammonium salts derived from long chain amines.
or What for are quaternary ammonium salts widely used ?
Answer:
The quaternary ammonium salts derived from long chain aliphatic amines are used as detergents, i.e.,
CH₃(CH₂)₁₅N⁺(CH₃)₃Cl⁻, cetyltrimethylammonium chloride.

Teacher's Note:
1. These are cationic detergents: the long hydrocarbon tail is attached to a positively charged N.

Question 7. What is Sandmeyer reaction ?
Answer:
The reaction of a diazonium salt with CuCl/HCl or CuBr/HBr or CuCN/KCN to give the corresponding chloroarene or bromoarene or arenenitrile is called Sandmeyer reaction. For example,
C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) C₆H₅Cl + N₂
C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuBr/HBr}} \) C₆H₅Br + N₂
C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN + N₂

Teacher's Note:
1. Sandmeyer uses cuprous salts; Gattermann reaction does the same job with copper powder and the halogen acid.

Question 8. How will you convert aniline into iodobenzene. (CBSE 2011)
Answer:
C₆H₅NH₂ (Aniline) \( \xrightarrow[\mathrm{273\text{-}278\ K\ (Diazotisation)}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N⁺≡NCl⁻ (Benzenediazonium chloride) \( \xrightarrow[\mathrm{\Delta}]{\mathrm{Aq.\ KI}} \) C₆H₅–I (Iodobenzene)

Teacher's Note:
1. Iodine is introduced simply with KI; no copper salt is needed, unlike Cl and Br.

Question 9. What is a coupling reaction ?
Answer:
The reaction of diazonium salts with phenols in basic medium (pH 9-10) or with 1° amines in acidic medium (pH 4-5) to give the corresponding azo dyes is called the coupling reaction. For example,
C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{\mathrm{OH^-}} \) p-HO–C₆H₄–N=N–C₆H₅ (p-Hydroxyazobenzene, orange dye) + Cl⁻ + H₂O
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ \( \xrightarrow{\mathrm{H^+}} \) p-H₂N–C₆H₄–N=N–C₆H₅ (p-Aminoazobenzene, yellow dye) + Cl⁻ + H₂O

Teacher's Note:
1. In coupling the N=N group is retained; this is the only common diazonium reaction in which nitrogen is not lost.

Question 10. Complete the following :
CH₃CH₂CN \( \xrightarrow{\mathrm{LiAlH_4}} \) A \( \xrightarrow{\mathrm{0^\circ C/HNO_2}} \) B
Answer:
CH₃CH₂CN (Propanenitrile) \( \xrightarrow{\mathrm{LiAlH_4}} \) CH₃CH₂CH₂NH₂ (Propanamine, A) \( \xrightarrow[\mathrm{HCl}]{\mathrm{0^\circ C/HNO_2}} \) [CH₃CH₂CH₂–N⁺≡NCl⁻] (n-Propyldiazonium chloride, Unstable) \( \xrightarrow[\mathrm{-N_2,\ -HCl}]{\mathrm{H_2O}} \) CH₃CH₂CH₂OH (1-Propanol, B)

Teacher's Note:
1. Aliphatic diazonium salts are unstable even at 0°C, so the 1° alkylamine ends up as an alcohol with evolution of N₂.

Question 11. Give an example of a Balz-Schiemann reaction.
Answer:
The thermal decomposition of diazonium fluoroborates to give fluoroarenes and BF₃ is called Balz-Schiemann reaction. For example,
C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{HBF_4}} \) C₆H₅N₂⁺BF₄⁻ (Benzenediazonium fluoroborate) \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅F (Fluorobenzene) + BF₃ + N₂

Teacher's Note:
1. Fluorine cannot be introduced with KF like iodine with KI; the fluoroborate salt route is used instead.

Question 12. How will you convert aniline into nitrobenzene without the intermediacy of benzene ? (CBSE 2011)
Answer:
C₆H₅NH₂ (Aniline) \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{HNO_2/HBF_4}} \) C₆H₅N₂⁺BF₄⁻ (Benzenediazonium tetrafluoroborate) \( \xrightarrow[\mathrm{\Delta}]{\mathrm{NaNO_2/Cu}} \) C₆H₅NO₂ (Nitrobenzene) + BF₃ + N₂

Teacher's Note:
1. Direct oxidation of NH₂ to NO₂ is not practical; replacing the diazo group by NO₂ with NaNO₂/Cu does the job.

Question 13. Give an example of Gomberg reaction.
Answer:
The reaction of benzenediazonium salts with benzene or bromobenzene or nitrobenzene in aqueous alkali to afford the corresponding biphenyl derivative is called Gomberg reaction. For example,
C₆H₅N₂⁺Cl⁻ + C₆H₆ \( \xrightarrow{\mathrm{NaOH}} \) C₆H₅–C₆H₅ (Biphenyl) + N₂ + NaCl + H₂O

Teacher's Note:
1. In Gomberg reaction the diazonium group is replaced by an aryl group, forming a new C–C bond between two rings.

Question 14. How is phenylhydrazine prepared from aniline ?
Answer:
By reduction of benzenediazonium chloride with SnCl₂/HCl or Na₂SO₃. For example,
C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{SnCl_2/HCl}} \) C₆H₅NHNH₂ (Phenylhydrazine)
C₆H₅N₂⁺Cl⁻ + 4[H] \( \xrightarrow{\mathrm{SnCl_2/HCl}} \) C₆H₅NHNH₂ + HCl

Teacher's Note:
1. Mild reducing agents (SnCl₂/HCl, Na₂SO₃) keep both N atoms (hydrazine); H₃PO₂ removes them completely (arene).

Short Answer Questions (2 Marks)

Question 15. Give equations for the reactions that occur when :
(i) Ethanamine reacts with acetyl chloride (Karnataka Board 2012)
(ii) Aniline is treated with CHCl₃ and ethanolic KOH solution. (CBSE 2022)
Answer:
(i) CH₃CH₂NH₂ + CH₃COCl \( \xrightarrow{\mathrm{Pyridine}} \) CH₃CH₂NHCOCH₃ (N-Ethylethanamide) + HCl
(ii) C₆H₅NH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅NC (Phenyl isocyanide) + 3KCl + 3H₂O

Teacher's Note:
1. Name the products in exam answers: N-ethylethanamide in (i) and phenyl isocyanide (offensive smell) in (ii).

Question 16. Electrophilic substitution reaction in case of aromatic amines takes place more readily as compared to benzene. (CBSE 2010 S)
Answer:
Due to +R-effect of the –NH₂ group, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, the electron density in the ring, particularly at o- and p-positions, increases as compared to benzene. Therefore, aromatic amines are activated and undergo electrophilic substitution more readily than benzene.

Teacher's Note:
1. This strong activation is why aniline reacts with bromine water at room temperature without any catalyst.

Question 17. Aniline undergoes electrophilic substitution reactions at ortho- and para-positions only.
Answer:
Aniline is a resonance hybrid of five structures. In three of these structures, the lone pair of N is delocalised into the ring and a negative charge appears at the two ortho and the para positions (H₂N⁺=C₆H₅⁻ with the charge at C-2, C-4 and C-6).
Thus, the electron density is higher at o- and p-positions than at m-position. Hence, electrophiles attack mainly the o- and p-positions.

Teacher's Note:
1. Note the exception: in strongly acidic medium aniline becomes the anilinium ion, which is m-directing.

Question 18. Electrophilic substitution in aniline takes place more readily than in benzene.
Answer:
In aniline, the –NH₂ group donates its lone pair of electrons to the benzene ring by resonance (+R-effect). This increases the electron density of the ring as compared to benzene. Since electrophiles are attracted by electron-rich centres, electrophilic substitution in aniline takes place more readily than in benzene.

Teacher's Note:
1. An activating group makes the ring more electron-rich than benzene; a deactivating group makes it less so.

Question 19. (a) Write complete chemical reactions for the conversion of aniline into sulphanilic acid.
(b) Mention two important uses of sulphanilic acid.
Answer:
(a) C₆H₅NH₂ + H₂SO₄ (conc.) → C₆H₅NH₃⁺HSO₄⁻ (Anilinium hydrogensulphate) \( \xrightarrow[\mathrm{-H_2O}]{\mathrm{453\text{-}473\ K}} \) p-H₂NC₆H₄SO₃H → p-H₃N⁺C₆H₄SO₃⁻ (Sulphanilic acid, zwitterion)
(b) Sulphanilic acid is used in the manufacture of (i) dyes (e.g., methyl orange) and (ii) sulpha drugs.

Teacher's Note:
1. Sulphanilic acid exists as a zwitterion, which explains its high melting point and insolubility in organic solvents.

Question 20. How will you separate a mixture of primary and secondary amines ? (Manipur Board 2011)
Answer:
The mixture is treated with benzenesulphonyl chloride (Hinsberg's reagent) and then with aqueous KOH.
RNH₂ + C₆H₅SO₂Cl → C₆H₅SO₂NHR + HCl ; C₆H₅SO₂NHR + KOH → C₆H₅SO₂N⁻RK⁺ (soluble) + H₂O
R₂NH + C₆H₅SO₂Cl → C₆H₅SO₂NR₂ (insoluble in KOH) + HCl
The insoluble sulphonamide of the 2° amine is filtered off. The filtrate is acidified to get the sulphonamide of the 1° amine. The two sulphonamides are separately hydrolysed with acid to recover the 1° and 2° amines.

Teacher's Note:
1. The N–H left in the 1° sulphonamide is acidic because of the strongly electron-withdrawing SO₂ group; that is why it dissolves in alkali.

Question 21. Explain why arenediazonium salts are much more stable than alkanediazonium salts. or Aromatic diazonium salts are more stable than aliphatic diazonium salts. Give reasons. (CBSE 2018)
Answer:
In arenediazonium salts, the positive charge on the diazo group is delocalised over the benzene ring by resonance (C₆H₅–N⁺≡N ↔ C₆H₅–N=N⁺ ↔ structures with the positive charge at o- and p-positions of the ring). Hence, they are stabilised and can be kept in solution at 273-278 K.
In alkanediazonium salts, no such resonance is possible. They readily lose N₂ (a very stable molecule) to form carbocations and therefore decompose even at low temperatures.

Teacher's Note:
1. This is why aliphatic 1° amines with HNO₂ give alcohols and N₂, while aromatic 1° amines give diazonium salts that can be used for further reactions.

Question 22. Why is it difficult to carry out the diazotisation of aniline in which electron withdrawing group is present on the ring at the para-position ?
Answer:
Diazotisation involves the attack of the lone pair of electrons of the N-atom of the amino group on the electrophile NO⁺. An electron-withdrawing group (e.g., NO₂) at the para-position decreases the electron density on the N-atom of the –NH₂ group by −I and −R effects. As a result, the amine becomes a weaker nucleophile and its attack on NO⁺ becomes difficult. Hence, diazotisation of such amines is difficult and needs more drastic conditions (e.g., stronger acid).

Teacher's Note:
1. Electron-withdrawing groups make an amine both less basic and less nucleophilic; the two properties go together.

Question 23. Suggest a reason for the use of excess of mineral acid in diazotisation reaction.
Answer:
In general, about three moles of mineral acid are used per mole of amine : (i) one mole to dissolve the amine as its salt, (ii) one mole to liberate HNO₂ from NaNO₂ and (iii) the excess to maintain the acidity of the reaction mixture so that the diazonium salt formed does not couple with the undiazotised (free) amine to form diazoamino compounds or azo dyes.

Teacher's Note:
1. If the medium is not acidic enough, C₆H₅N₂⁺ reacts with free aniline to give diazoaminobenzene, C₆H₅N=N–NHC₆H₅.

Question 24. Starting from benzenediazonium chloride, how will you prepare the following compounds :
(i) Iodobenzene (ii) Benzene (iii) Phenol ?
Answer:
(i) C₆H₅N₂⁺Cl⁻ + KI \( \xrightarrow{\mathrm{\Delta}} \) C₆H₅I + KCl + N₂
(ii) C₆H₅N₂⁺Cl⁻ + H₃PO₂ + H₂O \( \xrightarrow{\mathrm{Cu^+}} \) C₆H₆ + N₂ + H₃PO₃ + HCl
(iii) C₆H₅N₂⁺Cl⁻ + H₂O \( \xrightarrow{\mathrm{Warm}} \) C₆H₅OH + N₂ + HCl

Teacher's Note:
1. In all three reactions N₂ gas is evolved; the formation of very stable N₂ drives these replacement reactions.

Question 25. How can you get benzonitrile from aniline ?
Answer:
Aniline is diazotised and the benzenediazonium chloride so formed is treated with CuCN/KCN (Sandmeyer reaction) :
C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN (Benzonitrile) + N₂

Teacher's Note:
1. Direct substitution of NH₂ by CN is not possible; the diazonium salt is the key intermediate.

Question 26. Why does benzenediazonium chloride react with aniline in weakly acidic medium ?
Answer:
Coupling of benzenediazonium chloride with aniline is carried out in weakly acidic medium (pH 4-5). In strongly acidic medium, aniline is protonated to the anilinium ion (C₆H₅NH₃⁺) which does not couple, since the ring is deactivated. In alkaline medium, the diazonium ion is converted into diazohydroxide (C₆H₅N=N–OH) or diazotate ion which does not couple. In weakly acidic medium, sufficient concentrations of both the free amine and the diazonium ion are present and coupling occurs.
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ \( \xrightarrow{\mathrm{H^+\ (pH\ 4\text{-}5)}} \) p-H₂N–C₆H₄–N=N–C₆H₅ (p-Aminoazobenzene, yellow) + HCl

Teacher's Note:
1. Remember the pairs: amines couple at pH 4-5, phenols at pH 9-10.

Question 27. Give one reaction of a diazonium salt in which (i) diazo group is removed (ii) diazo group is retained.
Answer:
(i) Sandmeyer reaction : C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) C₆H₅Cl + N₂
(ii) Coupling reaction : C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{\mathrm{OH^-}} \) p-HO–C₆H₄–N=N–C₆H₅ (p-Hydroxyazobenzene, orange dye) + Cl⁻ + H₂O

Teacher's Note:
1. Removal reactions evolve N₂; in coupling reactions both N atoms stay in the product as the –N=N– (azo) group.

Question 28. How will you convert p-toluidine to 2-bromo-4-methylaniline ?
Answer:
p-CH₃C₆H₄NH₂ (p-Toluidine) \( \xrightarrow{\mathrm{(CH_3CO)_2O/pyridine}} \) p-CH₃C₆H₄NHCOCH₃ \( \xrightarrow{\mathrm{Br_2/CH_3COOH}} \) 2-Bromo-4-methylacetanilide \( \xrightarrow{\mathrm{H_2O/H^+\ or\ OH^-}} \) 2-Bromo-4-methylaniline
Acetylation reduces the activating effect of –NH₂ so that only one Br atom enters, at the position ortho to the NHCOCH₃ group (the para position being occupied by CH₃).

Teacher's Note:
1. Without protection, bromine water would brominate both ortho positions of p-toluidine.

Short Answer Questions (3 Marks)

Question 29. Write the method for the identification of primary, secondary and tertiary amines. Also write the equation of chemical reactions involved. (Uttarakhand Board 2012)
Answer:
Primary, secondary and tertiary amines can be identified by Hinsberg's test (or by their reaction with nitrous acid) :
(i) 1° amine : C₆H₅SO₂Cl + RNH₂ → C₆H₅SO₂NHR (soluble in KOH) + HCl. Also, a 1° amine gives the carbylamine test : RNH₂ + CHCl₃ + 3KOH → RNC (offensive smell) + 3KCl + 3H₂O. With HNO₂, a 1° aliphatic amine evolves N₂ gas : RNH₂ + HNO₂ → ROH + N₂ + H₂O.
(ii) 2° amine : C₆H₅SO₂Cl + R₂NH → C₆H₅SO₂NR₂ (insoluble in KOH) + HCl. With HNO₂, a 2° amine gives a yellow oily N-nitrosamine : R₂NH + HONO → R₂N–N=O + H₂O.
(iii) 3° amine : does not react with Hinsberg's reagent; with HNO₂ an aliphatic 3° amine merely forms a soluble salt : R₃N + HNO₂ → [R₃NH]⁺NO₂⁻.

Teacher's Note:
1. Learn one complete test (Hinsberg) for all three classes; the carbylamine test identifies only 1° amines.

Question 30. Give chemical tests to distinguish between the following compounds (one test in each case :
(i) Aniline and ethylamine (CBSE 2010, 2010 S ; Hr. Board 2011 ; CBSE (Foreign) 2017, 2019) (ii) Aniline and N-ethylaniline (iii) Methylamine and dimethylamine (Assam Board 2012 ; Hr. Board 2012, 2019) (iv) N-Methylpropan-2-amine and N-Ethyl-N-methylethanamine (v) Ethylamine and aniline (Hr. Board 2011 ; Assam Board 2013) (vi) CH₃CH₂NH₂ and CH₃NHCH₃ (CBSE 2007)
Answer:
(i) Azo dye test : Aniline is dissolved in dil. HCl, cooled to 273-278 K and treated with NaNO₂ solution; the diazonium salt formed gives an orange-red dye with an alkaline solution of β-naphthol. Ethylamine under these conditions gives effervescence of N₂ but no dye.
C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{\beta\text{-}naphthol/NaOH}} \) orange-red azo dye
(ii) Carbylamine test : Aniline (1°) on heating with CHCl₃ and alc. KOH gives phenyl isocyanide with an offensive smell; N-ethylaniline (2°) does not.
(iii) Carbylamine test : Methylamine (1°) gives methyl isocyanide with an offensive smell : CH₃NH₂ + CHCl₃ + 3KOH → CH₃NC + 3KCl + 3H₂O ; dimethylamine (2°) does not.
(iv) Hinsberg's test : N-Methylpropan-2-amine (2°) reacts with C₆H₅SO₂Cl to give a sulphonamide insoluble in KOH; N-ethyl-N-methylethanamine (3°) does not react.
(CH₃)₂CHNHCH₃ + C₆H₅SO₂Cl → C₆H₅SO₂N(CH₃)CH(CH₃)₂ (insoluble in alkali) + HCl
(v) Bromine water test : Aniline gives a white precipitate of 2, 4, 6-tribromoaniline with bromine water while ethylamine does not.
(vi) Carbylamine test : CH₃CH₂NH₂ (1°) gives ethyl isocyanide with an offensive smell on heating with CHCl₃ and alc. KOH; CH₃NHCH₃ (2°) does not.

Teacher's Note:
1. Pick the test that only one compound passes: carbylamine for 1° vs 2°/3°, Hinsberg for 2° vs 3°, azo dye or bromine water for aromatic vs aliphatic.

Question 31. What is diazotisation ? Discuss its mechanism. (Jharkhand Board 2012)
Answer:
The conversion of an aromatic primary amine into a diazonium salt by treatment with nitrous acid (NaNO₂ + HCl) at 273-278 K is called diazotisation.
C₆H₅NH₂ + NaNO₂ + 2HCl \( \xrightarrow{\mathrm{273\text{-}278\ K}} \) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
Mechanism : Step 1. Formation of nitrosonium ion : NaNO₂ + HCl → HO–N=O + NaCl ; HO–N=O + H⁺ ⇌ H₂O⁺–N=O → N⁺=O (nitrosonium ion) + H₂O
Step 2. Attack of the amine on NO⁺ and loss of a proton gives N-nitrosamine : C₆H₅NH₂ + NO⁺ → C₆H₅N⁺H₂–N=O \( \xrightarrow{\mathrm{-H^+}} \) C₆H₅NH–N=O
Step 3. Tautomerisation to diazohydroxide : C₆H₅NH–N=O → C₆H₅N=N–OH
Step 4. Protonation and loss of water : C₆H₅N=N–OH + H⁺ → C₆H₅N=N–O⁺H₂ \( \xrightarrow{\mathrm{-H_2O}} \) C₆H₅–N⁺≡N

Teacher's Note:
1. The electrophile in diazotisation is NO⁺, formed from nitrous acid in acid medium; the amine nitrogen acts as the nucleophile.

Question 32. Give a brief account of Sandmeyer reaction. In what respect, does it differ from Gattermann reaction ?
Answer:
Sandmeyer reaction : When a diazonium salt is treated with cuprous chloride in HCl, cuprous bromide in HBr or cuprous cyanide in KCN, the diazo group is replaced by –Cl, –Br or –CN respectively.
C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) C₆H₅Cl + N₂ ; C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuBr/HBr}} \) C₆H₅Br + N₂ ; C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN + N₂
Gattermann reaction : Chlorine or bromine is introduced by treating the diazonium salt with the corresponding halogen acid in presence of copper powder.
C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{Cu/HCl}} \) C₆H₅Cl + N₂ + CuCl ; C₆H₅N₂⁺Br⁻ \( \xrightarrow{\mathrm{Cu/HBr}} \) C₆H₅Br + N₂ + CuBr
Difference : Sandmeyer reaction uses cuprous salts (CuCl, CuBr, CuCN) while Gattermann reaction uses copper powder with the halogen acid. The yield in Sandmeyer reaction is better than in Gattermann reaction.

Teacher's Note:
1. A quick way to remember: Sandmeyer = cuprous salt; Gattermann = copper powder.

Question 33. Discuss the importance of pH in coupling reactions.
Answer:
(i) Coupling with phenols is carried out in mildly alkaline medium (pH 9-10). In this medium phenol is converted into phenoxide ion which is more strongly activating than phenol and couples readily. If the medium is too alkaline (pH > 10), the diazonium ion is converted into diazohydroxide/diazotate ion which does not couple.
C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{\mathrm{OH^-\ (pH\ 9\text{-}10)}} \) p-HOC₆H₄N=NC₆H₅ + Cl⁻ + H₂O
(ii) Coupling with amines is carried out in weakly acidic medium (pH 4-5). In strongly acidic medium, the amine is protonated to the anilinium ion which does not couple; in alkaline medium the diazonium ion is converted into diazohydroxide. At pH 4-5 enough free amine and diazonium ion are present.
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ \( \xrightarrow{\mathrm{H^+\ (pH\ 4\text{-}5)}} \) p-H₂NC₆H₄N=NC₆H₅ + HCl
Thus, the pH must be controlled so that both the electrophile (ArN₂⁺) and the reactive form of the coupling component are present together.

Teacher's Note:
1. The diazonium ion is a weak electrophile, so it needs a strongly activated ring (phenoxide or free amine) to react.

Question 34. How will you prepare
(i) o-Chlorobenzoic acid from o-aminobenzoic acid. (ii) Benzylamine from benzenediazonium chloride. (iii) m-Bromotoluene from toluene?
Answer:
(i) o-H₂NC₆H₄COOH \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) o-HOOCC₆H₄N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) o-ClC₆H₄COOH + N₂
(ii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN \( \xrightarrow{\mathrm{LiAlH_4\ or\ Na/C_2H_5OH}} \) C₆H₅CH₂NH₂ (Benzylamine)
(iii) C₆H₅CH₃ \( \xrightarrow{\mathrm{Conc.\ HNO_3/H_2SO_4}} \) p-CH₃C₆H₄NO₂ (separated from o-isomer) \( \xrightarrow{\mathrm{Sn/HCl}} \) p-CH₃C₆H₄NH₂ \( \xrightarrow{\mathrm{(CH_3CO)_2O}} \) p-CH₃C₆H₄NHCOCH₃ \( \xrightarrow{\mathrm{Br_2/CH_3COOH}} \) 2-Bromo-4-methylacetanilide \( \xrightarrow{\mathrm{H_2O/H^+}} \) 2-Bromo-4-methylaniline \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) \( \xrightarrow{\mathrm{H_3PO_2}} \) m-Bromotoluene (3-Bromotoluene)

Teacher's Note:
1. In (iii) the NH₂ group is put in only to direct Br to the position meta to CH₃, and is then removed through the diazonium salt.

Question 35. Identify A and B in the following reaction sequences :
(i) CH₃CH₂CN \( \xrightarrow[\mathrm{Partial\ hydrolysis}]{\mathrm{OH^-}} \) A \( \xrightarrow{\mathrm{NaOH + Br_2}} \) B (CBSE Sample Paper 2019-20) (Pb. Board 2012)
(ii) C₆H₅N₂Cl \( \xrightarrow[\mathrm{\Delta}]{\mathrm{+H_2O}} \) A \( \xrightarrow[\mathrm{water}]{\mathrm{Br_2}} \) B (Raj. Board 2012)
(iii) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2,\ HCl}} \) A \( \xrightarrow{\mathrm{CuBr/HBr}} \) B (Manipur Board 2012)
(iv) C₆H₅NH₂ \( \xrightarrow{\mathrm{NaNO_2/HCl/273\ K}} \) A \( \xrightarrow{\mathrm{H_2O/H^+}} \) B (CBSE 2018 S)
Answer:
(i) A = CH₃CH₂CONH₂ (Propanamide) ; B = CH₃CH₂NH₂ (Ethanamine)
(ii) A = C₆H₅OH (Phenol) ; B = 2, 4, 6-Tribromophenol (white ppt.)
(iii) A = C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride) ; B = C₆H₅Br (Bromobenzene)
(iv) A = C₆H₅N₂⁺Cl⁻ (Benzenediazonium chloride) ; B = C₆H₅OH (Phenol)

Teacher's Note:
1. In (i) partial hydrolysis stops at the amide, which then loses one carbon in the Hofmann reaction.

Question 36. Complete the following reactions.
(i) C₆H₅N₂Cl + C₆H₅NH₂ \( \xrightarrow[\mathrm{pH\ 4\text{-}5}]{\mathrm{Ice\ cold}} \) (CBSE 2010 ; HP Board 2011)
(ii) C₆H₅N₂Cl + H₂O \( \xrightarrow{\mathrm{warm}} \) (HP Board 2011 ; CBSE 2013)
(iii) C₆H₅N₂Cl \( \xrightarrow[\mathrm{HCl}]{\mathrm{Cu_2Cl_2}} \) (HP Board 2011, 2013 ; CBSE (Foreign) 2017)
(iv) C₆H₅N₂Cl + H₃PO₂ + H₂O → (CBSE 2009, 2011, 2012, 2013)
(v) C₆H₅N₂Cl + CH₃CH₂OH → (CBSE 2010 ; CBSE (Foreign) 2017, 2018)
(vi) C₆H₅N₂Cl \( \xrightarrow{\mathrm{Cu/HCl}} \) (CBSE (Foreign) 2017)
(vii) C₆H₅N₂Cl \( \xrightarrow{\mathrm{CuCN}} \) (CBSE (Foreign) 2017)
Answer:
(i) C₆H₅N₂Cl + C₆H₅NH₂ \( \xrightarrow{\mathrm{pH\ 4\text{-}5}} \) p-H₂NC₆H₄N=NC₆H₅ (p-Aminoazobenzene) + HCl
(ii) C₆H₅N₂Cl + H₂O \( \xrightarrow{\mathrm{Warm}} \) C₆H₅OH (Phenol) + N₂ + HCl
(iii) C₆H₅N₂Cl \( \xrightarrow{\mathrm{Cu_2Cl_2/HCl}} \) C₆H₅Cl (Chlorobenzene) + N₂
(iv) C₆H₅N₂Cl + H₃PO₂ + H₂O → C₆H₆ (Benzene) + N₂ + H₃PO₃ + HCl
(v) C₆H₅N₂Cl + CH₃CH₂OH → C₆H₆ (Benzene) + N₂ + CH₃CHO + HCl
(vi) C₆H₅N₂Cl \( \xrightarrow{\mathrm{Cu/HCl}} \) C₆H₅Cl (Chlorobenzene) + N₂ (Gattermann reaction)
(vii) C₆H₅N₂Cl \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN (Benzonitrile) + N₂ (Sandmeyer reaction)

Teacher's Note:
1. In (v) ethanol acts as a reducing agent and is itself oxidised to ethanal.

Question 37. Write giving chemical equations for each of the following reactions :
(i) Hofmann bromamide reaction. (CBSE 2008, 2009, 2012, 2016, 2018) (ii) Carbylamine reaction (HP Board 2011 ; CBSE 2010 S, 2012, 2016) (iii) Gabriel phthalimide reaction (CBSE 2008, 2018) (iv) Diazotisation reaction (Hr. Board 2011, 2013 ; HP Board 2011 ; Jharkhand Board 2012 ; CBSE 2018) (v) Coupling reaction (CBSE 2009 ; Manipur Board 2011 ; Hr. Board 2011, 2013) (vi) Mendius reaction (vii) Gattermann reaction (CBSE 2009 ; HP Board 2011) (viii) Sandmeyer reaction (CBSE 2008, 2009)
Answer:
(i) RCONH₂ + Br₂ + 4KOH \( \xrightarrow{\mathrm{\Delta}} \) RNH₂ + K₂CO₃ + 2KBr + 2H₂O ; e.g., CH₃CONH₂ → CH₃NH₂
(ii) RNH₂ + CHCl₃ + 3KOH (alc.) \( \xrightarrow{\mathrm{\Delta}} \) RNC + 3KCl + 3H₂O ; e.g., C₆H₅NH₂ → C₆H₅NC (offensive smell)
(iii) Phthalimide \( \xrightarrow{\mathrm{KOH\ (ethanolic)}} \) Potassium phthalimide \( \xrightarrow{\mathrm{R\text{-}X}} \) N-Alkylphthalimide \( \xrightarrow[\mathrm{hydrolysis}]{\mathrm{NaOH\ (aq)}} \) R–NH₂ + Sodium phthalate
C₆H₄(CO)₂NH + KOH → C₆H₄(CO)₂N⁻K⁺ + H₂O ; C₆H₄(CO)₂N⁻K⁺ + RX → C₆H₄(CO)₂NR + KX ; C₆H₄(CO)₂NR + 2NaOH → C₆H₄(COONa)₂ + RNH₂
(iv) C₆H₅NH₂ + NaNO₂ + 2HCl \( \xrightarrow{\mathrm{273\text{-}278\ K}} \) C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O
(v) C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{\mathrm{OH^-}} \) p-HOC₆H₄N=NC₆H₅ + Cl⁻ + H₂O ; C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ \( \xrightarrow{\mathrm{H^+}} \) p-H₂NC₆H₄N=NC₆H₅ + HCl
(vi) R–C≡N + 4[H] \( \xrightarrow{\mathrm{Na/C_2H_5OH}} \) R–CH₂NH₂ ; e.g., CH₃CN → CH₃CH₂NH₂
(vii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{Cu/HCl}} \) C₆H₅Cl + N₂ + CuCl
(viii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) C₆H₅Cl + N₂ ; C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuBr/HBr}} \) C₆H₅Br + N₂ ; C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN + N₂

Teacher's Note:
1. Gabriel synthesis gives only 1° aliphatic amines; aryl halides do not react with potassium phthalimide, so aniline cannot be made this way.

Question 38. How will you convert ?
(i) Aniline to nitrobenzene (CBSE 2011, 2012) (ii) Aniline to iodobenzene (CBSE 2011) (iii) Nitrobenzene to benzenediazonium chloride (HP Board 2011) (iv) Aniline to benzyl alcohol (CBSE 2012 ; West Bengal Board 2012) (v) Aniline to chlorobenzene (Hr. Board 2012) (vi) Aniline to benzoic acid (Tamil Nadu Board 2012) (vii) Aniline to phenol (CBSE 2015)
Answer:
(i) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HBF_4}} \) C₆H₅N₂⁺BF₄⁻ \( \xrightarrow[\mathrm{\Delta}]{\mathrm{NaNO_2/Cu}} \) C₆H₅NO₂ + BF₃ + N₂
(ii) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{KI}} \) C₆H₅I + KCl + N₂
(iii) C₆H₅NO₂ \( \xrightarrow{\mathrm{Sn/HCl}} \) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻
(iv) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN \( \xrightarrow{\mathrm{LiAlH_4}} \) C₆H₅CH₂NH₂ \( \xrightarrow{\mathrm{HNO_2}} \) C₆H₅CH₂OH + N₂ + H₂O
(v) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) C₆H₅Cl + N₂
(vi) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCN/KCN}} \) C₆H₅CN \( \xrightarrow{\mathrm{H_3O^+}} \) C₆H₅COOH
(vii) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{H_2O,\ warm}} \) C₆H₅OH + N₂ + HCl

Teacher's Note:
1. Benzylamine is an aralkyl amine (NH₂ on CH₂), so like aliphatic amines it gives an alcohol with HNO₂ in (iv).

Question 39. Write the structure of methyl orange. Give its synthesis.
Answer:
Structure : (CH₃)₂N–C₆H₄–N=N–C₆H₄–SO₃⁻Na⁺ (both groups at para-positions), i.e., sodium salt of p-dimethylaminoazobenzene-p'-sulphonic acid.
Synthesis : Sulphanilic acid is diazotised and the diazonium salt is coupled with N, N-dimethylaniline in weakly acidic medium; the red acid form (helianthin) so obtained is converted into its sodium salt.
p-HO₃SC₆H₄NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) p-HO₃SC₆H₄N₂⁺Cl⁻ \( \xrightarrow[\mathrm{H^+}]{\mathrm{C_6H_5N(CH_3)_2}} \) p-HO₃SC₆H₄N=NC₆H₄N(CH₃)₂-p (Helianthin, red) \( \xrightarrow{\mathrm{NaOH}} \) Methyl orange (orange-yellow)

Teacher's Note:
1. Methyl orange is an acid-base indicator: red in acid, yellow in alkali.

Question 40. How will you convert nitrobenzene into (i) p-Hydroxyazobenzene (ii) Iodobenzene (iii) Bromobenzene (iv) Chlorobenzene (v) Phenol (vi) m-Bromochlorobenzene (vii) m-Nitrophenol (viii) m-Bromophenol (ix) m-Nitroaniline ?
Answer:
First, nitrobenzene is converted into benzenediazonium chloride : C₆H₅NO₂ \( \xrightarrow{\mathrm{Sn/HCl}} \) C₆H₅NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) C₆H₅N₂⁺Cl⁻
(i) C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{\mathrm{OH^-}} \) p-HOC₆H₄N=NC₆H₅
(ii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{KI}} \) C₆H₅I + KCl + N₂
(iii) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuBr/HBr}} \) C₆H₅Br + N₂
(iv) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) C₆H₅Cl + N₂
(v) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\mathrm{H_2O,\ warm}} \) C₆H₅OH + N₂ + HCl
(vi) C₆H₅NO₂ \( \xrightarrow{\mathrm{Br_2/FeBr_3}} \) m-BrC₆H₄NO₂ \( \xrightarrow{\mathrm{Sn/HCl}} \) m-BrC₆H₄NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) m-BrC₆H₄N₂⁺Cl⁻ \( \xrightarrow{\mathrm{CuCl/HCl}} \) m-BrC₆H₄Cl
(vii) C₆H₅NO₂ \( \xrightarrow[\mathrm{363\ K}]{\mathrm{Conc.\ HNO_3/H_2SO_4}} \) m-C₆H₄(NO₂)₂ \( \xrightarrow{\mathrm{(NH_4)_2S}} \) m-O₂NC₆H₄NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) m-O₂NC₆H₄N₂⁺Cl⁻ \( \xrightarrow{\mathrm{H_2O,\ warm}} \) m-O₂NC₆H₄OH
(viii) C₆H₅NO₂ \( \xrightarrow{\mathrm{Br_2/FeBr_3}} \) m-BrC₆H₄NO₂ \( \xrightarrow{\mathrm{Sn/HCl}} \) m-BrC₆H₄NH₂ \( \xrightarrow[\mathrm{273\text{-}278\ K}]{\mathrm{NaNO_2/HCl}} \) m-BrC₆H₄N₂⁺Cl⁻ \( \xrightarrow{\mathrm{H_2O,\ warm}} \) m-BrC₆H₄OH
(ix) C₆H₅NO₂ \( \xrightarrow[\mathrm{363\ K}]{\mathrm{Conc.\ HNO_3/H_2SO_4}} \) m-C₆H₄(NO₂)₂ \( \xrightarrow[\mathrm{(selective\ reduction)}]{\mathrm{(NH_4)_2S}} \) m-O₂NC₆H₄NH₂

Teacher's Note:
1. For all meta products, carry out the substitution while the NO₂ group (a m-director) is still present, and only then reduce and diazotise.

Long Answer Questions (5 Marks)

Question 41. Complete the following reactions:
(i) C₆H₅NH₂ + H₂SO₄ (conc.) →
(ii) C₆H₅N₂Cl + C₂H₅OH →
(iii) C₆H₅NH₂ + (CH₃CO)₂O →
(iv) C₆H₅N₂Cl + H₃PO₂ + H₂O →
(v) C₆H₅NH₂ + CHCl₃ + 3KOH (alc.) →
Answer:
(i) C₆H₅NH₂ + H₂SO₄ → [C₆H₅NH₃]⁺HSO₄⁻ (anilinium hydrogensulphate) Δ
⟶ p-H₂NC₆H₄SO₃H (sulphanilic acid)
(ii) C₆H₆ + N₂ + HCl + CH₃CHO
(iii) C₆H₅NHCOCH₃ + CH₃COOH
(iv) C₆H₆ + H₃PO₃ + HCl + N₂
(v) C₆H₅NC + 3KCl + 3H₂O

Teacher's Note:
1. In (ii) and (iv) the diazonium group is replaced by H; ethanol is oxidised to ethanal and H₃PO₂ to H₃PO₃.

Question 42. Write A, B and C in the given reactions:
(i) C₆H₅N₂Cl CuCN
⟶ A H₂O/H⁺
⟶ B NH₃, Δ
⟶ C
(ii) CH₃CH₂Br KCN
⟶ A LiAlH₄
⟶ B HNO₂
⟶
0°C C
(iii) C₆H₅NO₂ Fe/HCl
⟶ A HNO₂
⟶
273 K B H₂O/H⁺
⟶
Δ C
(iv) CH₃COOH NH₃
⟶
Δ A NaOBr
⟶ B NaNO₂/HCl
⟶ C
(v) CH₃CH₂I NaCN
⟶ A OH⁻
⟶
partial hydrolysis B NaOH/Br₂
⟶ C
Answer:
(i) A = C₆H₅CN, B = C₆H₅COOH, C = C₆H₅CONH₂
(ii) A = CH₃CH₂CN, B = CH₃CH₂CH₂NH₂, C = CH₃CH₂CH₂OH
(iii) A = C₆H₅NH₂, B = C₆H₅N₂Cl, C = C₆H₅OH
(iv) A = CH₃CONH₂, B = CH₃NH₂, C = CH₃OH
(v) A = CH₃CH₂CN, B = CH₃CH₂CONH₂, C = CH₃CH₂NH₂

Teacher's Note:
1. Count carbon atoms at each step: KCN/NaCN adds one carbon, Hoffmann degradation removes one.
2. LiAlH₄ reduces R–CN to R–CH₂NH₂, so propanenitrile gives propan-1-amine, not ethylamine.

Question 43. Accomplish the following conversions:
(i) C₆H₅NO₂ → C₆H₅–COOH
(ii) Benzene → m-bromophenol
(iii) C₆H₅COOH → C₆H₅NH₂
(iv) Aniline → 2,4,6-tribromoaniline
(v) Benzyl chloride → 2-phenylethanamine
Answer:
(i) C₆H₅NO₂ (nitrobenzene) Sn/HCl
⟶
6[H] C₆H₅NH₂ NaNO₂/HCl
⟶
273 K C₆H₅N₂Cl KCN
⟶ C₆H₅CN H₂O
⟶ C₆H₅COOH (benzoic acid)
(ii) C₆H₆ Conc. HNO₃
⟶
Conc. H₂SO₄ C₆H₅NO₂ Br₂
⟶
Fe m-BrC₆H₄NO₂ Sn + HCl
⟶ m-BrC₆H₄NH₂ NaNO₂ + HCl
⟶
0–5°C m-BrC₆H₄N₂⁺Cl⁻ H⁺/H₂O
⟶ m-BrC₆H₄OH (m-bromophenol)
(iii) C₆H₅COOH (benzoic acid) NH₄OH
⟶ C₆H₅COONH₄ Δ
⟶
−H₂O C₆H₅CONH₂ Br₂
⟶
KOH C₆H₅NH₂ (aniline)
(iv) C₆H₅NH₂ (aniline) Br₂-water
⟶ 2,4,6-tribromoaniline
(v) C₆H₅CH₂Cl (benzyl chloride) KCN
⟶ C₆H₅CH₂CN LiAlH₄
⟶
4[H] C₆H₅CH₂CH₂NH₂ (2-phenylethanamine)

Teacher's Note:
1. In (ii) the order of steps matters: nitrate first so that the meta-directing NO₂ group sends Br to the meta position.
2. Write the diazotisation temperature (273–278 K) every time – examiners look for it.

Question 44. How is benzenediazonium chloride prepared in solution ? Give an account of its coupling reactions with phenols and amines.
Answer:
Preparation : Benzenediazonium chloride is prepared by diazotisation, i.e., by treating a solution of aniline in excess of dil. HCl with an aqueous solution of NaNO₂ at 273-278 K. Nitrous acid formed in situ reacts with aniline :
NaNO₂ + HCl → HNO₂ + NaCl ; C₆H₅NH₂ + HNO₂ + HCl \( \xrightarrow{\mathrm{273\text{-}278\ K}} \) C₆H₅N₂⁺Cl⁻ + 2H₂O
Being unstable, it is not isolated and its solution is used immediately.
Coupling reactions : Diazonium ion acts as a weak electrophile and attacks the electron-rich para-position of phenols and aromatic amines to form brightly coloured azo compounds (the –N=N– group is retained).
(i) With phenol in mildly alkaline medium (pH 9-10) :
C₆H₅N₂⁺Cl⁻ + C₆H₅OH \( \xrightarrow{\mathrm{OH^-}} \) p-HOC₆H₄N=NC₆H₅ (p-Hydroxyazobenzene, orange dye) + Cl⁻ + H₂O
With β-naphthol an orange-red dye is formed; this is used as a test for aromatic 1° amines.
(ii) With aniline in weakly acidic medium (pH 4-5) :
C₆H₅N₂⁺Cl⁻ + C₆H₅NH₂ \( \xrightarrow{\mathrm{H^+}} \) p-H₂NC₆H₄N=NC₆H₅ (p-Aminoazobenzene, yellow dye) + HCl
With N, N-dimethylaniline, p-(dimethylamino)azobenzene is formed. In alkaline medium phenol forms the more reactive phenoxide ion, whereas amines couple best at pH 4-5 where the free amine is not protonated. If the para-position is occupied, coupling occurs at the ortho-position.

Teacher's Note:
1. The extended conjugation through –N=N– between two rings is what makes azo compounds coloured; they are used as dyes and indicators.