Questions & explanations
1. Arrange the following in decreasing order of basicity: aniline, p-toluidine, p-nitroaniline, m-nitroaniline.
- p-toluidine > aniline > m-nitroaniline > p-nitroaniline
- aniline > p-toluidine > m-nitroaniline > p-nitroaniline
- p-toluidine > aniline > p-nitroaniline > m-nitroaniline
- p-nitroaniline > m-nitroaniline > aniline > p-toluidine
Answer: p-toluidine > aniline > m-nitroaniline > p-nitroaniline
p-Toluidine has an electron-donating -CH3 group, which increases electron density on nitrogen, making it more basic than aniline. p-Nitroaniline has a strong electron-withdrawing -NO2 group at para position, which greatly reduces basicity via resonance. m-Nitroaniline also withdraws electrons but only by inductive effect, so it is less basic than aniline but more basic than p-nitroaniline. Thus the order is p-toluidine > aniline > m-nitroaniline > p-nitroaniline.
2. Arrange in decreasing order of basicity: CH3NH2, (CH3)2NH, C6H5NH2, p-CH3C6H4NH2, p-O2NC6H4NH2.
- CH3NH2 > (CH3)2NH > C6H5NH2 > p-CH3C6H4NH2 > p-O2NC6H4NH2
- (CH3)2NH > CH3NH2 > p-CH3C6H4NH2 > C6H5NH2 > p-O2NC6H4NH2
- (CH3)2NH > CH3NH2 > C6H5NH2 > p-CH3C6H4NH2 > p-O2NC6H4NH2
- p-CH3C6H4NH2 > C6H5NH2 > (CH3)2NH > CH3NH2 > p-O2NC6H4NH2
Answer: (CH3)2NH > CH3NH2 > p-CH3C6H4NH2 > C6H5NH2 > p-O2NC6H4NH2
Aliphatic amines are stronger bases than aromatic amines due to resonance delocalisation of lone pair in aniline. Among aliphatic, aqueous basicity order is 2° > 1° ≈ 3° > NH3, so (CH3)2NH > CH3NH2. Among aromatic, electron-donating groups (CH3) increase basicity, electron-withdrawing (NO2) decrease, so p-toluidine > aniline > p-nitroaniline. Thus correct order: (CH3)2NH > CH3NH2 > p-toluidine > aniline > p-nitroaniline.
3. Three test tubes contain ethylamine, diethylamine, and triethylamine. Which test correctly identifies them?
- Carbylamine test: A gives foul smell, B and C do not; then Hinsberg test on B gives insoluble product.
- Hinsberg test: A no reaction, B dissolves in KOH, C insoluble; then carbylamine test on B gives foul smell.
- Hinsberg test: A dissolves in KOH, B insoluble, C no reaction; then carbylamine test on A gives foul smell.
- Hinsberg test: A insoluble, B dissolves in KOH, C no reaction; then carbylamine test on A gives foul smell.
Answer: Hinsberg test: A dissolves in KOH, B insoluble, C no reaction; then carbylamine test on A gives foul smell.
In Hinsberg test, primary amine (ethylamine) forms sulphonamide soluble in KOH, secondary (diethylamine) forms insoluble sulphonamide, tertiary (triethylamine) does not react. Then carbylamine test confirms primary amine: only ethylamine gives foul isocyanide odour. Thus sequence: A (soluble in KOH) is primary, B (insoluble) is secondary, C (no reaction) is tertiary. Carbylamine on A confirms.
4. Aniline is a much weaker base than methylamine because:
- aniline has a higher molecular weight than methylamine, reducing its basicity
- the benzene ring withdraws electrons by inductive effect, making nitrogen less basic
- the conjugate acid of aniline is stabilised by resonance, making protonation less favourable
- the lone pair on nitrogen in aniline is delocalised into the benzene ring, reducing electron density on nitrogen
Answer: the lone pair on nitrogen in aniline is delocalised into the benzene ring, reducing electron density on nitrogen
In aniline, the lone pair on nitrogen is delocalised into the benzene ring via resonance, which reduces electron density on nitrogen, making it less available for protonation. The conjugate acid, anilinium ion, does not have this resonance stabilisation, so protonation is less favourable. This explains why aniline (pKb ~9.4) is much weaker than methylamine (pKb ~3.4).
5. Aniline reacts with bromine water to give 2,4,6-tribromoaniline. What is the reason for this high reactivity?
- The -NH2 group withdraws electrons by resonance, deactivating the ring.
- The -NH2 group donates electrons by resonance, activating the ring.
- Bromine water contains FeBr3 as a catalyst.
- Aniline has a lone pair on nitrogen that is not available for donation.
Answer: The -NH2 group donates electrons by resonance, activating the ring.
The -NH2 group in aniline donates its lone pair into the benzene ring by resonance (+M effect), increasing electron density at ortho and para positions. This makes the ring highly activated, so bromine water (without any catalyst) can electrophilically substitute all three available positions to give 2,4,6-tribromoaniline.
6. Which route is best to prepare propan-1-amine from propan-1-ol?
- Propan-1-ol → propyl chloride → reduction with H2/Ni → propan-1-amine
- Propan-1-ol → propyl bromide → Gabriel phthalimide synthesis → propan-1-amine
- Propan-1-ol → propanal → reduction with LiAlH4 → propan-1-amine
- Propan-1-ol → propyl bromide → ammonolysis with NH3 → propan-1-amine
Answer: Propan-1-ol → propyl bromide → Gabriel phthalimide synthesis → propan-1-amine
Gabriel phthalimide synthesis yields pure primary amine without over-alkylation. Propan-1-ol is converted to propyl bromide using PBr3, then reacted with potassium phthalimide followed by hydrolysis to give propan-1-amine. This method avoids the mixture of secondary and tertiary amines formed in ammonolysis.
7. Which sequence correctly converts aniline to p-iodonitrobenzene?
- Acetylate, nitrate, hydrolyse, diazotise, treat with KI
- Nitrate, acetylate, hydrolyse, diazotise, treat with KI
- Acetylate, nitrate, diazotise, hydrolyse, treat with KI
- Nitrate, diazotise, acetylate, hydrolyse, treat with KI
Answer: Acetylate, nitrate, hydrolyse, diazotise, treat with KI
Aniline's -NH2 is strongly activating; direct nitration gives unwanted products. Acetylation protects -NH2, then nitration gives p-nitroacetanilide. Hydrolysis regenerates -NH2, diazotisation at 0-5°C gives diazonium salt, and KI replaces -N2+ with iodine. This sequence yields p-iodonitrobenzene.
8. Nitration of aniline with conc. HNO3/H2SO4 gives a mixture of meta- and para-nitroaniline. Why is meta product formed?
- The nitronium ion attacks the meta position due to steric hindrance.
- The -NH2 group is meta-directing in all conditions.
- Conc. H2SO4 oxidises -NH2 to -NO2, which is meta-directing.
- In strong acid, -NH2 is protonated to -NH3+, which is meta-directing.
Answer: In strong acid, -NH2 is protonated to -NH3+, which is meta-directing.
In the strongly acidic medium, aniline is protonated to form the anilinium ion (-NH3+). This group has no lone pair for resonance donation and is strongly deactivating and meta-directing. Therefore, the electrophile NO2+ attacks the meta position of the anilinium ion, giving meta-nitroaniline.
9. A student prepares benzenediazonium chloride at 0°C and stores it at 5°C. After 2 hours, the yield of azo dye formed upon coupling with phenol is low. What is the most likely reason?
- The diazonium salt decomposed to phenol and N2 at 5°C.
- The diazonium salt reacted with unreacted aniline to form a diazoamino compound.
- The diazonium salt was converted to the diazoate due to alkaline pH.
- The diazonium salt was destroyed by excess nitrous acid.
Answer: The diazonium salt reacted with unreacted aniline to form a diazoamino compound.
At 5°C, the diazonium salt is stable against decomposition to phenol, but any unreacted aniline present couples with the diazonium ion to form a diazoamino compound, reducing the amount available for azo dye formation. This side reaction is favored at slightly elevated temperatures.
10. What is the correct order of steps to synthesise methyl orange from benzene?
- Reduce, nitrate, sulphonate, diazotise, couple with N,N-dimethylaniline
- Nitrate, sulphonate, reduce, diazotise, couple with N,N-dimethylaniline
- Nitrate, reduce, sulphonate, diazotise, couple with N,N-dimethylaniline
- Nitrate, reduce, diazotise, sulphonate, couple with N,N-dimethylaniline
Answer: Nitrate, sulphonate, reduce, diazotise, couple with N,N-dimethylaniline
Methyl orange synthesis: benzene is nitrated to nitrobenzene, then sulphonated to m-nitrobenzenesulphonic acid. Reduction gives metanilic acid (m-aminobenzenesulphonic acid), which is diazotised and coupled with N,N-dimethylaniline. This order yields the correct azo dye.
11. Which compound is used for diazotisation in the synthesis of methyl orange?
- Benzenesulphonic acid
- Aniline
- N,N-Dimethylaniline
- Sulphanilic acid
Answer: Sulphanilic acid
In methyl orange synthesis, sulphanilic acid (4-aminobenzenesulphonic acid) is diazotised with NaNO2/HCl at 0-5°C to form the diazonium salt, which then couples with N,N-dimethylaniline. The -SO3H group prevents self-coupling and directs coupling to the para position.
12. Why does ammonolysis of alkyl halides give a mixture of amines?
- The reaction requires a catalyst that promotes side reactions
- The reaction follows SN1 mechanism
- Alkyl halides undergo elimination preferentially
- The product amines are more nucleophilic than the starting amine
Answer: The product amines are more nucleophilic than the starting amine
In ammonolysis, the primary amine formed is more nucleophilic than ammonia due to the +I effect of the alkyl group. It reacts further with alkyl halide to give secondary and tertiary amines, and finally quaternary salt. This stepwise alkylation leads to a mixture.