Questions & explanations
1. Aniline undergoes electrophilic substitution reactions. Which statement is correct?
- Bromination of aniline in bromine water yields 2,4,6‑tribromoaniline.
- Nitration of aniline with HNO3/H2SO4 gives p‑nitroaniline as major product.
- Sulphonation of aniline at room temperature yields m‑sulphanilic acid.
- Friedel‑Crafts alkylation of aniline is easily carried out.
Answer: Bromination of aniline in bromine water yields 2,4,6‑tribromoaniline.
The –NH2 group in aniline is strongly activating and ortho‑para directing. In bromine water (polar medium), the reaction is so fast that all three ortho and para positions are substituted, giving 2,4,6‑tribromoaniline as a white precipitate. Option B is false because under strongly acidic conditions (conc. HNO3/H2SO4) the amino group gets protonated to –NH3+, which is meta‑directing, so m‑nitroaniline is the major product. Option C is false: sulphonation of aniline at high temperature (180 °C) yields p‑sulphanilic acid as the major product. Option D is false because aniline forms a salt with the Lewis acid catalyst (AlCl3), deactivating the ring and also consuming the catalyst, so Friedel‑Crafts reactions are not feasible with free aniline.
2. Which of the following statements is correct regarding the alkylation and acylation of amines?
- Alkylation of a primary amine with an excess of alkyl halide yields the corresponding secondary amine as the final product.
- Acylation of an amine with an acyl chloride proceeds without the need of a base to neutralize the HCl formed.
- Both alkylation and acylation reactions of amines increase the basicity of the amine.
- Acylation of amines with acyl halides produces amides and the reaction is used to protect the amino group.
Answer: Acylation of amines with acyl halides produces amides and the reaction is used to protect the amino group.
Acylation of an amine with an acyl halide (or anhydride) gives an amide; the amino group is converted into an amide group, which is less reactive and can be later hydrolyzed back to the free amine. This is commonly used as a protection strategy. Option A is incorrect because excess alkyl halide leads to further alkylation, yielding tertiary amines and quaternary ammonium salts. Option B is incorrect because a base (e.g., pyridine or NaOH) is necessary to neutralize the HCl byproduct and prevent protonation of the amine. Option C is incorrect because acylation converts the amine into an amide, which has reduced basicity due to resonance interaction of the nitrogen lone pair with the carbonyl group.
3. Which of the following is correct regarding Hinsberg's test?
- Secondary amines react with benzenesulphonyl chloride to form a product that is soluble in alkali.
- Tertiary amines react with benzenesulphonyl chloride to give a sulphonamide that is insoluble in alkali.
- Primary amines give a sulphonamide that is soluble in alkali due to the presence of an acidic hydrogen on the nitrogen.
- All amines give a precipitate with benzenesulphonyl chloride.
Answer: Primary amines give a sulphonamide that is soluble in alkali due to the presence of an acidic hydrogen on the nitrogen.
In Hinsberg's test, a primary amine reacts with benzenesulphonyl chloride (C6H5SO2Cl) to form an N‑substituted sulphonamide having one hydrogen on the nitrogen. This N–H bond is sufficiently acidic to dissolve in aqueous KOH, forming a salt. Secondary amines give a sulphonamide without an N–H hydrogen; hence it is insoluble in alkali. Tertiary amines do not react, and they are often identified by their insolubility and ability to be extracted into acid. Option A is incorrect because the sulphonamide from a secondary amine is insoluble. Option B is incorrect because tertiary amines do not react. Option D is incorrect because tertiary amines do not produce a precipitate.
4. Aniline does not undergo Friedel‑Crafts reaction (alkylation or acylation) because:
- the amino group is deactivating for the benzene ring
- aniline reacts with the Lewis acid AlCl3 to form a salt, making the amino group positively charged and strongly deactivating the ring
- the amino group is a strong deactivator for electrophilic substitution
- aniline is a liquid and not suitable for Friedel‑Crafts reaction
Answer: aniline reacts with the Lewis acid AlCl3 to form a salt, making the amino group positively charged and strongly deactivating the ring
In Friedel‑Crafts reactions, a Lewis acid (e.g., AlCl3) is used as a catalyst. Aniline, being basic, donates its lone pair to AlCl3, forming an anilinium–AlCl3 complex. The resulting –NH3+ group is strongly deactivating and meta‑directing, and it also removes the catalyst from the reaction. Hence, aniline does not undergo Friedel‑Crafts alkylation or acylation. Option A is false: free –NH2 is activating, not deactivating. Option C is false because the free amino group is activating; the problem arises only after salt formation. Option D is irrelevant; many liquids undergo Friedel‑Crafts reactions.
5. Acetylation of aniline is done to protect the amino group. Which statement is correct?
- The acetyl group activates the ring more than the amino group.
- Acetylation is carried out by reacting aniline with acetic anhydride in the presence of NaOH or pyridine.
- After protection, the ring becomes deactivated and only meta‑substitution occurs.
- The acetyl group cannot be removed after the desired reaction.
Answer: Acetylation is carried out by reacting aniline with acetic anhydride in the presence of NaOH or pyridine.
Acetylation of aniline (protection) is typically performed using acetic anhydride or acetyl chloride in the presence of a base such as pyridine or aqueous NaOH to neutralize the acid formed. The product is acetanilide. Option A is false: the acetamido group (–NHCOCH3) is activating but less so than the free –NH2 group. Option C is false: the acetamido group is still ortho‑para directing, though the activating effect is reduced, allowing mononitration etc. Option D is false: the acetyl group can be removed by acid or base hydrolysis (e.g., refluxing with dilute HCl or NaOH) to regenerate aniline.
6. Which of the following amines has the highest boiling point?
- Propylamine (CH3CH2CH2NH2)
- Trimethylamine ((CH3)3N)
- Ethylmethylamine (CH3CH2NHCH3)
- Methylamine (CH3NH2)
Answer: Propylamine (CH3CH2CH2NH2)
Boiling points of amines depend on hydrogen bonding and molecular weight. Propylamine, trimethylamine, and ethylmethylamine are isomers with the same molecular weight; propylamine has the highest boiling point because as a primary amine it has two N-H bonds enabling more extensive intermolecular hydrogen bonding than the secondary amine (one N-H) or tertiary amine (no N-H). and stronger intermolecular forces. Trimethylamine and ethylmethylamine have weaker hydrogen bonding due to fewer N–H bonds; methylamine has lower molecular weight.
7. Aniline reacts with benzenesulphonyl chloride in the presence of aqueous KOH to form a product that:
- is insoluble in alkali and precipitates out
- is soluble in alkali due to the presence of an acidic hydrogen
- is a neutral compound
- does not react
Answer: is soluble in alkali due to the presence of an acidic hydrogen
Aniline (a primary aromatic amine) reacts with benzenesulphonyl chloride to give N‑phenylbenzenesulphonamide. The sulphonamide still bears a hydrogen on the nitrogen, which is acidic enough to be deprotonated by KOH, forming a water‑soluble salt. Hence the product dissolves in alkali. Options A, C, and D are incorrect because the reaction does occur, the product is acidic rather than neutral, and it is soluble in alkali.
8. The basicity of aniline is affected by substituents on the benzene ring. Which of the following substituents will increase the basicity of aniline?
- A) -NO2
- B) -Cl
- C) -CH3
- D) -CN
Answer: C) -CH3
The methyl group (-CH3) is an electron-donating group through the +I (inductive) effect. It increases the electron density on the nitrogen atom, making the lone pair more available for protonation and thus increasing the basicity of aniline. In contrast, -NO2, -Cl, and -CN are electron-withdrawing groups (by -M or -I effects), which decrease the electron density on nitrogen, reducing the basicity.
9. When a primary aliphatic amine is treated with nitrous acid (HNO2) at low temperature (0–5°C), the product is:
- a stable diazonium salt
- an alcohol and nitrogen gas
- an N-nitrosoamine
- an isocyanide
Answer: an alcohol and nitrogen gas
Primary aliphatic amines react with nitrous acid to form highly unstable diazonium salts that rapidly decompose to give an alcohol (or alkene if the alcohol can eliminate) and nitrogen gas. In contrast, primary aromatic amines yield stable diazonium salts at 0–5°C. N-nitrosoamines are formed from secondary amines, and isocyanides are obtained in the carbylamine reaction (CHCl3 + KOH).
10. The basicity of an amine depends on the availability of the lone pair on nitrogen. Which factor increases the basic strength of aliphatic amines?
- Presence of electron-withdrawing groups on the alkyl chain
- Presence of electron-donating alkyl groups
- Increased steric hindrance around nitrogen
- Increased s-character of the nitrogen lone pair
Answer: Presence of electron-donating alkyl groups
Electron-donating alkyl groups increase the electron density on the nitrogen atom, making the lone pair more available for protonation and thereby increasing basicity. Electron-withdrawing groups decrease basicity, steric hindrance can hinder protonation, and higher s-character (e.g., sp2 vs sp3) holds the lone pair more tightly, reducing basicity.
11. The carbylamine reaction is a test for:
- secondary and tertiary amines
- primary amines only
- all aliphatic and aromatic amines
- primary amines and also gives a positive test with nitro compounds
Answer: primary amines only
The carbylamine reaction (also known as the isocyanide test) is specific for primary amines (both aliphatic and aromatic). When a primary amine is heated with chloroform and alcoholic KOH, a foul-smelling isocyanide is formed. Secondary and tertiary amines do not give this reaction, and nitro compounds do not react under these conditions.
12. Which of the following compounds is a secondary amine?
- N-methylpropan-1-amine
- N,N-dimethylpropan-1-amine
- Propan-1-amine
- N-ethyl-N-methylpropan-1-amine
Answer: N-methylpropan-1-amine
A secondary amine has the nitrogen bonded to two carbon atoms and one hydrogen atom. N-methylpropan-1-amine (CH3CH2CH2NHCH3) fits this description. The other options: N,N-dimethylpropan-1-amine and N-ethyl-N-methylpropan-1-amine are tertiary amines (three carbon attachments), and propan-1-amine is a primary amine (one carbon attachment).