NEET Chemistry: Amines – Practice Questions

I. Multiple Choice Questions (MCQs)

1. Question: Which of the following is a primary amine? a) N,N-Dimethylmethanamine b) N-Ethylethanamine c) Propan-2-amine d) N-Methylpropan-2-amine

2. Question: The correct increasing order of boiling points for primary, secondary, and tertiary amines of comparable molecular mass is: a) Primary < Secondary < Tertiary b) Tertiary < Primary < Secondary c) Tertiary < Secondary < Primary d) Secondary < Primary < Tertiary

3. Question: Which of the following reactions is used for the preparation of pure primary amines? a) Ammonolysis of alkyl halides b) Hoffmann bromamide degradation c) Reduction of nitriles d) Reduction of amides

4. Question: Aniline reacts with bromine water at room temperature to give: a) p-Bromoaniline b) o-Bromoaniline c) 2,4,6-Tribromoaniline d) m-Bromoaniline

5. Question: The product formed when primary aliphatic amine reacts with nitrous acid is: a) Alcohol and nitrogen gas b) N-nitrosamine c) Diazonium salt d) Azo dye

6. Question: Carbylamine reaction is given by: a) Primary amines b) Secondary amines c) Tertiary amines d) All types of amines

7. Question: Which reagent is used in Hinsberg’s test? a) HNO2​ b) CHCl3​ and alc. KOH c) Benzene sulphonyl chloride d) Anhydrous AlCl3​

8. Question: Which of the following will be soluble in aqueous KOH solution after reacting with benzene sulphonyl chloride? a) Diethylamine b) Triethylamine c) Propylamine d) Aniline

9. Question: The hybridization of nitrogen in amines is: a) sp b) sp$^2$ c) sp$^3$ d) dsp$^2$

10. Question: Which of the following is the strongest base in aqueous solution? a) Methylamine b) Dimethylamine c) Trimethylamine d) Ammonia

11. Question: Gabriel phthalimide synthesis is used for the preparation of: a) Primary aromatic amines b) Primary aliphatic amines c) Secondary amines d) Tertiary amines

12. Question: Which of the following is a step-down reaction for the preparation of amines? a) Reduction of nitro compounds b) Ammonolysis c) Reduction of nitriles d) Hoffmann bromamide degradation

13. Question: Aniline is less basic than ethylamine because: a) Aniline is an aromatic compound. b) The lone pair of electrons on nitrogen in aniline is delocalized over the benzene ring. c) Ethylamine has a larger alkyl group. d) Anilinium ion is less stable than ethylammonium ion.

14. Question: What is the product of Sandmeyer reaction if chlorobenzene is the desired product from benzenediazonium chloride? a) CuCl2​/HCl b) Cu2​Cl2​/HCl c) Cu/HCl d) Fe/HCl

15. Question: The reaction of benzenediazonium chloride with phenol in a weakly alkaline medium gives: a) p-Hydroxyazobenzene b) p-Aminoazobenzene c) Benzene d) Phenol

II. Assertion-Reason Type Questions

Directions: In the following questions, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option. a) Both A and R are true and R is the correct explanation of A. b) Both A and R are true but R is NOT the correct explanation of A. c) A is true but R is false. d) A is false but R is true.

16. Assertion (A): Hoffmann bromamide degradation reaction is used to prepare primary amines. Reason (R): The amine formed has one carbon atom less than the starting amide.

17. Assertion (A): Aniline does not undergo Friedel-Crafts reaction. Reason (R): The amino group coordinates with the Lewis acid catalyst (AlCl3​), forming a deactivated complex.

18. Assertion (A): Primary amines have higher boiling points than tertiary amines of comparable molecular mass. Reason (R): Primary amines have two hydrogen atoms bonded to nitrogen, allowing for extensive intermolecular hydrogen bonding.

19. Assertion (A): Aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis. Reason (R): Aryl halides do not undergo nucleophilic substitution reaction with potassium phthalimide.

20. Assertion (A): Aniline is a weaker base than ammonia. Reason (R): The lone pair of electrons on nitrogen in aniline is delocalized over the benzene ring due to resonance.

III. Short Answer / Conceptual Questions

21. Question: Classify the following amines as primary, secondary, or tertiary: a) (CH3​)2​CHNH2​ b) (CH3​)3​N c) C6​H5​−NH−CH3​

22. Question: Explain why methylamine is a stronger base than ammonia in aqueous solution.

23. Question: Write the chemical equation for the Hoffmann bromamide degradation reaction starting with ethanamide. What is the main product?

24. Question: Describe the Carbylamine reaction. What is its significance?

25. Question: How will you distinguish between primary, secondary, and tertiary amines using Hinsberg’s reagent?

26. Question: Give the structures of the reagents used for converting: a) Nitrobenzene to Aniline b) Propanenitrile to Propan-1-amine

27. Question: What is Diazotisation? Write the reaction for the formation of benzenediazonium chloride.

28. Question: Explain why direct nitration of aniline is not preferred for obtaining p-nitroaniline. How can p-nitroaniline be prepared?

29. Question: Complete the following reactions: a) C6​H5​NH2​+(CH3​CO)2​OPyridine​ b) CH3​CH2​NH2​+HNO2​NaNO2​/HCl​

30. Question: Write the general reactions for: a) Sandmeyer reaction to prepare bromobenzene from benzenediazonium chloride. b) Coupling reaction to form an azo dye with β-naphthol.

Answers and Explanations

I. Multiple Choice Questions (MCQs) – Answers

1. Answer: c) Propan-2-amine Explanation: Propan-2-amine (CH3​CH(NH2​)CH3​) has one alkyl group attached to the nitrogen atom, classifying it as a primary amine. N,N-Dimethylmethanamine ((CH3​)3​N) is tertiary, N-Ethylethanamine ((C2​H5​)2​NH) is secondary, and N-Methylpropan-2-amine is secondary.

2. Answer: c) Tertiary < Secondary < Primary Explanation: In amines, primary and secondary amines can form intermolecular hydrogen bonds, while tertiary amines cannot (as they lack hydrogen directly bonded to nitrogen). Primary amines form more extensive H-bonds than secondary. Therefore, the boiling point order is Primary > Secondary > Tertiary for comparable molecular masses.

3. Answer: b) Hoffmann bromamide degradation Explanation: The Hoffmann bromamide degradation reaction produces pure primary amines by degrading an amide, without forming a mixture of amines. Ammonolysis yields a mixture. Reduction of nitriles and amides yield primary amines but are not inherently “pure” in the same sense as Gabriel/Hoffmann which are designed to avoid over-alkylation.

4. Answer: c) 2,4,6-Tribromoaniline Explanation: The −NH2​ group is a powerful activating group and ortho-para directing. When aniline reacts with bromine water, the activation is so strong that substitution occurs at all available ortho and para positions, leading to the formation of 2,4,6-tribromoaniline as a white precipitate.

5. Answer: a) Alcohol and nitrogen gas Explanation: Primary aliphatic amines react with nitrous acid to form primary alcohols with the evolution of nitrogen gas. This reaction is often accompanied by rearrangements.

6. Answer: a) Primary amines Explanation: The carbylamine reaction (also known as isocyanide test) is a characteristic test for primary amines (both aliphatic and aromatic). Secondary and tertiary amines do not undergo this reaction.

7. Answer: c) Benzene sulphonyl chloride Explanation: Hinsberg’s reagent is benzene sulphonyl chloride (C6​H5​SO2​Cl). It is used to distinguish between primary, secondary, and tertiary amines based on their differing reactivity and solubility properties with the reagent.

8. Answer: c) Propylamine Explanation: Propylamine is a primary amine. Primary amines react with Hinsberg’s reagent to form N-alkylbenzenesulphonamides, which have an acidic hydrogen on the nitrogen and are therefore soluble in aqueous KOH solution. Diethylamine is secondary and forms an insoluble product, while triethylamine is tertiary and does not react. Aniline is primary aromatic, so it also forms a soluble product, but it’s not among the options (it would be N-phenylbenzenesulphonamide). Between the given options, propylamine is the only one that yields a KOH soluble product.

9. Answer: c) sp$^3$ Explanation: The nitrogen atom in amines has three bond pairs (or alkyl/aryl groups) and one lone pair of electrons. These four electron domains repel each other, leading to an sp$^3$ hybridization for the nitrogen atom and a pyramidal geometry.

10. Answer: b) Dimethylamine Explanation: In aqueous solution, the basicity of alkylamines depends on the +I effect, steric hindrance, and solvation of the ammonium ion. For methylamines, the order is secondary > primary > tertiary > ammonia due to a balance of these factors. Thus, dimethylamine is the strongest base among the given options in aqueous solution.

11. Answer: b) Primary aliphatic amines Explanation: Gabriel phthalimide synthesis is a highly effective method for preparing pure primary amines, particularly aliphatic ones. Aromatic primary amines generally cannot be prepared by this method due to the poor reactivity of aryl halides in SN2 reactions.

12. Answer: d) Hoffmann bromamide degradation Explanation: The Hoffmann bromamide degradation reaction is known as a step-down reaction because the amine formed contains one carbon atom less than the starting amide.

13. Answer: b) The lone pair of electrons on nitrogen in aniline is delocalized over the benzene ring. Explanation: In aniline, the lone pair of electrons on the nitrogen atom participates in resonance with the benzene ring. This delocalization makes the lone pair less available for donation to a proton, thereby reducing the basicity of aniline compared to aliphatic amines like ethylamine.

14. Answer: b) Cu2​Cl2​/HCl Explanation: The Sandmeyer reaction uses cuprous halides (e.g., Cu2​Cl2​ or CuCl for chlorobenzene) in the presence of HX (e.g., HCl) to replace the diazonium group with a halogen or cyanide group. Cu/HCl is used in the Gattermann reaction.

15. Answer: a) p-Hydroxyazobenzene Explanation: Benzenediazonium chloride undergoes a coupling reaction with phenol (an activating group) in a weakly alkaline medium to form p-hydroxyazobenzene, which is an orange-colored azo dye.

II. Assertion-Reason Type Questions – Answers

16. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: The Hoffmann bromamide degradation reaction is indeed a valuable method for preparing primary amines, and its distinct feature is that the product amine has one carbon atom less than the starting amide, which makes R the correct explanation for A.

17. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Aniline does not undergo Friedel-Crafts reaction (alkylation or acylation) because the −NH2​ group, being basic, forms a complex with the Lewis acid catalyst (AlCl3​). This complex formation makes the nitrogen positively charged (−N+H2​AlCl3−​), which acts as a strong deactivating group and prevents further electrophilic substitution reactions on the ring.

18. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Primary amines have two hydrogen atoms on the nitrogen, allowing them to form more extensive intermolecular hydrogen bonds compared to secondary amines (one H) and tertiary amines (no H). This stronger intermolecular attraction leads to higher boiling points for primary amines of comparable molecular mass.

19. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Gabriel phthalimide synthesis involves the nucleophilic substitution of an alkyl halide by potassium phthalimide. Aryl halides are much less reactive towards nucleophilic substitution reactions due to resonance stabilization of the C-X bond and steric hindrance, hence they do not react readily with potassium phthalimide, preventing the formation of aromatic primary amines by this method.

20. Answer: a) Both A and R are true and R is the correct explanation of A. Explanation: Aniline is a weaker base than ammonia because the lone pair of electrons on the nitrogen atom in aniline is delocalized over the benzene ring through resonance. This delocalization makes the lone pair less available for protonation, reducing its basicity.

III. Short Answer / Conceptual Questions – Answers

21. Answer: a) (CH3​)2​CHNH2​: Primary amine (1°) – Only one alkyl group (isopropyl) attached to the nitrogen atom. b) (CH3​)3​N: Tertiary amine (3°) – Three alkyl groups (methyl) attached to the nitrogen atom. c) C6​H5​−NH−CH3​: Secondary amine (2°) – Two groups (phenyl and methyl) attached to the nitrogen atom, with one hydrogen atom on nitrogen.

22. Answer: Methylamine (CH3​NH2​) is a stronger base than ammonia (NH3​) in aqueous solution due to a combination of factors:

1. +I effect: The methyl group (CH3​) is an electron-donating group (+I effect). It increases the electron density on the nitrogen atom, making the lone pair more available for protonation and thus increasing basicity.
2. Solvation: The methylammonium ion (CH3​NH3+​) formed after protonation is stabilized by hydrogen bonding with water molecules. While ammonia’s conjugate acid (NH4+​) can form four H-bonds, methylammonium can still form three, which contributes to its stability. The combined effect of the +I effect and solvation (though slightly reduced compared to NH4+​) makes methylamine a stronger base.

23. Answer: The Hoffmann bromamide degradation reaction starting with ethanamide (CH3​CONH2​): CH3​CONH2​+Br2​+4NaOHHeat​CH3​NH2​+Na2​CO3​+2NaBr+2H2​O The main product is Methylamine (CH3​NH2​).

24. Answer:

• Carbylamine reaction (Isocyanide test): This reaction is a characteristic test for primary amines (aliphatic and aromatic). When a primary amine is heated with chloroform (CHCl3​) and an alcoholic solution of potassium hydroxide (KOH), it produces an isocyanide (carbylamine) which has a highly unpleasant and offensive smell. R−NH2​+CHCl3​+3KOH(alc.)Heat​R−NC+3KCl+3H2​O
• Significance: The distinctive foul odor of the isocyanide is a strong positive test, allowing for the easy distinction of primary amines from secondary and tertiary amines, which do not give this reaction.

25. Answer: Hinsberg’s reagent (benzene sulphonyl chloride, C6​H5​SO2​Cl) is used to distinguish between primary, secondary, and tertiary amines based on their reactivity and solubility in alkali:

1. Primary Amine (e.g., R-NH$_2$): Reacts with Hinsberg’s reagent to form N-alkylbenzenesulphonamide (C6​H5​SO2​NHR). This product has an acidic hydrogen attached to the nitrogen atom, making it soluble in aqueous KOH solution.
2. Secondary Amine (e.g., R$_2$NH): Reacts with Hinsberg’s reagent to form N,N-dialkylbenzenesulphonamide (C6​H5​SO2​NR2​). This product does not have any acidic hydrogen attached to the nitrogen atom, so it is insoluble in aqueous KOH solution.
3. Tertiary Amine (e.g., R$_3$N): Does not react with Hinsberg’s reagent at all due to the absence of any hydrogen atom attached to the nitrogen. Therefore, it remains insoluble in aqueous KOH.

26. Answer: a) Nitrobenzene to Aniline: Reagents: Sn/HCl (or Fe/HCl or H2​/Pd) Reaction: C6​H5​NO2​+6[H]Sn/HCl​C6​H5​NH2​+2H2​O b) Propanenitrile to Propan-1-amine: Reagents: LiAlH4​ (Lithium Aluminium Hydride) followed by hydrolysis with H2​O. (Alternatively, H2​/Ni, Pt, or Pd) Reaction: CH3​CH2​C≡N+4[H]LiAlH4​ or H2​/Ni​CH3​CH2​CH2​NH2​

27. Question:

• Diazotisation: It is the process of converting a primary aromatic amine into an arenediazonium salt by reacting it with nitrous acid (HNO2​) at a low temperature, typically 0−5∘C (273-278 K). Nitrous acid is usually generated in situ from sodium nitrite (NaNO2​) and a dilute mineral acid like hydrochloric acid (HCl).
• Reaction for the formation of benzenediazonium chloride: C6​H5​NH2​+NaNO2​+2HCl273−278K​C6​H5​N2+​Cl−+NaCl+2H2​O (Aniline) (Benzenediazonium chloride)

28. Answer: Direct nitration of aniline is not preferred for obtaining p-nitroaniline due to two main reasons:

1. Formation of Anilinium Ion: In the strongly acidic medium required for nitration (conc. HNO3​ and conc. H2​SO4​), aniline, being a base, gets protonated to form the anilinium ion (C6​H5​NH3+​). The anilinium ion is a meta-directing and deactivating group. This leads to a significant amount of the meta-isomer (m-nitroaniline) along with ortho and para products.
2. Oxidation: Concentrated nitric acid is a strong oxidizing agent. Aniline, being easily oxidizable, undergoes oxidation, leading to the formation of undesirable tarry products and reduced yield.

Preparation of p-nitroaniline: To prepare p-nitroaniline selectively and in good yield, the amino group must first be protected by acetylation.

1. Acetylation: Aniline is reacted with acetic anhydride or acetyl chloride to form acetanilide. The acetyl group (-$NHCOCH$_3) is a less powerful activating group than -$NH$_2, and it also protects the amino group from oxidation. C6​H5​NH2​+(CH3​CO)2​OPyridine​C6​H5​NHCOCH3​+CH3​COOH
2. Nitration: Acetanilide is then nitrated using a nitrating mixture (conc. HNO3​ + conc. H2​SO4​). The nitration predominantly occurs at the para position due to steric hindrance at the ortho positions, yielding p-nitroacetanilide. C6​H5​NHCOCH3​Conc.HNO3​,Conc.H2​SO4​​p−NO2​−C6​H4​−NHCOCH3​
3. Hydrolysis: Finally, the acetyl group is removed by acid or base hydrolysis of p-nitroacetanilide to obtain p-nitroaniline. p−NO2​−C6​H4​−NHCOCH3​+H2​OH+ or OH−​p−NO2​−C6​H4​−NH2​+CH3​COOH

29. Answer: a) C6​H5​NH2​+(CH3​CO)2​OPyridine​C6​H5​NHCOCH3​+CH3​COOH (Aniline) (Acetic anhydride) (Acetanilide) (Acetic acid) This is an acylation reaction.

b) CH3​CH2​NH2​+HNO2​NaNO2​/HCl​CH3​CH2​OH+N2​↑+H2​O (Ethanamine) (Nitrous acid) (Ethanol) (Nitrogen gas) This is the reaction of a primary aliphatic amine with nitrous acid.

30. Answer: a) Sandmeyer reaction to prepare bromobenzene from benzenediazonium chloride: C6​H5​N2+​Cl−Cu2​Br2​/HBr​C6​H5​Br+N2​↑ (Benzenediazonium chloride) (Bromobenzene)

b) Coupling reaction to form an azo dye with β-naphthol: C6​H5​N2+​Cl−+C10​H7​OHWeakly alkaline medium​C6​H5​−N=N−C10​H6​−OH+HCl (Benzenediazonium chloride) (β-naphthol) (1-Phenylazo-2-naphthol, an orange-red dye)