NEET Chemistry: Amines questions with solutions

Q1. Which of the following is \( 3^{0} \) amine?

1. 1-methylcyclohexylamine
2. Triethylamine
3. Tert-butylamine
4. N-methyl aniline

Answer: Triethylamine

Triethylamine is a tertiary amine because the nitrogen is attached to three ethyl groups. The other choices each have nitrogen bonded to only one or two carbon groups, so they are not 3° amines.

Q2. The electrolytic reduction of nitrobenzene in strongly acidic medium produces:

1. Azoxybenzene
2. Azobenzene
3. Aniline
4. p-Aminophenol

Answer: Aniline

In a strongly acidic medium, the electrolytic reduction of nitrobenzene leads to the formation of aniline as the major product due to the complete reduction of the nitro group (-NO2) to an amino group (-NH2).

Q3. Which of the following reactions is appropriate for converting acetamide to methanamine?

1. Hoffmann hypobromamide reaction
2. Stephens reaction
3. Gabriel's phthalimide synthesis
4. Carbylamine reaction

Answer: Hoffmann hypobromamide reaction

The Hoffmann hypobromamide reaction is used to convert amides like acetamide into primary amines like methanamine by treating them with bromine and a strong base.

Q4. Which of the following reagents will convert p-methylbenzenediazonium chloride into p-cresol?

1. Cu powder
2. H2O
3. H3PO2
4. C6H5OH

Answer: H2O

When p-methylbenzenediazonium chloride reacts with water (H2O), it undergoes hydrolysis to form p-cresol. This is a standard reaction of diazonium salts.

Q5. Diazo coupling is useful to prepare some

1. Pesticides
2. Dyes
3. Proteins
4. Vitamins

Answer: Dyes

Diazo coupling reactions are widely used in the synthesis of azo dyes, which are important in the textile and pigment industries.

Q6. Method by which aniline cannot be prepared is:

1. hydrolysis of phenylisocyanide with acidic solution
2. degradation of benzamide with bromine in alkaline solution
3. reduction of nitrobenzene with H2/Pd in ethanol
4. potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution

Answer: potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution

The Gabriel phthalimide synthesis (option D) is not suitable for preparing aniline because it specifically produces primary aliphatic amines, not aromatic amines like aniline.

Q7. Some reactions of amines are given. Which one is not correct?

1. (CH3)2NH + NaNO2 + HCl → (CH3)2N–N=O
2. (CH3)2N– + NaNO2 + HCl → (CH3)2N–N=NCl
3. CH3CH2NH2 + HNO2 → CH3CH2OH + N2
4. CH3NH2 + C6H5SO2Cl → CH3NHSO2C6H5

Answer: (CH3)2N– + NaNO2 + HCl → (CH3)2N–N=NCl

Option B is incorrect because secondary amines react with nitrous acid to form N-nitrosoamines, not diazonium salts. Diazonium salts are formed only with primary aromatic amines.

Q8. The major product of the following reaction is: COOH + NH3 → strong heating

1. COOH
2. COOH
3. COOH
4. CONH

Answer: CONH

When an ammonium salt of a carboxylic acid is strongly heated, it undergoes dehydration to form an amide (CONH).

Q9. An organic compound ‘A’ on treatment with NH₃ gives ‘B’ which on heating gives ‘C’. ‘C’ when treated with Br₂ in the presence of KOH produces ethylamine. Compound ‘A’ is:

1. (a) CH₃COOH
2. (b) CH₃CH₂CH₂COOH
3. (c) CH₃–CHCOOH | CH₃
4. (d) CH₃CH₂COOH

Answer: (d) CH₃CH₂COOH

Compound 'A' is CH₃CH₂COOH (propanoic acid). It reacts with NH₃ to form CH₃CH₂CONH₂ (amide 'B'), which on heating gives CH₃CH₂CN ('C'). When 'C' undergoes Hoffmann bromamide degradation with Br₂ and KOH, it produces ethylamine (CH₃CH₂NH₂).

Q10. In a set of reactions propionic acid yielded a compound D. CH3CH2COOH → SOCl2 → B → NH3 → C → KOH → D

1. CH3CH2CONH2
2. CH3CH2NHCH3
3. CH3CH2NH2
4. CH3CH2CH2NH2

Answer: CH3CH2CH2NH2

Propionic acid reacts with SOCl2 to form CH3CH2COCl (B). Reaction with NH3 gives CH3CH2CONH2 (C), which upon Hoffmann bromamide degradation with KOH forms CH3CH2CH2NH2 (D).

Q11. Strong heating of ammonium salts of dicarboxylic acids leads to the formation of cyclic imides.

1. COOH + NH3
2. COONH4
3. CONH2
4. CONH2 + NH3

Answer: CONH2

When ammonium salts of dicarboxylic acids are strongly heated, they undergo dehydration to form cyclic imides, which have the functional group -CONH-. This matches option C.

Q12. CH3CH2COOH reacts with SOCl2 to form CH3CH2COCl. The product reacts with NH3, KOH, and Br2 to form CH3CH2NH2. The correct option is:

1. CH3CH2COOH + SOCl2 → CH3CH2COCl → CH3CH2CONH2 → CH3CH2NH2
2. CH3CH2COOH + SOCl2 → CH3CH2COCl → CH3CH2CONH2 → CH3CH2NH2 + Br2
3. CH3CH2COOH + SOCl2 → CH3CH2COCl → CH3CH2CONH2 → CH3CH2NH2 + KOH
4. CH3CH2COOH + SOCl2 → CH3CH2COCl → CH3CH2CONH2 → CH3CH2NH2 + NH3

Answer: CH3CH2COOH + SOCl2 → CH3CH2COCl → CH3CH2CONH2 → CH3CH2NH2

The reaction sequence involves the conversion of CH3CH2COOH to CH3CH2COCl using SOCl2, followed by the formation of CH3CH2CONH2 with NH3, and finally the Hoffmann bromamide reaction with Br2 and KOH to yield CH3CH2NH2.

Q13. The amine that reacts with Hinsberg’s reagent to give an alkali insoluble product is

1. CH3 | C — CH — NH2 | CH3 CH3
2. CH3 — CH — NH — CH — CH3 | CH3 CH3
3. CH3 — CH2 — N — CH2CH3
4. CH3 — C — CH2CH2CH3 | NH2 CH3

Answer: CH3 | C — CH — NH2 | CH3 CH3

Hinsberg's reagent reacts with primary amines to form alkali-insoluble products. Option A is a primary amine, so it reacts as described.

Q14. Which of the following amine will give the carbylamine test?

1. NHCH3
2. N(CH3)2
3. NHC2H5
4. NH3

Answer: NHC2H5

The carbylamine test is specific for primary amines. Among the options, NHC2H5 is a primary amine and will give the test.

Q15. The correct statement regarding the basicity of arylamines is

1. Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system.
2. Arylamines are generally more basic than alkylamines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring π electron system.
3. Arylamines are generally more basic than alkylamines because of aryl group.
4. Arylamines are generally more basic than alkylamines, because the nitrogen atom in arylamines is sp-hybridized.

Answer: Arylamines are generally less basic than alkylamines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system.

Arylamines are less basic than alkylamines because the lone pair of electrons on the nitrogen atom in arylamines is delocalized into the aromatic ring, reducing its availability for protonation.

Q16. The correct order of the basic strength of methyl substituted amines in aqueous solution is:

1. (CH3)2NH > CH3NH2 > (CH3)3N
2. (CH3)3N > CH3NH2 > (CH3)2NH
3. (CH3)3N > (CH3)2NH > CH3NH2
4. CH3NH2 > (CH3)2NH > (CH3)3N

Answer: (CH3)2NH > CH3NH2 > (CH3)3N

In aqueous solution, the basic strength of amines depends on both inductive effects and solvation. Secondary amines like (CH3)2NH are more basic than primary amines (CH3NH2) due to stronger inductive effects and better solvation. Tertiary amines ((CH3)3N) are less basic due to steric hindrance reducing solvation.

Q17. Which of the following compounds is most basic?

1. O2N—NH2
2. CH2NH2
3. N—COCH3
4. NH2

Answer: CH2NH2

CH2NH2 is the most basic because the amino group (-NH2) is attached to a methyl group, which slightly increases electron density on nitrogen, enhancing its basicity. Other options have electron-withdrawing groups or resonance effects that reduce basicity.

Q18. Which of the following is more basic than aniline?

1. Triphenylamine
2. p-Nitroaniline
3. Benzylamine
4. Diphenylamine

Answer: Benzylamine

Benzylamine is more basic than aniline because the benzyl group does not delocalize the lone pair of electrons on nitrogen into the aromatic ring, making the lone pair more available for protonation.

Q19. Predict the product: NHCH3 + NaNO2 + HCl → Product

1. (i) CH3CH(OH)CH3 (iv) with Lucas reagent cloudiness appears after 5 minutes
2. (ii) CH3N—NO2
3. (iii) CH3N=N—O
4. (iv) CH3CH(OH)CH3

Answer: (iii) CH3N=N—O

The reaction involves methylamine (NHCH3) reacting with NaNO2 and HCl, leading to the formation of a diazonium ion intermediate, which rearranges to form CH3N=N—O (option C).

Q20. Which of the following statements about primary amines is ‘False’?

1. Alkyl amines are stronger bases than aryl amines.
2. Alkyl amines react with nitrous acid to produce alcohols.
3. Aryl amines react with nitrous acid to produce phenols.
4. Alkyl amines are stronger bases than ammonia.

Answer: Aryl amines react with nitrous acid to produce phenols.

Aryl amines react with nitrous acid to form diazonium salts, not phenols directly. Phenols are obtained only after hydrolysis of diazonium salts.