Nitrobenzene is treated with Sn / HCl to give a compound (X) which on treatment with NaNO2 and HCl at 278 K gives compound (Y). When the compound (Y) is treated with Cu / HBr , compound 'Z' is obtained. The compound 'Z' is

Compound Z is Bromobenzene.The sequence is three standard steps:Sn and HCl reduce nitrobenzene to aniline: X is C6H5NH2.NaNO2 and HCl at 273-278 K diazotise aniline: Y is benzenediazonium chloride, C6H5N2+Cl-.Copper powder with HBr replaces the diazonium group by bromine with loss of nitrogen (the Gattermann reaction): Z is C6H5Br.Benzene diazonium bromide is not formed; the diazonium salt is the intermediate Y, and it is decomposed in the last step. Phenol would come from warming Y with water, benzene from reduction with H3PO2, and chlorobenzene from Cu/HCl rather than Cu/HBr.

Amines and Nitrogen Compounds Questions and Answers

Q: Which of the following technique is used to separate chloroform and aniline?

The technique is Distillation.Simple distillation separates liquids whose boiling points differ widely. Chloroform boils at about 334 K and aniline at about 457 K, so chloroform distils over first and aniline stays behind. This is the standard textbook example.Fractional distillation is needed only for close boiling points; distillation under reduced pressure is for liquids that decompose near their boiling point; steam distillation is for steam-volatile substances immiscible with water; continuous extraction is a solvent method, not distillation.

Q: The order of basic strength of following amines is(i) CH3NH2 (ii) (C2H5)2NH (iii)C6H5NH2 (iv)C6H5NHCH3

The order is (iii) < (iv) < (i) < (ii).In the aryl amines the nitrogen lone pair is delocalised into the benzene ring, so aniline (iii) and N-methylaniline (iv) are much weaker bases than the alkyl amines. The extra methyl group on N makes N-methylaniline slightly more basic than aniline.Among the alkyl amines, diethylamine (ii), a secondary amine with two electron-releasing ethyl groups, is more basic in water than methylamine (i). Options putting an aryl amine above an alkyl amine reverse the effect of resonance.

Q: Acetanilide is prepared by treating acetic anhydride/pyridine with

Acetanilide is prepared from benzenamine (aniline).Acetylation of aniline with acetic anhydride, in the presence of pyridine to remove the acid formed, replaces one N–H hydrogen with an acetyl group: C6H5NH2 + (CH3CO)2O → C6H5NHCOCH3 + CH3COOH.Ethanamine and methanamine would give N-alkyl acetamides, not an anilide. N-Methylaniline would give N-methylacetanilide, and N-methylethanamine an aliphatic amide.

Q: Which of the following amine does not react with Hinsberg's reagent?

The amine that does not react with Hinsberg's reagent is N,N-dimethylethanamine.Hinsberg's reagent is benzenesulphonyl chloride. Primary and secondary amines replace an N–H hydrogen to form sulphonamides. N,N-Dimethylethanamine is a tertiary amine with no hydrogen on nitrogen, so it does not react.Methanamine, 1-propanamine and 2-propanamine are primary amines (their sulphonamides dissolve in alkali), and N-methylethanamine is secondary (its sulphonamide is insoluble in alkali); all of them react.

Q: Aryl fluorides are prepared from diazonium salts using

Aryl fluorides are prepared using HBF4 (fluoroboric acid).In the Balz-Schiemann reaction, the diazonium salt is treated with fluoroboric acid to precipitate arenediazonium fluoroborate, which on heating decomposes to the aryl fluoride, nitrogen and BF3.Simple fluoride salts such as NaF, KF or AlF3 do not give aryl fluorides from diazonium salts in the way CuCl or KI give chlorides and iodides, and BF3 is a by-product, not the reagent.

Q: Carbylamine is obtained when aniline is

The carbylamine is obtained when aniline is heated with \(\mathrm{CHCl_{3}}\) and ethanolic KOH.This is the carbylamine reaction (isocyanide test). A primary amine heated with chloroform and alcoholic potassium hydroxide gives an isocyanide, which has a very unpleasant smell:\[ \begin{aligned} &\mathrm{C_{6}H_{5}NH_{2} + CHCl_{3} + 3KOH} \\ &\mathrm{\rightarrow C_{6}H_{5}NC + 3KCl + 3H_{2}O} \end{aligned} \]The base is essential, since it generates dichlorocarbene from chloroform, so chloroform with HCl does nothing of the kind. \(\mathrm{NaNO_{2} / HCl}\) diazotises aniline, concentrated \(\mathrm{H_{2}SO_{4}}\) on heating sulphonates it (sulphanilic acid), and HCl alone just forms anilinium chloride.

Q: The following amines are having same molecular masses.(i) n-C4H9NH2 (ii) (C2H5)2NH (iii) C2H5N(CH3)2The correct order of boiling point of the above amines is

The order is (i) > (ii) > (iii).The three amines have the same molar mass (73), so the difference in boiling point comes from intermolecular hydrogen bonding through N-H bonds. The primary amine, n-butylamine, has two N-H hydrogens per molecule and forms the most hydrogen bonds. The secondary amine, diethylamine, has one. The tertiary amine, ethyldimethylamine, has no hydrogen on nitrogen and cannot hydrogen bond with itself at all.Their boiling points are about 351 K, 329 K and 310 K respectively. Every other option puts the tertiary amine above at least one of the others, or the secondary above the primary, which reverses the hydrogen bonding trend.

Q: Which of the following is Hinsberg's reagent?

Hinsberg's reagent is benzenesulphonyl chloride.C6H5SO2Cl distinguishes primary, secondary and tertiary amines: primary amines give a sulphonamide soluble in alkali, secondary amines give one insoluble in alkali, and tertiary amines do not react.Nitrous acid is also used to test amines but is not Hinsberg's reagent, 2,4-dinitrophenylhydrazine tests for carbonyl groups, ammoniacal silver nitrate is Tollens' reagent for aldehydes, and ethanolic KOH is used for dehydrohalogenation.

Q: The IUPAC name of allylamine is

The IUPAC name is Prop-2-en-1-amine.The allyl group is CH2=CH-CH2-, so allylamine is CH2=CH-CH2-NH2. Number the three-carbon chain from the end bearing the amino group, which takes priority: NH2 on C-1 and the double bond between C-2 and C-3.Prop-1-en-2-amine would put the amino group on the middle carbon, and the butene names have one carbon too many. '2-Amino 1-propene' is not IUPAC style, which names the amine as the suffix.

11.

Which of the following technique is used to separate chloroform and aniline?

A.Fractional distillation
B.Distillation under reduced pressure
C.Steam distillation
D.Continuous extraction
E.Distillation

Exam Year(s)

KEAM - 2025,

Answer: E

The technique is Distillation.

Simple distillation separates liquids whose boiling points differ widely. Chloroform boils at about 334 K and aniline at about 457 K, so chloroform distils over first and aniline stays behind. This is the standard textbook example.

Fractional distillation is needed only for close boiling points; distillation under reduced pressure is for liquids that decompose near their boiling point; steam distillation is for steam-volatile substances immiscible with water; continuous extraction is a solvent method, not distillation.

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12.

The order of basic strength of following amines is
(i) CH₃NH₂ (ii) (C₂H₅)₂NH (iii)C₆H₅NH₂ (iv)C₆H₅NHCH₃

A.(ii) < (i) < (iv) < (iii)
B.(iii) < (iv) < (ii) < (i)
C.(ii) < (iii) < (iv) < (i)
D.(i) < (ii) < (iii) < (iv)
E.(iii) < (iv) < (i) < (ii)

Exam Year(s)

KEAM - 2025,

Answer: E

The order is (iii) < (iv) < (i) < (ii).

In the aryl amines the nitrogen lone pair is delocalised into the benzene ring, so aniline (iii) and N-methylaniline (iv) are much weaker bases than the alkyl amines. The extra methyl group on N makes N-methylaniline slightly more basic than aniline.

Among the alkyl amines, diethylamine (ii), a secondary amine with two electron-releasing ethyl groups, is more basic in water than methylamine (i). Options putting an aryl amine above an alkyl amine reverse the effect of resonance.

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13.

Acetanilide is prepared by treating acetic anhydride/pyridine with

A.Ethanamine
B.Methanamie
C.Benzenamine
D.N-Methylaniline
E.N-Methylethanamine

Exam Year(s)

KEAM - 2025,

Answer: C

Acetanilide is prepared from benzenamine (aniline).

Acetylation of aniline with acetic anhydride, in the presence of pyridine to remove the acid formed, replaces one N–H hydrogen with an acetyl group: C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH.

Ethanamine and methanamine would give N-alkyl acetamides, not an anilide. N-Methylaniline would give N-methylacetanilide, and N-methylethanamine an aliphatic amide.

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14.

Which of the following amine does not react with Hinsberg's reagent?

A.Methanamine
B.N-Methylethanamine
C.N,N-Dimethylethanamine
D.1-Propanamine
E.2-Propanamine

Exam Year(s)

KEAM - 2025,

Answer: C

The amine that does not react with Hinsberg's reagent is N,N-dimethylethanamine.

Hinsberg's reagent is benzenesulphonyl chloride. Primary and secondary amines replace an N–H hydrogen to form sulphonamides. N,N-Dimethylethanamine is a tertiary amine with no hydrogen on nitrogen, so it does not react.

Methanamine, 1-propanamine and 2-propanamine are primary amines (their sulphonamides dissolve in alkali), and N-methylethanamine is secondary (its sulphonamide is insoluble in alkali); all of them react.

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15.

Aryl fluorides are prepared from diazonium salts using

A.\(\mathrm{NaF}\)
B.\(\mathrm{KF}\)
C.\(\mathrm{BF_{3}}\)
D.\(\mathrm{AlF_{3}}\)
E.\(\mathrm{HBF_{4}}\)

Exam Year(s)

KEAM - 2025,

Answer: E

Aryl fluorides are prepared using HBF₄ (fluoroboric acid).

In the Balz-Schiemann reaction, the diazonium salt is treated with fluoroboric acid to precipitate arenediazonium fluoroborate, which on heating decomposes to the aryl fluoride, nitrogen and BF₃.

Simple fluoride salts such as NaF, KF or AlF₃ do not give aryl fluorides from diazonium salts in the way CuCl or KI give chlorides and iodides, and BF₃ is a by-product, not the reagent.

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16.

Nitrobenzene is treated with Sn / HCl to give a compound (X) which on treatment with NaNO₂ and HCl at 278 K gives compound (Y). When the compound (Y) is treated with Cu / HBr , compound 'Z' is obtained. The compound 'Z' is

A.Benzene
B.Benzene diazonium bromide
C.Phenol
D.Bromobenzene
E.Chlorobenzene

Exam Year(s)

KEAM - 2025,

Answer: D

Compound Z is Bromobenzene.

The sequence is three standard steps:

Sn and HCl reduce nitrobenzene to aniline: X is C₆H₅NH₂.
NaNO₂ and HCl at 273-278 K diazotise aniline: Y is benzenediazonium chloride, C₆H₅N₂⁺Cl^-.
Copper powder with HBr replaces the diazonium group by bromine with loss of nitrogen (the Gattermann reaction): Z is C₆H₅Br.

Benzene diazonium bromide is not formed; the diazonium salt is the intermediate Y, and it is decomposed in the last step. Phenol would come from warming Y with water, benzene from reduction with H₃PO₂, and chlorobenzene from Cu/HCl rather than Cu/HBr.

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17.

Carbylamine is obtained when aniline is

A.heated with Con.\(\mathrm{H_{2}SO_{4}}\)
B.treated with \(\mathrm{NaNO_{2} / HCl}\)
C.heated with \(\mathrm{CHCl_{3}}\) and ethanolic \(\mathrm{KOH}\)
D.treated with \(\mathrm{CHCl_{3}}\) and \(\mathrm{HCl}\)
E.treated with \(\mathrm{HCl}\)

Exam Year(s)

KEAM - 2025,

Answer: C

The carbylamine is obtained when aniline is heated with \(\mathrm{CHCl_{3}}\) and ethanolic KOH.

This is the carbylamine reaction (isocyanide test). A primary amine heated with chloroform and alcoholic potassium hydroxide gives an isocyanide, which has a very unpleasant smell:

\[ \begin{aligned} &\mathrm{C_{6}H_{5}NH_{2} + CHCl_{3} + 3KOH} \\ &\mathrm{\rightarrow C_{6}H_{5}NC + 3KCl + 3H_{2}O} \end{aligned} \]

The base is essential, since it generates dichlorocarbene from chloroform, so chloroform with HCl does nothing of the kind. \(\mathrm{NaNO_{2} / HCl}\) diazotises aniline, concentrated \(\mathrm{H_{2}SO_{4}}\) on heating sulphonates it (sulphanilic acid), and HCl alone just forms anilinium chloride.

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18.

The following amines are having same molecular masses.
(i) n-C₄H₉NH₂ (ii) (C₂H₅)₂NH (iii) C₂H₅N(CH₃)₂
The correct order of boiling point of the above amines is

A.(i) > (ii) > (iii)
B.(ii) > (iii) > (i)
C.(iii) > (i) > (ii)
D.(ii) > (i) > (iii)
E.(i) > (iii) > (ii)

Exam Year(s)

KEAM - 2025,

Answer: A

The order is (i) > (ii) > (iii).

The three amines have the same molar mass (73), so the difference in boiling point comes from intermolecular hydrogen bonding through N-H bonds. The primary amine, n-butylamine, has two N-H hydrogens per molecule and forms the most hydrogen bonds. The secondary amine, diethylamine, has one. The tertiary amine, ethyldimethylamine, has no hydrogen on nitrogen and cannot hydrogen bond with itself at all.

Their boiling points are about 351 K, 329 K and 310 K respectively. Every other option puts the tertiary amine above at least one of the others, or the secondary above the primary, which reverses the hydrogen bonding trend.

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19.

Which of the following is Hinsberg's reagent?

A.Nitrous acid
B.Ethanolic potassium hydroxide
C.Benzenesulphonyl chloride
D.2,4-Dinitrophenyl hydrazine
E.Ammoniacal silver nitrate

Exam Year(s)

KEAM - 2025,

Answer: C

Hinsberg's reagent is benzenesulphonyl chloride.

C₆H₅SO₂Cl distinguishes primary, secondary and tertiary amines: primary amines give a sulphonamide soluble in alkali, secondary amines give one insoluble in alkali, and tertiary amines do not react.

Nitrous acid is also used to test amines but is not Hinsberg's reagent, 2,4-dinitrophenylhydrazine tests for carbonyl groups, ammoniacal silver nitrate is Tollens' reagent for aldehydes, and ethanolic KOH is used for dehydrohalogenation.

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20.

The IUPAC name of allylamine is

A.But-2-en-1-amine
B.But-1-en-2-amine
C.Prop-2-en-1-amine
D.Prop-1-en-2-amine
E.2-Amino 1-propene

Exam Year(s)

KEAM - 2025,

Answer: C

The IUPAC name is Prop-2-en-1-amine.

The allyl group is CH₂=CH-CH₂-, so allylamine is CH₂=CH-CH₂-NH₂. Number the three-carbon chain from the end bearing the amino group, which takes priority: NH₂ on C-1 and the double bond between C-2 and C-3.

Prop-1-en-2-amine would put the amino group on the middle carbon, and the butene names have one carbon too many. '2-Amino 1-propene' is not IUPAC style, which names the amine as the suffix.

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