Chemistry — Class 12, Chapter 16: Amines

2079 Jestha · 1 mark

The number of possible structural isomers of 1° amines of molecular formula $C_4H_{11}N$ give
(a) 1 (b) 2 (c) 3 (d) 4

Answer

(d) 4. The four primary amines of $C_4H_{11}N$ are:

butan-1-amine, $CH_3CH_2CH_2CH_2NH_2$
butan-2-amine, $CH_3CH_2CH(NH_2)CH_3$
2-methylpropan-1-amine, $(CH_3)_2CHCH_2NH_2$
2-methylpropan-2-amine, $(CH_3)_3CNH_2$

These come from the two butyl skeletons (n-butane and isobutane), each having two different positions for $-NH_2$ .

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2079 Jestha · 5 marks

How would you apply Hoffmann's method for the separation of 1°, 2° and 3° amine from their mixture?

Answer

Hoffmann's method separates a mixture of 1°, 2° and 3° amines using diethyl oxalate.

Principle: Hoffmann's method uses diethyl oxalate, $(COOC_2H_5)_2$ , which reacts differently with the three classes of amines.

Step 1 - Treat the mixture with diethyl oxalate.

1° amine:  (COOC2H5)2 + 2 RNH2  →  (CONHR)2  +  2 C2H5OH      (solid dialkyl oxamide)
2° amine:  (COOC2H5)2 +   R2NH  →  R2N-CO-COOC2H5 + C2H5OH   (liquid dialkyl oxamic ester)
3° amine:  (COOC2H5)2 +   R3N   →  no reaction

Step 2 - Fractional distillation. The 3° amine (unreacted, low boiling) distils over first and is collected.

Step 3 - Cool/filter. The residue contains solid dialkyl oxamide (from 1° amine) and liquid oxamic ester (from 2° amine). The solid is filtered off; the liquid is the filtrate (or separated by distillation).

Step 4 - Hydrolysis with KOH regenerates the amines.

(CONHR)2 + 2 KOH            →  (COOK)2 + 2 RNH2        (1° amine recovered)
R2N-CO-COOC2H5 + 2 KOH      →  (COOK)2 + R2NH + C2H5OH   (2° amine recovered)

Thus all three amines are obtained separately: 3° by distillation, 1° from the solid oxamide, 2° from the liquid oxamic ester.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2080 Baishakh · 1 mark

Which one of the following is most basic in nature?
(a) $RNH_2$ (b) $C_6H_5NH_2$ (c) $(CH_3)_2NH$ (d) $NH_3$

Answer

(c) $(CH_3)_2NH$ . Dimethylamine is a secondary amine. Two electron-releasing ( $+I$ ) methyl groups raise the electron density on nitrogen, and in aqueous solution the resulting ion is well stabilised by solvation and hydrogen bonding. It is therefore more basic than ammonia and than aniline, whose lone pair is delocalised into the benzene ring. $RNH_2$ (with only one alkyl group) is less basic in water than a secondary amine.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2080 Baishakh · 1 mark

Write an example of carbylamine reaction.

Answer

The carbylamine (isocyanide) test: a primary amine heated with chloroform and alcoholic KOH gives a foul-smelling isocyanide.

CH3NH2 + CHCl3 + 3 KOH (alc.)  --heat-->  CH3-NC + 3 KCl + 3 H2O
(methylamine)                              (methyl isocyanide)

C6H5NH2 + CHCl3 + 3 KOH (alc.)  --heat-->  C6H5-NC + 3 KCl + 3 H2O
(aniline)                                   (phenyl isocyanide)

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2080 Baishakh · 2+1+2 marks

Benzoic --Reaction 1--> Benzene --Conc.HNO3 / Conc.H2SO4--> X --Sn/HCl--> Y
acid                                 (Reaction 2)          (Reaction 3)   |
                                                                   Reaction 4
                                                                          v
                                         Z <--H2O, boil-- BDC --Reaction 5--> A

a) Write down the reaction and condition of reaction 1 and reaction 4. [2]
b) Draw the structural formula of x and y. [1]
c) Identify the compound Z and A, where A is formed by reacting BDC with compound Z. Write chemical reaction and also give name of reaction. [2]
Figure: Reaction flow chart: Benzoic acid goes by Reaction 1 to Benzene; Benzene with Conc. HNO3/Conc. H2SO4 (Reaction 2) gives X; X with Sn/HCl (Reaction 3) gives Y; an arrow down from Y labelled Reaction 4 gives BDC; BDC with H2O on boiling gives Z (arrow pointing left); BDC by Reaction 5 (arrow pointing right) gives A.

Answer

a) Reaction 1 and Reaction 4

Reaction 1 (decarboxylation): benzoic acid is heated with soda lime (NaOH + CaO).

C6H5COOH + 2 NaOH  --CaO, heat-->  C6H6 + Na2CO3 + H2O
(i.e. C6H5COONa + NaOH  --CaO, heat-->  C6H6 + Na2CO3)

Reaction 4 (diazotisation): aniline (Y) is treated with NaNO $_2$ and HCl at 0-5 °C.

C6H5NH2 + NaNO2 + 2 HCl  --0-5 °C-->  C6H5N2+Cl- (BDC) + NaCl + 2 H2O

b) Structures

X = nitrobenzene, $C_6H_5-NO_2$ (a benzene ring bearing $-NO_2$ ).
Y = aniline, $C_6H_5-NH_2$ (a benzene ring bearing $-NH_2$ ).

c) Z and A

BDC = benzenediazonium chloride. Boiling with water gives Z = phenol:

C6H5N2Cl + H2O  --boil-->  C6H5OH + N2 + HCl

BDC reacts with phenol (Z) in alkaline medium at 0-5 °C to give A = p-hydroxyazobenzene (an orange azo dye). This is the azo coupling (coupling) reaction.

C6H5N2+Cl- + C6H5OH  --NaOH, 0-5 °C-->  C6H5-N=N-C6H4-OH (p) + HCl

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2080 Baishakh · 3+1+4 marks

There are three possible isomeric amine of $C_3H_6N$ .
i) Draw the structural formula of the isomer and compare their boiling points. [3]
ii) Predict the order of their increasing basic strength. [1]
iii) Separate these isomer by using Hoffmann's Method. [4]

Answer

The formula should be $C_3H_9N$ (three isomeric amines).

i) Isomers and boiling points

Isomer	Structure	Class	b.p.
Propan-1-amine	$CH_3CH_2CH_2NH_2$	1°	about 48 °C
N-methylethanamine	$CH_3CH_2NHCH_3$	2°	about 37 °C
N,N-dimethylmethanamine	$(CH_3)_3N$	3°	about 3 °C

Order of boiling point: 1° > 2° > 3°. A primary amine has two N-H bonds and forms the most extensive intermolecular hydrogen bonding; a secondary has one N-H, so less; a tertiary amine has no N-H and cannot form hydrogen bonds with its own molecules (only weak dipole forces).

ii) Basic strength (increasing)

$(CH_3)_3N < CH_3CH_2CH_2NH_2 < CH_3CH_2NHCH_3$

(tertiary < primary < secondary in aqueous solution, balancing the $+I$ effect, solvation of the ion and steric hindrance).

iii) Separation by Hoffmann's method

Principle: Hoffmann's method uses diethyl oxalate, $(COOC_2H_5)_2$ , which reacts differently with the three classes of amines.

Step 1 - Treat the mixture with diethyl oxalate.

1° amine:  (COOC2H5)2 + 2 RNH2  →  (CONHR)2  +  2 C2H5OH      (solid dialkyl oxamide)
2° amine:  (COOC2H5)2 +   R2NH  →  R2N-CO-COOC2H5 + C2H5OH   (liquid dialkyl oxamic ester)
3° amine:  (COOC2H5)2 +   R3N   →  no reaction

Step 2 - Fractional distillation. The 3° amine (unreacted, low boiling) distils over first and is collected.

Step 3 - Cool/filter. The residue contains solid dialkyl oxamide (from 1° amine) and liquid oxamic ester (from 2° amine). The solid is filtered off; the liquid is the filtrate (or separated by distillation).

Step 4 - Hydrolysis with KOH regenerates the amines.

(CONHR)2 + 2 KOH            →  (COOK)2 + 2 RNH2        (1° amine recovered)
R2N-CO-COOC2H5 + 2 KOH      →  (COOK)2 + R2NH + C2H5OH   (2° amine recovered)

For the isomers here: trimethylamine distils off; propylamine forms solid $(CONHCH_2CH_2CH_3)_2$ ; ethylmethylamine forms liquid $C_2H_5(CH_3)N-CO-COOC_2H_5$ ; hydrolysis with KOH regenerates the amines.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2081 Baishakh · 1 mark

$CH_3Cl \xrightarrow{Alc.\ AgCN} (A) \xrightarrow[\Delta]{LiAlH_4} (B)$ , what would be the product B?
(a) Methanamine (b) Ethanamine (c) N-methyl methanamine (d) N-methyl ethanamine

Answer

(c) N-methyl methanamine. Alcoholic AgCN with CH $_3$ Cl gives methyl isocyanide, $CH_3-NC$ (A), because AgCN is covalent and the nitrogen attacks. Reduction of an isocyanide with LiAlH $_4$ (or H $_2$ /Ni) gives a secondary amine with an N-methyl group:

CH3-NC + 4[H]  --LiAlH4-->  CH3-NH-CH3   (dimethylamine = N-methylmethanamine)

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2081 Baishakh · 1+2 marks

Starting from nitrobenzene, how would you prepare aniline? Convert aniline into:
a) phenyl isocyanide
b) azo-dye
[1+2]

Answer

Preparation of aniline from nitrobenzene (1 mark): reduction with Sn/conc. HCl (or Fe/HCl, or H $_2$ /Ni).

C6H5NO2 + 6[H]  --Sn/HCl-->  C6H5NH2 + 2 H2O

a) Aniline to phenyl isocyanide (carbylamine reaction):

C6H5NH2 + CHCl3 + 3 KOH (alc.)  --heat-->  C6H5NC + 3 KCl + 3 H2O

b) Aniline to azo dye:

Diazotisation: $C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5\,°C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O$
Coupling with phenol (alkaline) or aniline (weakly acidic) or $\beta$ -naphthol:

C6H5N2+Cl- + C6H5OH  --NaOH, 0-5 °C-->  C6H5-N=N-C6H4-OH (p-hydroxyazobenzene, orange dye) + HCl

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2081 Baishakh · 1+1+2+1 marks

A sequence of chemical reactions is represented as:
$A \xrightarrow[\text{Ice cold}]{NaNO_2 + dil.\,HCl} B \xrightarrow[\lbrack O\rbrack]{KMnO_4/H^+} CH_3CHO$ a) Identify the compounds (A) and (B). [1]
b) Write the laboratory test of compound (B). [1]
c) Starting from compound (B), how would you obtain nitrolic acid? [2]
d) Predict the product when compound (A) is heated with chloroform in alcoholic medium. [1]

Answer

a) A = ethanamine (ethylamine), $C_2H_5NH_2$ ; B = ethanol, $C_2H_5OH$ .

C2H5NH2 + HNO2 (NaNO2 + HCl) → C2H5OH + N2 + H2O
C2H5OH + [O] (KMnO4/H+) → CH3CHO + H2O

b) Test for ethanol (iodoform test): warm with iodine and NaOH; a yellow precipitate of iodoform with an antiseptic smell forms.

C2H5OH + 4 I2 + 6 NaOH → CHI3 (yellow ppt) + HCOONa + 5 NaI + 5 H2O

c) Ethanol to nitrolic acid:

C2H5OH + PI3 (or red P + I2) → C2H5I
C2H5I + AgNO2 → C2H5NO2 (nitroethane) + AgI
CH3CH2NO2 + HNO2 → CH3-C(=NOH)-NO2 (ethylnitrolic acid) + H2O

Nitroethane (a primary nitroalkane) reacts with nitrous acid to give nitrolic acid, which dissolves in NaOH to give a blood-red solution.

d) Ethylamine heated with chloroform and alcoholic KOH gives ethyl isocyanide (carbylamine reaction), a foul-smelling compound.

C2H5NH2 + CHCl3 + 3 KOH → C2H5NC + 3 KCl + 3 H2O

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2081 Baishakh · 4 marks

How do you separate the isomers of $C_3H_9N$ by Hoffmann's method? Explain with suitable chemical equations. [4]

Answer

The isomers of $C_3H_9N$ are propan-1-amine (1°), N-methylethanamine (2°) and N,N-dimethylmethanamine (3°). They are separated by treating the mixture with diethyl oxalate (Hoffmann's method).

1°:  (COOC2H5)2 + 2 CH3CH2CH2NH2 → (CONHC3H7)2 (solid) + 2 C2H5OH
2°:  (COOC2H5)2 + CH3CH2NHCH3 → CH3(C2H5)N-CO-COOC2H5 (liquid) + C2H5OH
3°:  (CH3)3N → no reaction

Distil the mixture: the 3° amine (trimethylamine, b.p. 3 °C) distils off.
Filter/cool the residue: solid dipropyloxamide (from 1° amine) is separated from the liquid oxamic ester (2° amine).
Hydrolyse each with KOH:

(CONHC3H7)2 + 2 KOH → (COOK)2 + 2 C3H7NH2
CH3(C2H5)N-CO-COOC2H5 + 2 KOH → (COOK)2 + CH3NHC2H5 + C2H5OH

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2082 Baishakh · 1 mark

Which of the following compounds gives positive carbylamine reaction?
(a) $\mathrm{(CH_3)_2NH}$ (b) $\mathrm{(CH_3)_3N}$ (c) $\mathrm{(CH_3)_4N^+}$ (d) $\mathrm{CH_3NH_2}$

Answer

(d) $CH_3NH_2$ . Only primary amines give the carbylamine test (foul-smelling isocyanide with $CHCl_3$ + alc. KOH). $(CH_3)_2NH$ is secondary, $(CH_3)_3N$ is tertiary and $(CH_3)_4N^+$ is a quaternary ion with no N-H, so none of these respond.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2082 Baishakh · 5 marks

A list of compounds is given as follows:

p-Aminoazobenzene, Benzenediazonium chloride, Aniline, Nitrobenzene, Phenol and Benzene

From the above list of compounds, prepare a sequence of reaction chain with suitable conditions and reagents.

Answer

Phenol is not required (it is the distractor; phenol would give p-hydroxyazobenzene). Sequence:

benzene → nitrobenzene → aniline → benzenediazonium chloride → p-aminoazobenzene.

1. C6H6 + HNO3  --conc. H2SO4, 60 °C-->  C6H5NO2 + H2O                (nitration)
2. C6H5NO2 + 6[H]  --Sn/HCl-->  C6H5NH2 + 2 H2O                       (reduction)
3. C6H5NH2 + NaNO2 + 2 HCl  --0-5 °C-->  C6H5N2+Cl- + NaCl + 2 H2O    (diazotisation)
4. C6H5N2+Cl- + C6H5NH2  --pH 4-5, 0-5 °C-->  C6H5-N=N-C6H4-NH2 (p) + HCl   (coupling)

The product p-aminoazobenzene (aniline yellow) is a yellow azo dye; coupling occurs at the para position of aniline.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2082 Baishakh · 3+2 marks

You are given the mixture of two compounds A and B.
$\mathrm{A = CH_3CH_2CH_2\text{-}NH_2}\quad\text{and}\quad \mathrm{B = CH_3CH_2NHCH_3}$ How would you separate their mixture by applying Hoffmann's method? Give the nitrous acid test to distinguish between compounds A and B.

Answer

Separation (Hoffmann's method, 3 marks)

A is a primary amine (propan-1-amine); B is a secondary amine (N-methylethanamine). Add diethyl oxalate to the mixture.

A: (COOC2H5)2 + 2 CH3CH2CH2NH2 → (CONHCH2CH2CH3)2 (solid dipropyloxamide) + 2 C2H5OH
B: (COOC2H5)2 + CH3CH2NHCH3 → CH3(C2H5)N-CO-COOC2H5 (liquid oxamic ester) + C2H5OH

Filter off the solid (from A); separate the liquid (from B) by distillation. Hydrolyse each with KOH:

(CONHC3H7)2 + 2 KOH → (COOK)2 + 2 C3H7NH2   (A recovered)
CH3(C2H5)N-CO-COOC2H5 + 2 KOH → (COOK)2 + CH3NHC2H5 + C2H5OH   (B recovered)

Nitrous acid test (2 marks)

	A (1° amine)	B (2° amine)
With NaNO $_2$ + HCl	Brisk effervescence of N $_2$ gas; propan-1-ol formed	Yellow oily liquid (N-nitrosoamine) forms; no gas
Equation	$C_3H_7NH_2 + HNO_2 \rightarrow C_3H_7OH + N_2 + H_2O$	$CH_3(C_2H_5)NH + HNO_2 \rightarrow CH_3(C_2H_5)N-N=O + H_2O$

The nitrosoamine gives a green colour with phenol and conc. H $_2$ SO $_4$ (Liebermann's test).

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2082 Ashoj · Supplementary · 1 mark

An organic compound (A) on reduction gives compound (B). (B) on treatment with $\mathrm{CHCl_3}$ and alcoholic KOH gives (C). (C) on catalytic reduction gives N-methylaniline. What compound is (A)?
(a) Methylamine (b) Aniline (c) Nitrobenzene (d) Nitromethane

Answer

(c) Nitrobenzene.

C6H5NO2  --Sn/HCl-->  C6H5NH2 (B, aniline)
C6H5NH2 + CHCl3 + 3 KOH  --alc., heat-->  C6H5NC (C, phenyl isocyanide)
C6H5NC + 4[H]  --Ni/H2-->  C6H5NHCH3 (N-methylaniline)

A is nitrobenzene: its reduction gives aniline, a primary amine that gives the carbylamine test, and catalytic reduction of the isocyanide gives N-methylaniline.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2082 Ashoj · Supplementary · 2+1+1+1 marks

Starting from benzene, how will you prepare aniline? What happens when aniline is treated with:
(a) benzene diazonium chloride
(b) aqueous bromine
(c) benzaldehyde

Answer

Preparation of aniline from benzene (2 marks)

C6H6 + HNO3  --conc. H2SO4, 60 °C-->  C6H5NO2 + H2O            (nitration)
C6H5NO2 + 6[H]  --Sn/conc. HCl (or Fe/HCl)-->  C6H5NH2 + 2 H2O  (reduction)

Reactions of aniline

(a) With benzenediazonium chloride (weakly acidic, pH 4-5, 0-5 °C): azo coupling at the para position gives yellow p-aminoazobenzene.

C6H5N2+Cl- + C6H5NH2 → C6H5-N=N-C6H4-NH2 (p) + HCl

(b) With aqueous bromine: a white precipitate of 2,4,6-tribromoaniline forms immediately.

C6H5NH2 + 3 Br2 (aq) → C6H2Br3NH2 + 3 HBr

(c) With benzaldehyde: condensation gives the Schiff base benzylideneaniline (N-benzylideneaniline).

C6H5NH2 + OHC-C6H5 → C6H5-N=CH-C6H5 + H2O

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2082 Ashoj · Supplementary · 4 marks

Write down two isomeric amines and their IUPAC name for $\mathrm{C_2H_7N}$ . How would you apply Hoffmann's method to separate them from their mixture?

Answer

Isomers

Structure	IUPAC name	Class
$CH_3CH_2NH_2$	Ethanamine	1°
$CH_3NHCH_3$	N-methylmethanamine	2°

Separation by Hoffmann's method

Treat the mixture with diethyl oxalate.

CH3CH2NH2 (1°): (COOC2H5)2 + 2 C2H5NH2 → (CONHC2H5)2 (solid diethyloxamide) + 2 C2H5OH
(CH3)2NH (2°):  (COOC2H5)2 + (CH3)2NH → (CH3)2N-CO-COOC2H5 (liquid oxamic ester) + C2H5OH

The solid is filtered off and separated from the liquid by distillation. Each is hydrolysed with KOH:

(CONHC2H5)2 + 2 KOH → (COOK)2 + 2 C2H5NH2
(CH3)2N-CO-COOC2H5 + 2 KOH → (COOK)2 + (CH3)2NH + C2H5OH

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2082 Ashoj · Supplementary · 5 marks

An amine with molecular formula C $_4$ H $_9$ N has three isomers. Name those isomers and explain the method of their separation by using Hoffmann's method.

Answer

The formula $C_4H_9N$ is not a saturated amine; the three isomers (primary, secondary, tertiary) are those of $C_4H_{11}N$ . Representative isomers:

Class	Isomer (IUPAC name)
1°	Butan-1-amine, $CH_3CH_2CH_2CH_2NH_2$
2°	N-ethylethanamine (diethylamine), $(C_2H_5)_2NH$
3°	N,N-dimethylethanamine, $(CH_3)_2NC_2H_5$

Hoffmann's method of separation

Principle: Hoffmann's method uses diethyl oxalate, $(COOC_2H_5)_2$ , which reacts differently with the three classes of amines.

Step 1 - Treat the mixture with diethyl oxalate.

1° amine:  (COOC2H5)2 + 2 RNH2  →  (CONHR)2  +  2 C2H5OH      (solid dialkyl oxamide)
2° amine:  (COOC2H5)2 +   R2NH  →  R2N-CO-COOC2H5 + C2H5OH   (liquid dialkyl oxamic ester)
3° amine:  (COOC2H5)2 +   R3N   →  no reaction

Step 2 - Fractional distillation. The 3° amine (unreacted, low boiling) distils over first and is collected.

Step 3 - Cool/filter. The residue contains solid dialkyl oxamide (from 1° amine) and liquid oxamic ester (from 2° amine). The solid is filtered off; the liquid is the filtrate (or separated by distillation).

Step 4 - Hydrolysis with KOH regenerates the amines.

(CONHR)2 + 2 KOH            →  (COOK)2 + 2 RNH2        (1° amine recovered)
R2N-CO-COOC2H5 + 2 KOH      →  (COOK)2 + R2NH + C2H5OH   (2° amine recovered)

Here N,N-dimethylethanamine distils off first, butan-1-amine comes from the solid dibutyloxamide, and diethylamine comes from the liquid diethyloxamic ester.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2083 Baishakh · 1 mark

You are given (I) aniline, (II) o-nitro aniline, (III) p-nitro aniline and (IV) m-nitroaniline. What is the order of increasing basic nature?
(a) II < I < III < IV (b) I < II < III < IV (c) II < III < I < IV (d) II < III < IV < I

Answer

(d) II < III < IV < I

Aniline is the most basic. The $-NO_2$ group is electron-withdrawing, which lowers the electron density on nitrogen and reduces basicity.

m-nitroaniline (IV): only the $-I$ effect operates, so it is a little weaker than aniline.
p-nitroaniline (III): both $-I$ and the $-R$ (resonance) effect withdraw electron density, so it is weaker.
o-nitroaniline (II): $-I$ , $-R$ and the ortho effect (steric/intramolecular H-bonding) make it the weakest base.

Increasing basicity: o-nitro < p-nitro < m-nitro < aniline.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

2083 Baishakh · 4 marks

Explain Hoffmann's method to separate ethanamine and N-methylmethanamine from their mixture. [4]

Answer

Ethanamine ( $C_2H_5NH_2$ , 1°) and N-methylmethanamine ( $(CH_3)_2NH$ , 2°) are separated with diethyl oxalate.

2 C2H5NH2 + (COOC2H5)2 → (CONHC2H5)2 (solid N,N'-diethyloxamide) + 2 C2H5OH
(CH3)2NH + (COOC2H5)2 → (CH3)2N-CO-COOC2H5 (liquid ethyl N,N-dimethyloxamate) + C2H5OH

The solid diethyloxamide is separated by filtration (or from the liquid by fractional distillation).
Each product is hydrolysed with aqueous KOH:

(CONHC2H5)2 + 2 KOH → (COOK)2 + 2 C2H5NH2        (ethanamine recovered)
(CH3)2N-CO-COOC2H5 + 2 KOH → (COOK)2 + (CH3)2NH + C2H5OH   (dimethylamine recovered)

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

2083 Baishakh · 1 mark

What happens when aniline is added to ice-cold solution of sodium nitrite and hydrochloric acid?

Answer

Aniline reacts with ice-cold NaNO $_2$ + HCl (0-5 °C) to give benzenediazonium chloride (diazotisation). The solution stays clear and no gas is evolved.

C6H5NH2 + NaNO2 + 2 HCl  --0-5 °C-->  C6H5N2+Cl- + NaCl + 2 H2O

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

CDC Grade 12 Chemistry Model Question 2078 · 1+1+1+1+1 marks

A list of compounds are given as follows:

p-hydroxyazobenzene, $C_6H_5N_2Cl$ , $C_6H_5NH_2$ , $C_6H_5NO_2$ , $C_6H_6$
From the above list of compounds, prepare a sequence of reaction chain with suitable conditions and reactions. (1+1+1+1+1)

Answer

Reaction chain: benzene → nitrobenzene → aniline → benzenediazonium chloride → p-hydroxyazobenzene.

C6H6  --conc.HNO3 + conc.H2SO4, 50-60 °C-->  C6H5NO2
C6H5NO2  --Sn/HCl (or Fe/HCl), reduction-->  C6H5NH2
C6H5NH2  --NaNO2 + HCl, 0-5 °C (diazotisation)-->  C6H5N2Cl   [C6H5-N≡N]+ Cl-
C6H5N2Cl + C6H5OH  --NaOH, 0-5 °C (coupling)-->  C6H5-N=N-C6H4-OH (p)  + HCl

Benzene to nitrobenzene: nitration with the nitrating mixture (conc. HNO $_3$ + conc. H $_2$ SO $_4$ ) at about 60 °C.
Nitrobenzene to aniline: reduction with Sn/HCl (or Fe/HCl or H $_2$ /Ni).
Aniline to benzenediazonium chloride: diazotisation with NaNO $_2$ + HCl at 0-5 °C.
Coupling: the diazonium salt couples with phenol in mildly alkaline medium at the para position to give the orange p-hydroxyazobenzene (an azo dye).

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

CDC Grade 12 Chemistry Model Question 2078 · 2 marks

An aromatic compound known as oil of mirabane is prepared from benzene. Give the complete reaction for the conversion of the compound into yellow dye.

Answer

Oil of mirabane is nitrobenzene ( $C_6H_5NO_2$ ), obtained by nitrating benzene. It is converted to the yellow dye p-aminoazobenzene as follows.

(i)   C6H5NO2 + 6[H]  --Sn/HCl-->  C6H5NH2 + 2 H2O                 (aniline)
(ii)  C6H5NH2 + NaNO2 + 2 HCl  --0-5 °C-->  C6H5N2+Cl- + NaCl + 2 H2O   (benzenediazonium chloride)
(iii) C6H5N2+Cl- + C6H5NH2  --weakly acidic (pH 4-5)-->  C6H5-N=N-C6H4-NH2 (p) + HCl

The final product, p-aminoazobenzene (aniline yellow), is the yellow azo dye formed by coupling at the para position of aniline.

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

CDC Grade 12 Chemistry Model Question 2078 · 5 marks

Starting from compound P, how do the reactions proceed ahead to obtain T which gives benzene where R is aniline? Complete the reaction sequence with suitable conditions. [5x1=5]

      T  ---->  P
      ^         |
      |         v
      S         Q
       ^       /
        \     v
           R

Figure: Cyclic arrow diagram: P -> Q -> R -> S -> T -> P (T at top-left, P at top-right, Q at right, R at bottom, S at left), with curved arrows going clockwise.

Answer

Since R is aniline, the cycle is: P = benzene, Q = nitrobenzene, R = aniline, S = benzenediazonium chloride, T = phenol, and T gives back benzene (P).

P → Q:  C6H6 + HNO3  --conc. H2SO4, 60 °C-->  C6H5NO2 + H2O          (nitration)
Q → R:  C6H5NO2 + 6[H]  --Sn/HCl-->  C6H5NH2 + 2 H2O                  (reduction)
R → S:  C6H5NH2 + NaNO2 + 2 HCl  --0-5 °C-->  C6H5N2Cl + NaCl + 2 H2O (diazotisation)
S → T:  C6H5N2Cl + H2O  --warm (boil)-->  C6H5OH + N2 + HCl           (hydrolysis)
T → P:  C6H5OH + Zn  --distil-->  C6H6 + ZnO                          (reduction of phenol with zinc dust)

Compound	Name
P	Benzene
Q	Nitrobenzene
R	Aniline
S	Benzenediazonium chloride
T	Phenol

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

NEB Class 12 Model Question 2079 · 1 mark

What are the isomers of formula $C_2H_7N$ ? (1)

Answer

Two isomers:

$CH_3CH_2NH_2$ , ethanamine (ethylamine), a primary amine
$CH_3NHCH_3$ , N-methylmethanamine (dimethylamine), a secondary amine

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 1 MCQ

Loading…

NEB Class 12 Model Question 2079 · 2+3+1+2 marks

Amines are formally derivatives of ammonia, wherein one or more hydrogen atoms have been replaced by a substituent such as an alkyl or aryl group which may respectively be called alkyl amines and aryl amines.
a) Give a test to distinguish alkyl amine and aryl amine. (2)
b) How can you separate ethylamine and dimethylamine present in mixture by Hoffmann's method? (3)
c) Arrange the following amines in terms of increasing order of basic strength. (1)
Propyl amine, ethyl methylamine, trimethylamine.
d) Convert: ethanamine to methanamine (2)

Answer

a) Test to distinguish alkyl and aryl amines (azo dye test)

Treat the primary amine with NaNO $_2$ + HCl at 0-5 °C, then add alkaline $\beta$ -naphthol solution.

Aryl amine (e.g. aniline) forms a stable diazonium salt, which couples with $\beta$ -naphthol to give an orange-red azo dye.
Alkyl amine (e.g. ethylamine) gives an unstable diazonium salt that decomposes at once, with effervescence of N $_2$ and formation of alcohol; no dye forms.

C6H5NH2 → C6H5N2+Cl- → (β-naphthol, NaOH) → orange-red dye
C2H5NH2 + HNO2 → C2H5OH + N2↑ + H2O

b) Separation of ethylamine and dimethylamine (Hoffmann's method)

2 C2H5NH2 + (COOC2H5)2 → (CONHC2H5)2 (solid) + 2 C2H5OH
(CH3)2NH + (COOC2H5)2 → (CH3)2N-CO-COOC2H5 (liquid) + C2H5OH

Filter off the solid and separate the liquid; hydrolyse each with KOH:

(CONHC2H5)2 + 2 KOH → (COOK)2 + 2 C2H5NH2
(CH3)2N-CO-COOC2H5 + 2 KOH → (COOK)2 + (CH3)2NH + C2H5OH

c) Increasing basic strength

$(CH_3)_3N < CH_3CH_2CH_2NH_2 < CH_3CH_2NHCH_3$

(trimethylamine < propylamine < ethylmethylamine; 3° < 1° < 2° in water).

d) Ethanamine to methanamine (Hoffmann bromamide degradation)

C2H5NH2 --HNO2--> C2H5OH --[O], KMnO4/H+--> CH3COOH
CH3COOH + NH3 → CH3COONH4 --heat--> CH3CONH2 + H2O
CH3CONH2 + Br2 + 4 KOH → CH3NH2 + K2CO3 + 2 KBr + 2 H2O

Answer checked by Opus 5.5 · Oct 18, 2026

Test yourself · 2 MCQs

Loading…

NEB Class 12 Chemistry 3021 Model Question, Set A · 1 mark

Identify Z in the following reaction.
$Z \xrightarrow[\Delta]{CHCl_3/\text{alc. }KOH} X \xrightarrow[\Delta]{LiAlH_4} C_6H_5\text{-}NH\text{-}CH_3$ (a) aniline (b) benzene diazonium chloride (c) phenyl isocyanide (d) nitrobenzene

Answer

(a) aniline

$CHCl_3$ with alcoholic KOH on heating is the carbylamine reaction, given only by primary amines. So Z is a primary amine and X is the isocyanide. $LiAlH_4$ reduces an isocyanide ( $R-NC$ ) to a secondary amine ( $R-NH-CH_3$ ). Since the final product is $C_6H_5NHCH_3$ , X is phenyl isocyanide and Z is aniline:

C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O

C_6H_5NC + 4[H] \xrightarrow{LiAlH_4} C_6H_5NHCH_3

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

NEB Class 12 Chemistry 3021 Model Question, Set A · 5+1+2 marks

In the following reaction sequence:
$P \xrightarrow{KMnO_4/H^+} Q \xrightarrow[\Delta]{NH_3} R \xrightarrow{Br_2 + \text{aq. }KOH} S \xrightarrow[\text{cold}]{NaNO_2 + HCl} T$ a) Identify P, Q, R, S and T with the necessary chemical reactions. [5]
b) What happens when Q is heated with soda lime? [1]
c) Convert compound S into p-hydroxyazobenzene. [2]

Answer

a) Identification

The last step gives a diazonium salt, and part (c) shows S is aniline. So the compounds are:

P = toluene, $C_6H_5CH_3$
Q = benzoic acid, $C_6H_5COOH$
R = benzamide, $C_6H_5CONH_2$
S = aniline, $C_6H_5NH_2$
T = benzene diazonium chloride, $C_6H_5N_2^+Cl^-$

Reactions:

C_6H_5CH_3 \xrightarrow{KMnO_4/H^+} C_6H_5COOH

C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O

C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O

C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{0-5\,^\circ C} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

b) Q heated with soda lime

Benzoic acid undergoes decarboxylation to give benzene:

C_6H_5COOH + 2NaOH(\text{soda lime}) \xrightarrow{CaO,\ \Delta} C_6H_6 + Na_2CO_3 + H_2O

c) Aniline to p-hydroxyazobenzene

Aniline is first diazotised (as in part a), and the cold diazonium salt is then coupled with phenol in alkaline medium (aq. NaOH, 0–5 °C). The coupling takes place at the para position of phenol:

C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH,\ 0-5\,^\circ C} C_6H_5\text{-}N{=}N\text{-}C_6H_4\text{-}OH\,(p) + HCl

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

NEB Class 12 Chemistry 3021 Model Question, Set B · 1 mark

Give an example of diazotization reaction.

Answer

Diazotization is the conversion of a primary aromatic amine to a diazonium salt by treatment with nitrous acid at 0-5 $^\circ C$ . For example, aniline gives benzenediazonium chloride:

C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\ K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 1 MCQ

Loading…

NEB Class 12 Chemistry 3021 Model Question, Set B · 5 marks

A list of compounds is given as follows:
benzamide, aniline, benzene, benzoic acid, toluene
Prepare a correct sequence of reactions with suitable conditions and reagents. [5]

Answer

A correct sequence is benzene $\rightarrow$ toluene $\rightarrow$ benzoic acid $\rightarrow$ benzamide $\rightarrow$ aniline.

Step 1. Benzene to toluene (Friedel-Crafts alkylation):

C_6H_6 + CH_3Cl \xrightarrow{anhyd.\ AlCl_3} C_6H_5CH_3 + HCl

Step 2. Toluene to benzoic acid (oxidation of the side chain):

C_6H_5CH_3 \xrightarrow{\text{alk. } KMnO_4,\ \Delta;\ then\ H_3O^+} C_6H_5COOH

Step 3. Benzoic acid to benzamide:

C_6H_5COOH + NH_3 \rightarrow C_6H_5COONH_4 \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O

Step 4. Benzamide to aniline (Hofmann bromamide degradation, one carbon is lost):

C_6H_5CONH_2 + Br_2 + 4NaOH \rightarrow C_6H_5NH_2 + Na_2CO_3 + 2NaBr + 2H_2O

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

NEB Class 12 Chemistry 3021 Model Question, Set B · 4+1+3 marks

a) Explain Hofmann's method for the separation of 1° and 2° amines from their mixture. [4]
What is meant by Hofmann's ammonolysis? [1]
b) What happens when ethanal reacts with: [3]
i) 2,4-DNPH (2,4-dinitrophenylhydrazine)
ii) HCN
iii) Tollens' reagent

Answer

(a) Hofmann's method of separation. The mixture of amines is treated with diethyl oxalate, $(COOC_2H_5)_2$ :

The primary amine gives a solid dialkyl oxamide: $(COOC_2H_5)_2 + 2RNH_2 \rightarrow (CONHR)_2 + 2C_2H_5OH$
The secondary amine gives a liquid dialkyl oxamic ester: $(COOC_2H_5)_2 + R_2NH \rightarrow R_2NCOCOOC_2H_5 + C_2H_5OH$

The solid and liquid products are separated by distillation or filtration. Each is then hydrolysed by boiling with KOH, which regenerates the pure primary and secondary amines respectively.

Hofmann's ammonolysis is the reaction of an alkyl halide with an alcoholic solution of ammonia, heated in a sealed tube. The halogen is replaced by $-NH_2$ and a mixture of amines results:

RX + NH_3 \rightarrow RNH_2 \xrightarrow{RX} R_2NH \xrightarrow{RX} R_3N \xrightarrow{RX} R_4N^+X^-

(b) Reactions of ethanal.

i) With 2,4-DNPH it forms an orange-yellow precipitate of ethanal 2,4-dinitrophenylhydrazone:

CH_3CHO + H_2NNHC_6H_3(NO_2)_2 \rightarrow CH_3CH=NNHC_6H_3(NO_2)_2 + H_2O

ii) With HCN it forms ethanal cyanohydrin (2-hydroxypropanenitrile):

CH_3CHO + HCN \rightarrow CH_3CH(OH)CN

iii) With Tollens' reagent it is oxidised to ethanoate, and a silver mirror is deposited:

CH_3CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3COO^- + 2Ag\downarrow + 4NH_3 + 2H_2O

Answer checked by Claude Sonnet · Oct 21, 2026

Test yourself · 2 MCQs

Loading…

Past questions and answers

1.The number of possible structural isomers of 1° amines of molecular formula C4H11NC_4H_{11}NC4​H11​N give (a) 1 (b) 2 (c) 3 (d) 4

2.How would you apply Hoffmann's method for the separation of 1°, 2° and 3° amine from their mixture?

3.Which one of the following is most basic in nature? (a) RNH2RNH_2RNH2​ (b) C6H5NH2C_6H_5NH_2C6​H5​NH2​ (c) (CH3)2NH(CH_3)_2NH(CH3​)2​NH (d) NH3NH_3NH3​

4.Write an example of carbylamine reaction.

6.There are three possible isomeric amine of C3H6NC_3H_6NC3​H6​N. i) Draw the structural formula of the isomer and compare their boiling points. [3] ii) Predict the order of their increasing basic strength. [1] iii) Separate these isomer by using Hoffmann's Method. [4]

7.CH3Cl→Alc. AgCN(A)→ΔLiAlH4(B)CH_3Cl \xrightarrow{Alc.\ AgCN} (A) \xrightarrow[\Delta]{LiAlH_4} (B)CH3​ClAlc. AgCN​(A)LiAlH4​Δ​(B), what would be the product B? (a) Methanamine (b) Ethanamine (c) N-methyl methanamine (d) N-methyl ethanamine

8.Starting from nitrobenzene, how would you prepare aniline? Convert aniline into: a) phenyl isocyanide b) azo-dye [1+2]

10.How do you separate the isomers of C3H9NC_3H_9NC3​H9​N by Hoffmann's method? Explain with suitable chemical equations. [4]

11.Which of the following compounds gives positive carbylamine reaction? (a) (CH3)2NH\mathrm{(CH_3)_2NH}(CH3​)2​NH (b) (CH3)3N\mathrm{(CH_3)_3N}(CH3​)3​N (c) (CH3)4N+\mathrm{(CH_3)_4N^+}(CH3​)4​N+ (d) CH3NH2\mathrm{CH_3NH_2}CH3​NH2​

12.A list of compounds is given as follows: p-Aminoazobenzene, Benzenediazonium chloride, Aniline, Nitrobenzene, Phenol and Benzene From the above list of compounds, prepare a sequence of reaction chain with suitable conditions and reagents.

14.An organic compound (A) on reduction gives compound (B). (B) on treatment with CHCl3\mathrm{CHCl_3}CHCl3​ and alcoholic KOH gives (C). (C) on catalytic reduction gives N-methylaniline. What compound is (A)? (a) Methylamine (b) Aniline (c) Nitrobenzene (d) Nitromethane

15.Starting from benzene, how will you prepare aniline? What happens when aniline is treated with: (a) benzene diazonium chloride (b) aqueous bromine (c) benzaldehyde

16.Write down two isomeric amines and their IUPAC name for C2H7N\mathrm{C_2H_7N}C2​H7​N. How would you apply Hoffmann's method to separate them from their mixture?

17.An amine with molecular formula C4_44​H9_99​N has three isomers. Name those isomers and explain the method of their separation by using Hoffmann's method.

18.You are given (I) aniline, (II) o-nitro aniline, (III) p-nitro aniline and (IV) m-nitroaniline. What is the order of increasing basic nature? (a) II < I < III < IV (b) I < II < III < IV (c) II < III < I < IV (d) II < III < IV < I

19.Explain Hoffmann's method to separate ethanamine and N-methylmethanamine from their mixture. [4]

20.What happens when aniline is added to ice-cold solution of sodium nitrite and hydrochloric acid?

22.An aromatic compound known as oil of mirabane is prepared from benzene. Give the complete reaction for the conversion of the compound into yellow dye.

24.What are the isomers of formula C2H7NC_2H_7NC2​H7​N? (1)

28.Give an example of diazotization reaction.

29.A list of compounds is given as follows: benzamide, aniline, benzene, benzoic acid, toluene Prepare a correct sequence of reactions with suitable conditions and reagents. [5]

30.a) Explain Hofmann's method for the separation of 1° and 2° amines from their mixture. [4] What is meant by Hofmann's ammonolysis? [1] b) What happens when ethanal reacts with: [3] i) 2,4-DNPH (2,4-dinitrophenylhydrazine) ii) HCN iii) Tollens' reagent

The number of possible structural isomers of 1° amines of molecular formula $C_4H_{11}N$ give
(a) 1 (b) 2 (c) 3 (d) 4

How would you apply Hoffmann's method for the separation of 1°, 2° and 3° amine from their mixture?

Which one of the following is most basic in nature?
(a) $RNH_2$ (b) $C_6H_5NH_2$ (c) $(CH_3)_2NH$ (d) $NH_3$

Write an example of carbylamine reaction.

There are three possible isomeric amine of $C_3H_6N$ .
i) Draw the structural formula of the isomer and compare their boiling points. [3]
ii) Predict the order of their increasing basic strength. [1]
iii) Separate these isomer by using Hoffmann's Method. [4]

$CH_3Cl \xrightarrow{Alc.\ AgCN} (A) \xrightarrow[\Delta]{LiAlH_4} (B)$ , what would be the product B?
(a) Methanamine (b) Ethanamine (c) N-methyl methanamine (d) N-methyl ethanamine

Starting from nitrobenzene, how would you prepare aniline? Convert aniline into:
a) phenyl isocyanide
b) azo-dye
[1+2]

How do you separate the isomers of $C_3H_9N$ by Hoffmann's method? Explain with suitable chemical equations. [4]

Which of the following compounds gives positive carbylamine reaction?
(a) $\mathrm{(CH_3)_2NH}$ (b) $\mathrm{(CH_3)_3N}$ (c) $\mathrm{(CH_3)_4N^+}$ (d) $\mathrm{CH_3NH_2}$

A list of compounds is given as follows:

p-Aminoazobenzene, Benzenediazonium chloride, Aniline, Nitrobenzene, Phenol and Benzene

From the above list of compounds, prepare a sequence of reaction chain with suitable conditions and reagents.

An organic compound (A) on reduction gives compound (B). (B) on treatment with $\mathrm{CHCl_3}$ and alcoholic KOH gives (C). (C) on catalytic reduction gives N-methylaniline. What compound is (A)?
(a) Methylamine (b) Aniline (c) Nitrobenzene (d) Nitromethane

Starting from benzene, how will you prepare aniline? What happens when aniline is treated with:
(a) benzene diazonium chloride
(b) aqueous bromine
(c) benzaldehyde

Write down two isomeric amines and their IUPAC name for $\mathrm{C_2H_7N}$ . How would you apply Hoffmann's method to separate them from their mixture?

An amine with molecular formula C $_4$ H $_9$ N has three isomers. Name those isomers and explain the method of their separation by using Hoffmann's method.

You are given (I) aniline, (II) o-nitro aniline, (III) p-nitro aniline and (IV) m-nitroaniline. What is the order of increasing basic nature?
(a) II < I < III < IV (b) I < II < III < IV (c) II < III < I < IV (d) II < III < IV < I

Explain Hoffmann's method to separate ethanamine and N-methylmethanamine from their mixture. [4]

What happens when aniline is added to ice-cold solution of sodium nitrite and hydrochloric acid?

An aromatic compound known as oil of mirabane is prepared from benzene. Give the complete reaction for the conversion of the compound into yellow dye.

What are the isomers of formula $C_2H_7N$ ? (1)

Give an example of diazotization reaction.

A list of compounds is given as follows:
benzamide, aniline, benzene, benzoic acid, toluene
Prepare a correct sequence of reactions with suitable conditions and reagents. [5]

a) Explain Hofmann's method for the separation of 1° and 2° amines from their mixture. [4]
What is meant by Hofmann's ammonolysis? [1]
b) What happens when ethanal reacts with: [3]
i) 2,4-DNPH (2,4-dinitrophenylhydrazine)
ii) HCN
iii) Tollens' reagent