The document contains 12 multiple choice questions about amines. It tests knowledge of the properties of primary, secondary, and tertiary amines as well as reactions such as diazotization. It also contains short answer questions testing understanding of amine acid-base properties, conversions between amine derivatives, and distinguishing chemical tests.

1. Multiple Choice Type Questions
1. Amine is a derivative of-
(a) Hydrocarbon (b) Ammonia (c) Alcohol (d) Aldehyde. Ans.(b)
2. Common aryl amine is-
(a) CH3NH2 (b) C6H5–NH2 (c) C2H5–NH2 (d) (CH3)2NH Ans. (b)
3. Aniline is-
(a) More basic than NH3 (b) Less basic than NH3 (c) Equal to basic strength of NH3 (d) None of these. Ans. (b)
4. C2H5NH2 + HONO
NaN O2+HCl
Product, the ‘Product’ is-
(a) CH3OH, (b) C2H5OH, (c) C2H5NC, (d) None of these. Ans. (b)
5. C6H5NH2
NaN O2+HCl
Product, the ‘Product’ is-
(a) C6H5N2
+
Cl-
, (b) C6H5-I, (c) C6H5NO2, (d) None of these. Ans. (a)
6. Basic strength of alkyl amine is-
(a) 10>20>30 (b) 20>10>30 (c) 30>20>10 (d) 10>30>20 Ans. (b)
7. R3N is known as-
(a) 10
amine (b) 20
amine (c) 30
amine (d) None of these. Ans. (c)
8. Structure of (CH3)3 N is-
(a) Square planar (b) Pyramidal (c) Bipyramidal (d) Hexagonal. Ans. (b)
9. General formula of primary amine is-
(a) R2NH (b) R-NH2 (c) R3N (d) R4N+ Ans. (b)
10. Bond angle in trimethyl amine is-
(a) 1090
(b) 1080
(c) 1090
(d) 1050
Ans. (b)
11. Organic compound containing N-atom Hylerization of N-atom in amine is-
(a) Sp2
(b) Sp3
(c) Sp (d) Sp3
d Ans. (b)
12. IUPAC name of the compound CH3 –CH –CH3
(a) Methyl amine (b) Prop-2 amine (c) Dimethyl (d) Ethyl amine. Ans. (b)
Very short Answer Type Questions
1. Name the amide which on reduction gives
Ans. Butanamide NH2
2. Complete the reaction
Ans:-
3. How is aniline converted to sulphanilicacid ?
4. What are ambidentnucleophiles ? Give two examples.
Ans. A nucleophile which can form bonds through more than one atom in it, e.g., NO-
2 and CN (through C and N).
5. What is the directive influence of the amino group in arylamines ?
Ans. o, p-directing.
6. Write down the IUPAC name of-
(a) (C2H5)2 NH,
(b) (CH3)3N.
Ans.(a) N-Ethylethanamine,
(b) N, N-Dimethylmethanamine.
7. What is a coupling reaction ?
Ans. The reaction of diazonium salts with phenols in basic medium (pH 9-10) and with amines in acidic medium (pH
4-
5) to give corresponding azo (- N = N -) dyes is called coupling reaction.
8. Aniline dissolves in aqueousHCl. Why ?
Ans. Aniline dissolves in aqueous HCl due to the formation of water soluble salt.
NH3
NH2
O
CHO + H2 N–R →
TEST SERIES 2023-24
CHEMISTRY XII :- AMINES

2. 9. How will you convert aniline to iodobenzene ?
10. Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained.
Short Answer Type Questions
1. Why is it necessary to add excess of mineral acids t diazotization of amines ?
Ans. During diazotization of arylamines, excess of mineral acids is used. In general, one mole of amine is treated with
approx three moles of acid (a) One mole to dissolve amine, (b) one mole to liberate HNO2 from NaNO2 and (c) one
mole to maintain proper acidity of the reaction mixture to prevent its coupling with diazotized salt.
2. Aniline does not undergo Friedel-Crafts reactions. Explain.
Ans. Aniline is a base while AlCl3 used as a catalyst in Friedel-Crafts reaction is a Lewis acid. They combine with each
other to form the salt.
C6H5NH2 + AlCl3 →C6H5N+
H2 AlCl3
-
Due to the presence of a positive charge on N-atom, the group N+
H2AlCl3 acts as strongly deactivating group. As a
result, it reduces the electron density in the benzene ring and hence aniline does not undergo Friedel craft raction.
3. Aniline is a weaker base than ethyl amine. Explain.
Ans. Aniline has lone electron pair at nitrogen atom which is attached with pi-electron of ring. Therefore, nitrogen
atom is positively charged and tends to attract electrons. Whereas, in ethyl amine nitrogenatom is sp3
hybridized
and tends to withdraw electrons. Therefore, ethyl amine more casily attracts electron pair and acts as a lewis base.
Thus, aniline is more weaker base than ethyl amine.
4. How will you convert-
(a) Benzene into aniline,
(b) Benzene into N, N-dimethylaniline?
5. Pyridine is more basic than pyrrole. Why ?
Ans. In case of pyridine, the lone pair of ‘N’ –atom is not involved in hextet formation and hence can be easiy
available for combination with proton (H+
) hence pyridine is more basic.
Pyridine can structurally represented as-
In pyrrole, the lone pairs of electrons of N atom are involved in hextet formation and therefore not readily available
for combination with H+
Hence pyrrole is less basic. From these fact it is clear that Pyridine is more basic than pyrrole.

3. 6. Boiling point of 10
-amine is more than 30
-amine. Why ?
Ans. This is because maximum intermolecular hydrogen bonding in primary amines (two H-atoms are available),
lesser in secondary amines (only one H-atom is available) and least in tertiary amines (no H-atom is available).
Therefore boiling point of 10
-amine is more than 30
-amine.
7. Trifluoro methyl amine is less basic than CH3 –NH2. Explain.
Ans. The structural formula of trifluoro methyl amine and methyl amine are given below-
We know that fluorine atom is electron withdrawing and CH3 group is electron releasing. So in the case of
trifluoromethyl amine there is –I effect and in case of methyl amine there is + I effect. Due to –I effect, the
availability of electron on ‘N’ atom will be less and hence it will be less basic and in case of CH3 –NH2, due to + I
effect, the electron density will be increased on ‘N’ atom hence it will be more basic. Thus it is clear that trifluoro
methyl amine is less basic than methyl amine.
8. Why aniline is less basic than ammonia ? Explain. Or, Why are aromatic amines weaker bases than aliphatic
amines ?
Ans. Aniline (Kb = 4.2 × 10-10
) is less basic than ammonia (Kb=1.8×10-5
) and aliphatic amines. This can be explained on
the basis of resonating structures of aniline.
It is clear from above resonating structures that the lone pair on nitrogen gets delocalized and less available for a
proton. Hence the aniline is less basic than ammonia or aliphatic amines, where electron pair is more available for
protonation.
Moreover, after accepting proton aniline is converted into anilinium ion which is less stable than aniline hence the
equilibrium is shifted towards aniline.
In alphatic amine + I alkyl group is attached with the nitrogen of amine. Due to which the electron density on
nitrogen is further increased which makes the aliphatic amines more basic.
Long Answer Type Questions
1. Write short noters on the following-
(a) Diazotisation Reaction,
(b) Acetylation,
(c) Hoffmann’s bromamide reaction,
(d) Gabriel phthalimide synthesis,
Ans. (a) Diazotisation Reaction- When Aniline react with nitrous acid under cold conditions (273-278K) to form
benzene dazonium salt. The reaction is known as diazotization reaction.
(b) Acetylation- The replacement of an active hydrogen of alcohols, phenols or eamines with an acyl group (RCO) to
form the corresponding esters or amides is called acetylation. Acetylation is carried out in the presence of a base like
pyridine, dimethyllanilineetc, by acetyl chloride or acetic anhydride.

4. CH3COCl + C2H5OH
Pyridine
CH3COOC2H5 + HCl
Acetyl anhydride ethanol Ethyl acetate
(CH3CO)2O + C2H5NH2 → CH3CONHCH2CH3 + CH3COOH
Acetylchloride EthylamineN-Ethylacetamide
(c) Hoffmann bromamide reaction- The reaction of a higher acid amide with Br2 and KOH to give a lower amine is
called
Hoffmann bromamide reaction. It gives an amine containing one carbon atom less than the original amide.
R –CONH2 + Br2 + 4KOH → R –NH2 + K2CO3 + 2KBr + 2H2O
(d) Gabrial’sphthalimide synthesis- In this reaction phthalimide is converted into its potassium salt by treating it with
alcoholic KOH. Then potassium phthalimide is heated with an alkyl halide to yield on N-alkyl phthalimide which is
hydrolysed to phthalic acid and primary amine by heating with HCl or KOH solution. This synthesis is very useful for
the preparation of pure aliphatic primary amines.
2. Give one chemical test to distinguish between the following pairs of compounds-
(a) Methylamine and dimethylamine,
(b) Secondary and tertiary amines,
(c) Aniline and benzylamine,
(d) Aniline and N-methlyaniline.
Ans. (a) Methylamine and dimethylamine- a primary amine responds to Carbylamine reaction whereas
dimethylamine does not.
CH3NH2 + CHCl3 + 3KOH(alc)
∆
→ CH3N=C + 3KCl + 3H2O
Chloroform Carbylamine (Four-small)
(b) Secondary and tertiary amines- A secondary amine reacts with benzene sulphonyl chloride to form N, N-dialkyl
benzene sulphonamide which does not dissolve in KOH. A tertiary amine does not react with benzene sulphonyl
chloride.
(c) Aniline and benzylamine- Nitrous acid test- Benzylamine on reaction with HNO2 decomposes even at low
temperature to liberate N2 gas.
Aniline, on the other hand, reacts with HNO2 + HCl to form benzene diazonium chloride which is stable at 273-278 K
and hence does not decompose to give N2 gas.

5. (d) Aniline and N-methylaniline- Aniline is a primary aromatic amine, N-methylaniline is a secondary aromatic amine.
They may be distinguished by Carbylamine test.
Aniline gives phenylisocyanide on heating with chloroform (CHCl3) and alcoholic KOH which is extremely foul
smelling while N-methylaniline does not give this test.
3. How will you convert-
(a) Ethanoic acid into methanamine,
(b) Hexanenitrtile into 1-aminopentane,
(c) Methanol to ethanoic acid,
(d) Ethanamine into methanamine,
(e) Ethanoic acid into propanoic acid,
(f) Methanamine into ethanamine,
(g) Nitromethaneinto dimethylamine,
(h) Propanoic acid into ethanoicacid ?
4. Complete the following reactions-
(a) C6H5N2Cl + H3PO2 + H2O →
(b) C6H5NH2 + H2SO4 (cone.) →
(c) C6H5NH2 + (CH3CO)2 O →
(d) C6H5N2Cl
i HB𝐹4
𝑖𝑖 𝑁𝑎𝑁𝑂3/𝐶𝑢,∆
Ans. (a) C6H5N2Cl + H3PO2 + H2O
𝐶𝑢
C6H6 + H3PO3 + HCl + N2↑
(b)C6H5NH2 + H2SO4 (conc.)
∆
→ SO2OH +H2O

6. (c) C6H5NH2 + (CH3CO2) O
∆
C6H5NHCOCH3 + CH3COOH
(d) C6H5N2Cl
𝑖 𝐻𝐵𝐹4
𝑖𝑖 𝑁𝑎𝑁𝑂3/𝐶𝑢3 ∆
C6H5 –NO2 + NaBF4
5. Complete the following reactions-
(a) C6H4NH2 + CHCl3 + alc. KOH →
(b) C6H5N2Cl + C2H5OH →
(c) C2H5NH2 + HNO2→
Ans. (a) C6H5NH2 + CHCl3 + alc. 3KOH
∆
C6H5N = C + 3 KCl + 3 H2O
(b) C6H5N2Cl + C2H5OH
∆
→ C6H6 + N2 + HCl + CH3CHO
(c) C2H5NH2 + HNO2
𝑁𝑎𝑁𝑂2/𝐻𝐶𝑙
273−278𝐾
C2H5OH+ H2O + N2↑
6. Complete the following reactions-
(a) CH3CHO + NH2OH →
(b) C6H5NH2 + Bromine water →
7. Accomplish the following conversions-
(a) Nitrobenzene to benzoic acid, (b) Benzene to m-bromophenol, (c) Benzoic acid to aniline,
(d) Aniline to p-bromoaniline, (e) Benzamide to toluene, (f) Aniline to benzyl alcohol.

7. 8. What happens when-
(a) Aniline reacts with Br2water ?
(b) Benzene diazonium chloride is reacted with HCl in presence of Cu ?
(c) Acetamide is reacted with with LiAiH4 in presence of ether ?
(d) Toluene is treated with cone. HNO3 and H2SO4.
(e) Benzene diazonium chloride treated with fluroboric acid followed with heating.
9. Describe the method for the identification of primary, secondary and tertiary amines. Also write chemical
equation of the reaction involved.
Ans. Hinsber’g Method- This is an excellent test for distinguishing primary, secondary and tertiary amines. The amine
is shaken with benzene sulphonyl chloride in the presence of aqueous KOH solution.
(i) A primary amine gives a clear solution on insoluble N-alkyl benzene sulphonamide.
(ii) A secondary amine gives an insoluble N, N-dialkyl benzene sulphonamide which remains unaffected on addition
of acid.
(iii) A tertiary amine does not react at all. Therefore it remains insoluble in the alkaline solution but dissolves on
acidification to give a clear solution.
C6H5SO2Cl + NR3
𝐾𝑂𝐻
No reaction
𝐻𝐶𝑙
R3N+
HNCl-
(3
0
amine) (Soluble in HCl)
10. Give plausible explanation for each of the following-
(a) Why are amines less acidic than alcohols of comparable molecular masses ?
(b) Why do primary amines have higher boiling than tertiary amines ?
(c) Why are allphatic amines stronger bases than aromatic amines ?
Ans. (a) Thealkoxide ion (RO) left after the removal of a proton from an alcohol ROH is more stable than RNH2. This is
due to the fact that oxygen atom in alcohols is more electronegative than N atom in amines and thus can
accommodate the negative charge better.
(b) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-
bonding while tertiary amines due to the absence of aH-atom on the N-atom, do not undergo hydrogen bonding. As

8. a result, primary amines have higher boiling points than tertiary amines of comparable molecular mass, e.g., B. pt of
n-Butylaniline (351 K) is much higher than that of tert-butylamine (319 K).
(c) Aliphatic amines are stronger bases than aromatic amines. In aliphatic amines, alkyl groups present exert + I
effect R → NH2 (electron releasing) thus make the lone pair of electrons more readily available for sharing by an acid,
whereas in aromatic amines, the presence of benzene ring exerts –I effect (electron withdrawing) due to which the
availability of lone pair of electrons on N in NH2 decreases thus making it less basic.