CBSE Class 12 Chemistry Amines Assignment

Level one

Question. Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
a) \( \mathrm{(CH_3)_2CHNH_2} \)
b) \( \mathrm{CH_3(CH_2)_2NH_2} \)
c) \( \mathrm{CH_3NHCH(CH_3)_2} \)
d) \( \mathrm{(CH_3)_3CNH_2} \)
e) \( \mathrm{C_6H_5NHCH_3} \)
f) \( \mathrm{m\text{-}BrC_6H_4NH_2} \)
Answer:
a) propan-2-amine (primary)
b) propan-1-amine (primary)
c) N-Methyl propan-2-amine (secondary)
d) 2-Methylpropan-2-amine (primary)
e) N-Methylaniline (secondary)
f) 3-Bromobenzenamine (primary)

Question. Give one chemical test to distinguish between the following pairs of compounds.
a) Methylamine and dimethylamine
b) Ethylamine and aniline
d) Aniline and benzylamine
e) Aniline and N-methylaniline
Answer:
a) Methyl amine being a \( 1^\circ \) amine will give carbyl amine test, while dimethylamine, a \( 2^\circ \) amine will not give this test.
b) Aniline will give brilliantly coloured orange dye on diazotization and coupling with phenol. Ethylamine on treatment with \( \mathrm{HNO_2} \) (\( \mathrm{NaNO_2 + HCl} \)) gives \( \mathrm{N_2} \) and ethanol, and fails to couple with phenol to give a dye.
d) Aniline (an aromatic primary amine) will give brilliantly coloured orange dye on diazotization and coupling with phenol. Benzylamine which is not an aromatic primary amine will not give this test. It only gives \( \mathrm{N_2} \) and benzyl alcohol with \( \mathrm{HNO_2} \).
e) Aniline is a primary amine it will give carbyl amine test while N-methylaniline is a secondary amine, and as such it will not give carbylamine test.

Question. Arrange the following:
a) In decreasing order of the \( \mathrm{pK_b} \) values:
\( \mathrm{C_2H_5NH_2, C_6H_5NHCH_3, (C_2H_5)_2NH} \) and \( \mathrm{C_6H_5NH_2} \)
b) In increasing order of basic strength:
\( \mathrm{C_6H_5NH_2, C_6H_5N(CH_3)_2, (C_2H_5)_2NH} \) and \( \mathrm{CH_3NH_2} \)
c) In increasing order of basic strength:
    a) Aniline, p-nitroaniline and p-toluidine
    b) \( \mathrm{C_6H_5NH_2, C_6H_5NHCH_3, C_6H_5CH_2NH_2} \)
d) In decreasing order of basic strength in gas phase:
\( \mathrm{C_2H_5NH_2, (C_2H_5)_2NH, (C_2H_5)_3N} \) and \( \mathrm{NH_3} \)
e) In increasing order of boiling point:
\( \mathrm{C_2H_5OH, (CH_3)_2NH, C_2H_5NH_2} \)
f) In increasing order of solubility in water:
\( \mathrm{C_6H_5NH_2, (C_2H_5)_2NH, C_2H_5NH_2} \)
Answer:
a) \( \mathrm{C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH} \)
b) \( \mathrm{C_6H_5NH_2 < C_6H_5N(CH_3)_2 < CH_3NH_2 < (C_2H_5)_2NH} \)
c)
    a) p-nitroaniline < aniline < p-toluidine
    b) \( \mathrm{C_6H_5NH_2 < C_6H_5NHCH_3 < C_6H_5CH_2NH_2} \)
d) \( \mathrm{(C_2H_5)_3N > (C_2H_5)_2NH > C_2H_5NH_2 > NH_3} \)
e) \( \mathrm{(CH_3)_2NH < C_2H_5NH_2 < C_2H_5OH} \)
f) \( \mathrm{C_6H_5NH_2 < (C_2H_5)_2NH < C_2H_5NH_2} \)

Question. Write short notes on the following:
i) Carbylamine reaction
ii) Diazotisation
iii) Hofmann’s bromamide reaction
iv) Coupling reaction
v) Ammonolysis
vi) Acetylation
vii) Gabriel phthalimide synthesis
Answer:
i) Carbylamine reaction: All primary amines on heating with chloroform and an alcoholic solution of potassium hydroxide produce isocyanide.
ii) Diazotisation: The conversion of aniline or other aromatic primary amines into diazonium salts by the action of nitrous acid and dilute hydrochloric acid is called diazotization.
iii) Hofmann’s bromamide reaction: The heating of an amide with bromine and caustic alkali to yield an amine with one carbon less than the original amide this reaction is called Hofmann’s bromamide reaction.
iv) Coupling reaction: Reaction of aromatic diazonium salt with phenols and amines to give azo compounds is called coupling reaction.
v) Ammonolysis: Alkyl halides react with alcoholic ammonia to primary, secondary and tertiary amines along with some quaternary ammonium salts.
vi) Acetylation: Introduction of acetyl group (\( \mathrm{CH_3CO-} \)) into any molecule is known as acetylation.
vii) Gabriel phthalimide synthesis: This is an excellent method to prepare pure primary aliphatic amine from corresponding alkyl halide and phthalimide.

Question. Gabriel Phthalimide synthesis is not suitable for preparation of secondary and tertiary amines.
Answer: Since the reaction involves \( \mathrm{S_N2} \) mechanism, only a primary alkyl group can approach the sterically hindered Phthalimide ring.

Question. \( \mathrm{pK_b} \) of aniline is more than that of methylamine.
Answer: Since the lone pair on the N-atom remains in conjugation with the benzene ring and is unavailable for donation. The basic character of methylamine is enhanced due to +I effect of \( \mathrm{-CH_3} \) group.

Question. Aniline does not undergo Friedel Crafts reaction.
Answer: Aniline is basic and reacts with anhyd. \( \mathrm{AlCl_3} \) to form a salt which deactivates the benzene ring towards electrophilic substitution.

Question. Give the IUPAC name of \( \mathrm{H_2N - CH_2 - CH_2 - CH = CH_2} \).
Answer: 4-amino-but-1-ene

Question. Write structure of methyl amine?
Answer: \( \mathrm{CH_3\text{-}NH_2} \)

Question. Write the structure of methyl isocyanides?
Answer: \( \mathrm{CH_3NC} \)

Question. Name the tests for Primary amine.
Answer: Carbylamines test

Question. Primary amines have higher b.p than tertiary amines.
Answer: Due to intermolecular hydrogen bonding.

Question. Why is alkyl amine more basic than ammonia?
Answer: Due to +I effect of alkyl group.

Question. Why do amine react as nucleophile.
Answer: Due to lone pair of electron on nitrogen.

Question. Why are aqueous solution of amine basic in nature?
Answer: Because of high electron density on nitrogen it gains \( \mathrm{H^+} \) from water.

Level Two Questions

Question. Name one test to distinguish between ethyl cyanide and ethyl isocyanide.
Answer: Ethyl cyanide on hydrolysis with acids forms propionic acid, whereas ethyl isocyanide with dilute \( \mathrm{HCl} \) forms ethylamine and formic acid.

Question. Identify A and B:
\( \mathrm{C_6H_5NH_2 \xrightarrow{NaNO_2/HCl} A \xrightarrow{CuBr} B } \)
Answer:
\( \mathrm{A} = \mathrm{C_6H_5N_2^+Cl^-} \) (Benzene diazonium chloride)
\( \mathrm{B} = \mathrm{C_6H_5Br} \) (Bromobenzene)

Question. Name the reaction in which amide is directly converted into amines.
Answer: Hofmann’s bromamide reaction.

Question. Complete the following:
\( \mathrm{RNH_2 + CHCl_3 + 3KOH \rightarrow ?} \)
Answer: \( \mathrm{RNC + 3KCl + 3H_2O} \)

Question. Complete the following:
\( \mathrm{RCON_2 + Br_2 + 4NaOH \rightarrow ?} \)
Answer: \( \mathrm{RNH_2 + 2NaBr + Na_2CO_3 + 2H_2O} \)

Question. Write the formula of hinsberg’s reagent.
Answer: Benzoyl chloride.

Question. What is meant by diazotization?
Answer: Conversion of primary aromatic amines into diazonium salts.

Question. In an increasing order of basic strength:
\( \mathrm{C_6H_5NH_2, C_6H_5N(CH_3)_2, (C_2H_5)_2NH} \) & \( \mathrm{CH_3NH_2} \)
Answer:
Basic strength: \( \mathrm{C_6H_5NH_2 < C_6H_5N(CH_3)_2 < CH_3NH_2 < (C_2H_5)_2NH} \)
Aliphatic amines are stronger bases than aromatic amines due to the presence of lone pair of \( \mathrm{e^-} \) on nitrogen atom. In case of aromatic amines the lone pair gets delocalised by resonance. Diethyl amine has greater +I effect. Hence, \( \mathrm{e^-} \) density over the nitrogen atom is more in this case. Similarly, N,N-dimethyl aniline has greater +I effect than aniline.

Question. In a decreasing order of basic strength: Aniline, p-nitroaniline & p-toluidine
Answer:
\( \text{p-toluidine} > \text{aniline} > \text{p-nitroaniline} \)
Methyl (\( \mathrm{-CH_3} \)) is an electron-donating group. It increases the electron density on the ring. Therefore, the lone pair of nitrogen is available for donation. Hence, it is most basic. On the other hand, nitro (\( \mathrm{-NO_2} \)) is an electron-withdrawing group. It decreases the electron density of the ring. Therefore, the lone pair is more delocalized in this case & is less available for donation. Thus, it will be least basic among the three.

Level Three Questions

Question. In an increasing order of \( \mathrm{pK_b} \) value:
\( \mathrm{C_2H_5NH_2, C_6H_5NHCH_3, (C_2H_5)_2NH} \) & \( \mathrm{C_6H_5NH_2} \)
Answer:
\( \mathrm{(C_2H_5)_2NH < C_2H_5NH_2 < C_6H_5NHCH_3 < C_6H_5NH_2} \)
Stronger the base is, the lesser is the \( \mathrm{pK_b} \) value. \( \mathrm{(C_2H_5)_2NH} \) is the strongest base due to two electron-releasing groups followed by \( \mathrm{C_2H_5NH_2} \) which has only one electron-releasing group. \( \mathrm{C_6H_5NHCH_3} \) is the next stronger base because of the presence of one electron-releasing alkyl group & electron-delocalising phenyl group. \( \mathrm{C_6H_5NH_2} \) is the least basic wherein the electrons get delocalised by resonance.

Question. Write a chemical reaction in which the iodide ion replaces the diazonium group in a diazonium salt.
Answer:
\( \mathrm{C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2} \)
(Benzenediazonium chloride) → (Iodobenzene)

Question. Why is an alkylamine more basic than ammonia?
Answer: An alkylamine is more basic than ammonia because of inductive effect (+I effect). Alkyl group or ‘R’ has an electron-releasing effect, which increases electron density over nitrogen atom. This increases its basicity.

4 marks questions

Question. Describe the Hofmann’s bromamide reaction.
Answer: It involves the reaction of bromine with an acid amide in the presence of an alkali. It results in the formation of a primary amine with one carbon less than the parent compound. Here, the alkyl group migrates from carbonyl, with the elimination of \( \mathrm{CO_2} \). For example:
\( \mathrm{RCONH_2 + Br_2 + 4KOH \rightarrow RNH_2 + 2KBr + K_2CO_3 + 2H_2O} \)
\( \mathrm{CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + 2KBr + K_2CO_3 + 2H_2O} \)

Question. Describe the Gatterman reaction.
Answer: This is a modification of Sandmeyer reaction in which benzenediazonium chloride is treated with copper powder & halogen acid to form aryl halides.
\( \mathrm{C_6H_5N_2^+Cl^- \xrightarrow{Cu/HCl, \Delta} C_6H_5Cl + N_2} \)

Question. Describe the coupling reaction.
Answer: It is the reaction of diazonium salts with phenols & aromatic amines to form azo compounds of the general formula \( \mathrm{Ar - N = N - Ar} \). The coupling of phenol takes place in a mildly alkaline medium.
\( \mathrm{C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{273-278\text{ K}, \text{ OH}^-} p\text{-HOC_6H_4-N=N-C_6H_5 + HCl} \)
(p-Hydroxyazobenzene, Orange dye)

Question. Why is \( \mathrm{pK_b} \) for aniline more than that for methylamine?
Answer: In aniline, the lone pair of electrons on the N atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen atom decreases. In contrast, in \( \mathrm{CH_3NH_2} \), the +I effect of \( \mathrm{-CH_3} \) increases the electron density on the N atom. Therefore, aniline is a weaker base than methylamine. Hence, its \( \mathrm{pK_b} \) value is higher than that of methylamine.

Question. Methylamine solution in water reacts with ferric chloride solution to give a precipitate of ferric hydroxide.
Answer: Being more basic than water, methylamine accepts a proton from water, liberating \( \mathrm{OH^-} \) ions.
\( \mathrm{CH_3NH_2 + H_2O \rightarrow CH_3NH_3^+ + OH^-} \)
These \( \mathrm{OH^-} \) ions combine with \( \mathrm{Fe^{3+}} \) ions present in water to form a brown precipitate of hydrated ferric oxide.
\( \mathrm{FeCl_3 \rightarrow Fe^{3+} + 3Cl^-} \)
\( \mathrm{2Fe^{3+} + 6OH^- \rightarrow 2Fe(OH)_3} \) (Hydrated ferric oxide, Brown ppt.)

Question. An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with \( \mathrm{Br_2} \) and \( \mathrm{KOH} \) forms a compound ‘C’ of molecular formula \( \mathrm{C_6H_7N} \). Write the structures and IUPAC names of compounds A, B and C.
Answer:
Compound A: Benzoic acid (\( \mathrm{C_6H_5COOH} \))
Compound B: Benzamide (\( \mathrm{C_6H_5CONH_2} \))
Compound C: Aniline (\( \mathrm{C_6H_5NH_2} \))

Question. Complete the following reactions:
(i) \( \mathrm{C_6H_5NH_2 + CHCl_3 + \text{alc. KOH} \rightarrow} \)
(ii) \( \mathrm{C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow} \)
(iii) \( \mathrm{C_6H_5NH_2 + H_2SO_4\text{(conc.)} \rightarrow} \)
(iv) \( \mathrm{C_6H_5N_2Cl + C_2H_5OH \rightarrow} \)
(v) \( \mathrm{C_6H_5NO_2 + Fe/HCl \rightarrow} \)
Answer:
(i) \( \mathrm{C_6H_5NH_2 + CHCl_3 + \text{alc. KOH} \rightarrow C_6H_5NC + 3KCl + 3H_2O} \)
(ii) \( \mathrm{C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow C_6H_6 + H_3PO_3 + N_2 + HCl} \)
(iii) \( \mathrm{C_6H_5NH_2 + H_2SO_4\text{(conc.)} \rightarrow \text{p-NH}_2\text{-C}_6\text{H}_4\text{-SO}_3\text{H} + H_2O} \)
(iv) \( \mathrm{C_6H_5N_2Cl + C_2H_5OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl} \)
(v) \( \mathrm{C_6H_5NO_2 + Fe/HCl \rightarrow C_6H_5NH_2} \)