AminesClass 12 Chemistry NCERT Solutions

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Q1Exercises

Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.

(i)

$(\text{CH}_3)_2\text{CHNH}_2$

(ii)

$\text{CH}_3(\text{CH}_2)_2\text{NH}_2$

(iii)

$\text{CH}_3\text{NHCH}(\text{CH}_3)_2$

(iv)

$(\text{CH}_3)_3\text{CNH}_2$

(v)

$\text{C}_6\text{H}_5\text{NHCH}_3$

(vi)

$(\text{CH}_3\text{CH}_2)_2\text{NCH}_3$

(vii)

$m-\text{BrC}_6\text{H}_4\text{NH}_2$

Solution

(i)

IUPAC Name: Propan-2-amine Classification: Primary (

1^\circ

) amine

(ii)

IUPAC Name: Propan-1-amine Classification: Primary (

1^\circ

) amine

(iii)

IUPAC Name: N-Methylpropan-2-amine Classification: Secondary (

2^\circ

) amine

(iv)

IUPAC Name: 2-Methylpropan-2-amine Classification: Primary (

1^\circ

) amine

(v)

IUPAC Name: N-Methylaniline or N-Methylbenzenamine Classification: Secondary (

2^\circ

) amine

(vi)

IUPAC Name: N-Ethyl-N-methylethanamine Classification: Tertiary (

3^\circ

) amine

(vii)

IUPAC Name: 3-Bromoaniline or 3-Bromobenzenamine Classification: Primary (

1^\circ

) amine

Q2Exercises

Give one chemical test to distinguish between the following pairs of compounds.

(i)

Methylamine and dimethylamine

(ii)

Secondary and tertiary amines

(iii)

Ethylamine and aniline

(iv)

Aniline and benzylamine

(v)

Aniline and N-methylaniline.

Solution

(i)

Methylamine and dimethylamine: Test: Carbylamine test. Procedure: When heated with chloroform and alcoholic potassium hydroxide, methylamine (a primary amine) gives a foul-smelling substance, methyl isocyanide. Dimethylamine (a secondary amine) does not give this test. Reaction:

\text{CH}_3\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{Heat}} \text{CH}_3\text{NC} \text{ (foul smell)} + 3\text{KCl} + 3\text{H}_2\text{O}

(\text{CH}_3)_2\text{NH} + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{Heat}} \text{No reaction}

(ii)

Secondary and tertiary amines: Test: Hinsberg's test. Procedure: When reacted with benzenesulphonyl chloride (Hinsberg's reagent), a secondary amine forms a precipitate (N,N-dialkylbenzenesulphonamide) which is insoluble in alkali (KOH or NaOH). A tertiary amine does not react with Hinsberg's reagent. Reaction:

\text{R}_2\text{NH} + \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{NR}_2 + \text{HCl}

(Insoluble in alkali)

\text{R}_3\text{N} + \text{C}_6\text{H}_5\text{SO}_2\text{Cl} \rightarrow \text{No reaction}

(iii)

Ethylamine and aniline: Test: Azo dye test. Procedure: Aniline, an aromatic primary amine, when treated with

\text{NaNO}_2

and HCl at 273-278 K followed by reaction with an alkaline solution of 2-naphthol (phenol), forms a brilliant orange-red dye. Ethylamine, an aliphatic primary amine, reacts with nitrous acid to form unstable diazonium salt which decomposes to give alcohol and liberates nitrogen gas, but does not form a dye. Reaction for Aniline:

\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{2-Naphthol, } \text{OH}^-} \text{Orange-red dye}

Reaction for Ethylamine:

\text{C}_2\text{H}_5\text{NH}_2 + \text{HNO}_2 \rightarrow [\text{C}_2\text{H}_5\text{N}_2^+\text{Cl}^-] \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{N}_2 \uparrow + \text{HCl}

(iv)

Aniline and benzylamine: Test: Nitrous acid test. Procedure: Aniline (aromatic primary amine) reacts with nitrous acid at low temperature (273-278 K) to form a stable benzenediazonium chloride. Benzylamine (aliphatic primary amine) reacts with nitrous acid to form an unstable diazonium salt which decomposes to form benzyl alcohol with the evolution of nitrogen gas. Reaction for Aniline:

\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \text{ (stable diazonium salt)}

Reaction for Benzylamine:

\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2 + \text{HNO}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{N}_2 \uparrow + \text{H}_2\text{O}

(v)

Aniline and N-methylaniline: Test: Carbylamine test. Procedure: Aniline (a primary amine) on heating with chloroform and alcoholic potassium hydroxide gives a foul-smelling isocyanide. N-methylaniline (a secondary amine) does not give this test. Reaction:

\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{Heat}} \text{C}_6\text{H}_5\text{NC} \text{ (foul smell)} + 3\text{KCl} + 3\text{H}_2\text{O}

\text{C}_6\text{H}_5\text{NHCH}_3 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\text{Heat}} \text{No reaction}

Q2Exercises

Give one chemical test to distinguish between the following pairs of compounds.

(i)

Methylamine and dimethylamine

(ii)

Secondary and tertiary amines

(iii)

Ethylamine and aniline

(iv)

Aniline and benzylamine

(v)

Aniline and N-methylaniline.

Solution

(i)

Methylamine and dimethylamine:

Test: Carbylamine test (Isocyanide test).
Procedure: Warm each compound with chloroform ( $CHCl_3$ ) and alcoholic potassium hydroxide (KOH).
Observation: Methylamine (a primary amine) will give a foul-smelling substance, methyl isocyanide. Dimethylamine (a secondary amine) will not give this test.
Reaction (Methylamine): $CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC + 3KCl + 3H_2O$

(ii)

Secondary and tertiary amines:

Test: Hinsberg's test.
Procedure: Shake each amine with Hinsberg's reagent (benzenesulphonyl chloride, $C_6H_5SO_2Cl$ ) in the presence of excess aqueous KOH.
Observation: A secondary amine will react to form a precipitate (N,N-dialkylbenzenesulphonamide) which is insoluble in KOH. A tertiary amine will not react.
Reaction (Secondary amine): $(R)_2NH + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2N(R)_2 + HCl$

(iii)

Ethylamine and aniline:

Test: Azo dye test.
Procedure: Dissolve each amine in dilute HCl and cool to 0-5°C. Add a cold solution of $NaNO_2$ . Then add a cold alkaline solution of 2-naphthol ( $\beta$ -naphthol).
Observation: Aniline (an aromatic amine) will form a bright orange-red dye. Ethylamine (an aliphatic amine) will not form a dye; instead, it will produce nitrogen gas.

(iv)

Aniline and benzylamine:

Test: Azo dye test.
Procedure: Same as above (iii).
Observation: Aniline will form a bright orange-red dye. Benzylamine, being an aliphatic amine, will react with nitrous acid to form an unstable diazonium salt which decomposes to give benzyl alcohol and liberates nitrogen gas, but no dye is formed.

(v)

Aniline and N-methylaniline:

Test: Carbylamine test.
Procedure: Same as above (i).
Observation: Aniline (a primary amine) will give a foul-smelling isocyanide. N-methylaniline (a secondary amine) will not react.

Q3Exercises

Account for the following:

(i)

$\mathrm{p} K_{b}$ of aniline is more than that of methylamine.

(ii)

Ethylamine is soluble in water whereas aniline is not.

(iii)

Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv)

Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.

(v)

Aniline does not undergo Friedel-Crafts reaction.

(vi)

Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii)

Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Solution

(i)

A higher

\mathrm{p} K_{b}

value indicates a weaker base. Aniline is a much weaker base than methylamine. In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring due to resonance. This makes the lone pair less available for protonation. In contrast, methylamine has an electron-donating methyl group (+

I

effect) which increases the electron density on the nitrogen atom, making it a stronger base.

(ii)

Ethylamine can form intermolecular hydrogen bonds with water molecules due to the presence of N-H bonds. This makes it soluble in water. Aniline has a large, non-polar, hydrophobic phenyl group (

C_6H_5-

). The hydrophobic nature of this group is more significant than the hydrogen bonding of the

-NH_2

group, making aniline insoluble in water.

(iii)

Methylamine is a base and accepts a proton from water to produce hydroxide ions (

OH^-

).

CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-

These hydroxide ions react with ferric chloride (

FeCl_3

) to form a brown precipitate of hydrated ferric oxide.

Fe^{3+} + 3OH^- \rightarrow Fe(OH)_3 \downarrow \text{ (or } Fe_2O_3 \cdot xH_2O)

(iv)

Nitration is carried out in a strongly acidic medium (conc.

HNO_3

+ conc.

H_2SO_4

). In this medium, aniline, being a base, gets protonated to form the anilinium ion (

C_6H_5NH_3^+

). The anilinium group (

-NH_3^+

) is strongly deactivating and a meta-directing group. Therefore, a significant amount of m-nitroaniline is formed along with ortho and para isomers.

(v)

Aniline is a Lewis base. The catalyst used in the Friedel-Crafts reaction, anhydrous aluminium chloride (

AlCl_3

), is a Lewis acid. Aniline reacts with

AlCl_3

to form a salt. This places a positive charge on the nitrogen atom, which strongly deactivates the benzene ring towards electrophilic substitution. Hence, aniline does not undergo the Friedel-Crafts reaction.

(vi)

The diazonium ion from an aromatic amine (arenediazonium ion) is stabilized by resonance. The positive charge is delocalized over the benzene ring, which increases its stability. Aliphatic diazonium ions lack such resonance stabilization. They readily decompose to form a carbocation and evolve nitrogen gas, making them highly unstable.

(vii)

Gabriel phthalimide synthesis yields pure primary amines. The reaction proceeds through nucleophilic substitution and does not produce mixtures of secondary and tertiary amines, which is a major disadvantage of ammonolysis of alkyl halides. This method ensures that only one alkyl group attaches to the nitrogen atom, leading to the exclusive formation of a primary amine.

Q3Exercises

Account for the following:

(i)

$\text{p}K_b$ of aniline is more than that of methylamine.

(ii)

Ethylamine is soluble in water whereas aniline is not.

(iii)

Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

(iv)

Although amino group is $o$ - and $p$ - directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of $m$ -nitroaniline.

(v)

Aniline does not undergo Friedel-Crafts reaction.

(vi)

Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

(vii)

Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Solution

(i)

A higher

\text{p}K_b

value indicates a weaker base. Aniline is a weaker base than methylamine. In aniline, the lone pair of electrons on the nitrogen atom is delocalized over the benzene ring due to resonance. This makes the lone pair less available for protonation. In contrast, in methylamine, the methyl group (

\text{CH}_3

) is an electron-donating group (+I effect), which increases the electron density on the nitrogen atom, making it more basic.

(ii)

Ethylamine can form intermolecular hydrogen bonds with water molecules due to the presence of N-H bonds and the lone pair on nitrogen. The energy released upon formation of these hydrogen bonds is sufficient to overcome the hydrophobic interactions of the small ethyl group. Aniline has a large hydrophobic aryl group (phenyl group), which outweighs the effect of hydrogen bonding. The energy required to break the hydrogen bonds in water and accommodate the large non-polar phenyl group is high, making aniline insoluble in water.

(iii)

Methylamine is a weak base and reacts with water to form methylammonium hydroxide, which is basic in nature.

\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-

This hydroxide ion concentration is sufficient to cause the precipitation of ferric ions (

\text{Fe}^{3+}

) from a ferric chloride solution as hydrated ferric oxide (

\text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O}

).

2\text{FeCl}_3 + 6\text{CH}_3\text{NH}_3^+\text{OH}^- \rightarrow \text{Fe}_2\text{O}_3 \cdot x\text{H}_2\text{O} \text{ (precipitate)} + 6\text{CH}_3\text{NH}_3^+\text{Cl}^-

(iv)

Nitration is carried out in a strongly acidic medium (conc.

\text{HNO}_3

and conc.

\text{H}_2\text{SO}_4

). In this medium, the basic amino group of aniline gets protonated to form the anilinium ion (

\text{C}_6\text{H}_5\text{NH}_3^+

). The anilinium ion is an electron-withdrawing group and is meta-directing. Therefore, a significant amount of the meta-isomer (m-nitroaniline) is formed along with ortho and para isomers.

(v)

Aniline does not undergo Friedel-Crafts reaction (alkylation or acylation) because it is a Lewis base and reacts with the Lewis acid catalyst, such as anhydrous aluminium chloride (

\text{AlCl}_3

), to form a salt.

\text{C}_6\text{H}_5\text{NH}_2 + \text{AlCl}_3 \rightarrow \text{C}_6\text{H}_5\text{NH}_2^+\text{AlCl}_3^-

The nitrogen atom of aniline acquires a positive charge, which makes the group strongly deactivating for electrophilic substitution. As a result, the Friedel-Crafts reaction does not occur.

(vi)

The stability of aromatic diazonium salts is due to the delocalization of the positive charge over the benzene ring through resonance. The resonating structures stabilize the arenediazonium ion. Aliphatic diazonium salts lack such resonance stabilization, making them highly unstable. They readily decompose to form a carbocation and liberate nitrogen gas even at low temperatures.

(vii)

Gabriel phthalimide synthesis is preferred for synthesizing primary amines because it yields pure primary amines without any contamination of secondary or tertiary amines. The reaction involves nucleophilic substitution of the phthalimide anion on an alkyl halide, followed by hydrolysis. This method avoids the problem of over-alkylation which is common in the ammonolysis of alkyl halides, a method that often produces a mixture of primary, secondary, and tertiary amines.

Q4Exercises

Arrange the following:

(i)

In decreasing order of the $\text{p}K_b$ values: $\text{C}_2\text{H}_5\text{NH}_2, \text{C}_6\text{H}_5\text{NHCH}_3, (\text{C}_2\text{H}_5)_2\text{NH}$ and $\text{C}_6\text{H}_5\text{NH}_2$

(ii)

In increasing order of basic strength: $\text{C}_6\text{H}_5\text{NH}_2, \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2, (\text{C}_2\text{H}_5)_2\text{NH}$ and $\text{CH}_3\text{NH}_2$

(iii)

In increasing order of basic strength:

(a)

Aniline,

p

-nitroaniline and

p

-toluidine

(b)

\text{C}_6\text{H}_5\text{NH}_2, \text{C}_6\text{H}_5\text{NHCH}_3, \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2

.

(iv)

In decreasing order of basic strength in gas phase: $\text{C}_2\text{H}_5\text{NH}_2, (\text{C}_2\text{H}_5)_2\text{NH}, (\text{C}_2\text{H}_5)_3\text{N}$ and $\text{NH}_3$

(v)

In increasing order of boiling point: $\text{C}_2\text{H}_5\text{OH}, (\text{CH}_3)_2\text{NH}, \text{C}_2\text{H}_5\text{NH}_2$

(vi)

In increasing order of solubility in water: $\text{C}_6\text{H}_5\text{NH}_2, (\text{C}_2\text{H}_5)_2\text{NH}, \text{C}_2\text{H}_5\text{NH}_2$ .

Solution

(i)

Decreasing order of $\text{p}K_b$ values: Decreasing

\text{p}K_b

means increasing basic strength. The order of basic strength is:

(\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{NHCH}_3 > \text{C}_6\text{H}_5\text{NH}_2

. Therefore, the decreasing order of

\text{p}K_b

values is:

\text{C}_6\text{H}_5\text{NH}_2 > \text{C}_6\text{H}_5\text{NHCH}_3 > \text{C}_2\text{H}_5\text{NH}_2 > (\text{C}_2\text{H}_5)_2\text{NH}

(ii)

Increasing order of basic strength: Basic strength is decreased by electron-withdrawing groups (phenyl) and increased by electron-donating groups (alkyl). Aromatic amines are weaker bases than aliphatic amines.

(\text{C}_2\text{H}_5)_2\text{NH}

is the strongest base due to two +I effect groups and solvation effects. Order:

\text{C}_6\text{H}_5\text{NH}_2 < \text{C}_6\text{H}_5\text{N}(\text{CH}_3)_2 < \text{CH}_3\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH}

(iii)

Increasing order of basic strength: (a) Electron-donating groups (like

-\text{CH}_3

) increase basic strength, while electron-withdrawing groups (like

-\text{NO}_2

) decrease it. Order:

p

-nitroaniline < Aniline <

p

-toluidine (b) Benzylamine (

\text{C}_6\text{H}_5\text{CH}_2\text{NH}_2

) is an aliphatic amine (Aralkylamine) and is a stronger base than aromatic amines because the lone pair on N is not delocalized into the ring. N-methylaniline is slightly stronger than aniline due to the +I effect of the methyl group. Order:

\text{C}_6\text{H}_5\text{NH}_2 < \text{C}_6\text{H}_5\text{NHCH}_3 < \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2

(iv)

Decreasing order of basic strength in gas phase: In the gaseous phase, solvation effects are absent, and basicity is determined solely by the inductive effect (+I) of the alkyl groups. More alkyl groups lead to higher electron density on nitrogen, making the amine more basic. Order:

(\text{C}_2\text{H}_5)_3\text{N} > (\text{C}_2\text{H}_5)_2\text{NH} > \text{C}_2\text{H}_5\text{NH}_2 > \text{NH}_3

(v)

Increasing order of boiling point: Boiling point depends on the extent of intermolecular hydrogen bonding. Alcohols form stronger H-bonds than amines. Primary amines (2 H-atoms for H-bonding) have higher boiling points than secondary amines (1 H-atom for H-bonding) of comparable mass. Order:

(\text{CH}_3)_2\text{NH} < \text{C}_2\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{OH}

(vi)

Increasing order of solubility in water: Solubility in water depends on the ability to form hydrogen bonds with water and the size of the hydrophobic alkyl/aryl part. A larger hydrophobic part decreases solubility. Order:

\text{C}_6\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH} < \text{C}_2\text{H}_5\text{NH}_2

Q4Exercises

Arrange the following:

(i)

In decreasing order of the $\mathrm{p} K_{b}$ values: $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}$

(ii)

In increasing order of basic strength: $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}(\mathrm{CH}_{3})_{2}, (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}$ and $\mathrm{CH}_{3} \mathrm{NH}_{2}$

(iii)

In increasing order of basic strength:

(a)

Aniline, p-nitroaniline and p-toluidine

(b)

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}

.

(iv)

In decreasing order of basic strength in gas phase: $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}, (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}, (\mathrm{C}_{2} \mathrm{H}_{5})_{3} \mathrm{N}$ and $\mathrm{NH}_{3}$

(v)

In increasing order of boiling point: $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}, (\mathrm{CH}_{3})_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}$

(vi)

In increasing order of solubility in water: $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}$ .

Solution

(i)

Decreasing order of $\mathrm{p} K_{b}$ values (Increasing order of basic strength): A lower

\mathrm{p} K_{b}

value corresponds to a stronger base. Aliphatic amines are stronger bases than aromatic amines. Order of basic strength:

(\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} > \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} > \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} > \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}

. Therefore, the decreasing order of

\mathrm{p} K_{b}

values is:

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} > \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} > \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} > (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}

(ii)

Increasing order of basic strength: Aliphatic amines are stronger than aromatic amines.

(\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}

is a strong secondary aliphatic amine. Order:

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}(\mathrm{CH}_{3})_{2} < \mathrm{CH}_{3} \mathrm{NH}_{2} < (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH}

(iii)

Increasing order of basic strength: (a) Electron-donating groups (like

-CH_3

) increase basicity, while electron-withdrawing groups (like

-NO_2

) decrease basicity. Order: p-nitroaniline < Aniline < p-toluidine (b) Benzylamine (

C_6H_5CH_2NH_2

) is an aliphatic amine and stronger than aromatic amines. N-methylaniline is slightly stronger than aniline due to the +I effect of the methyl group. Order:

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NHCH}_{3} < \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}

(iv)

Decreasing order of basic strength in gas phase: In the gas phase, basicity is primarily determined by the inductive effect (+I) of alkyl groups. More alkyl groups lead to greater electron density on nitrogen and thus higher basicity. Solvation effects are absent. Order:

(\mathrm{C}_{2} \mathrm{H}_{5})_{3} \mathrm{N} > (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} > \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} > \mathrm{NH}_{3}

(v)

Increasing order of boiling point: Boiling point depends on the extent of intermolecular hydrogen bonding. Alcohols form stronger H-bonds than amines. Primary amines form stronger H-bonds than secondary amines. The molar masses are comparable. Order:

(\mathrm{CH}_{3})_{2} \mathrm{NH} < \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} < \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}

(vi)

Increasing order of solubility in water: Solubility depends on hydrogen bonding with water and the size of the hydrophobic alkyl/aryl group. Aniline is least soluble due to its large hydrophobic phenyl group. Primary amines are generally more soluble than secondary amines of similar mass due to better H-bonding and smaller hydrophobic parts. Order:

\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2} < (\mathrm{C}_{2} \mathrm{H}_{5})_{2} \mathrm{NH} < \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}

Q5Exercises

How will you convert:

(i)

Ethanoic acid into methanamine

(ii)

Hexanenitrile into 1-aminopentane

(iii)

Methanol to ethanoic acid

(iv)

Ethanamine into methanamine

(v)

Ethanoic acid into propanoic acid

(vi)

Methanamine into ethanamine

(vii)

Nitromethane into dimethylamine

(viii)

Propanoic acid into ethanoic acid?

Solution

(i)

Ethanoic acid into methanamine

CH_3COOH \xrightarrow{NH_3, \Delta} CH_3CONH_2 \xrightarrow{Br_2, KOH} CH_3NH_2

(ii)

Hexanenitrile into 1-aminopentane

CH_3(CH_2)_4CN \xrightarrow{H_2O/H^+} CH_3(CH_2)_4COOH \xrightarrow{NH_3, \Delta} CH_3(CH_2)_4CONH_2 \xrightarrow{Br_2, KOH} CH_3(CH_2)_3CH_2NH_2

(iii)

Methanol to ethanoic acid

CH_3OH \xrightarrow{PCl_5} CH_3Cl \xrightarrow{KCN(alc.)} CH_3CN \xrightarrow{H_3O^+} CH_3COOH

(iv)

Ethanamine into methanamine

CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH \xrightarrow{K_2Cr_2O_7/H^+} CH_3COOH \xrightarrow{NH_3, \Delta} CH_3CONH_2 \xrightarrow{Br_2, KOH} CH_3NH_2

(v)

Ethanoic acid into propanoic acid

CH_3COOH \xrightarrow{LiAlH_4} CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br \xrightarrow{KCN(alc.)} CH_3CH_2CN \xrightarrow{H_3O^+} CH_3CH_2COOH

(vi)

Methanamine into ethanamine

CH_3NH_2 \xrightarrow{HNO_2} CH_3OH \xrightarrow{PCl_5} CH_3Cl \xrightarrow{KCN(alc.)} CH_3CN \xrightarrow{LiAlH_4} CH_3CH_2NH_2

(vii)

Nitromethane into dimethylamine

CH_3NO_2 \xrightarrow{Fe/HCl} CH_3NH_2 \xrightarrow{2CH_3Cl} (CH_3)_3N^+Cl^- \xrightarrow{NaOH} (CH_3)_2NH + CH_3NH_2 + (CH_3)_3N

This reaction yields a mixture. A more controlled method is:

CH_3NO_2 \xrightarrow{Fe/HCl} CH_3NH_2 \xrightarrow{CHCl_3, KOH} CH_3NC \xrightarrow{H_2/Ni} CH_3NHCH_3

This gives N-methylmethanamine (dimethylamine).

(viii)

Propanoic acid into ethanoic acid

CH_3CH_2COOH \xrightarrow{NH_3, \Delta} CH_3CH_2CONH_2 \xrightarrow{Br_2, KOH} CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH \xrightarrow{K_2Cr_2O_7/H^+} CH_3COOH

Q5Exercises

How will you convert:

(i)

Ethanoic acid into methanamine

(ii)

Hexanenitrile into 1-aminopentane

(iii)

Methanol to ethanoic acid

(iv)

Ethanamine into methanamine

(v)

Ethanoic acid into propanoic acid

(vi)

Methanamine into ethanamine

(vii)

Nitromethane into dimethylamine

(viii)

Propanoic acid into ethanoic acid?

Solution

(i)

Ethanoic acid into methanamine

\text{CH}_3\text{COOH} \xrightarrow{\text{SOCl}_2 \text{ or } \text{PCl}_5} \text{CH}_3\text{COCl} \xrightarrow{\text{NH}_3 (excess)} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2 + \text{NaOH}} \text{CH}_3\text{NH}_2

(Ethanoic acid

\rightarrow

Ethanoyl chloride

\rightarrow

Ethanamide

\rightarrow

Methanamine)

(ii)

Hexanenitrile into 1-aminopentane This requires degradation. First, hydrolyze the nitrile to a carboxylic acid, then use Hofmann bromamide degradation.

\text{CH}_3(\text{CH}_2)_4\text{CN} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3(\text{CH}_2)_4\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3(\text{CH}_2)_4\text{CONH}_2 \xrightarrow{\text{Br}_2 + \text{NaOH}} \text{CH}_3(\text{CH}_2)_3\text{CH}_2\text{NH}_2

(Hexanenitrile

\rightarrow

Hexanoic acid

\rightarrow

Hexanamide

\rightarrow

1-Aminopentane)

(iii)

Methanol to ethanoic acid

\text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{COOH}

(Methanol

\rightarrow

Chloromethane

\rightarrow

Ethanenitrile

\rightarrow

Ethanoic acid)

(iv)

Ethanamine into methanamine

\text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CONH}_2 \xrightarrow{\text{Br}_2 + \text{NaOH}} \text{CH}_3\text{NH}_2

(Ethanamine

\rightarrow

Ethanol

\rightarrow

Ethanoic acid

\rightarrow

Ethanamide

\rightarrow

Methanamine)

(v)

Ethanoic acid into propanoic acid

\text{CH}_3\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{PBr}_3} \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{KCN}} \text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{H}^+/\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{COOH}

(Ethanoic acid

\rightarrow

Ethanol

\rightarrow

Bromoethane

\rightarrow

Propanenitrile

\rightarrow

Propanoic acid)

(vi)

Methanamine into ethanamine

\text{CH}_3\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{OH} \xrightarrow{\text{PCl}_5} \text{CH}_3\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{NH}_2

(Methanamine

\rightarrow

Methanol

\rightarrow

Chloromethane

\rightarrow

Ethanenitrile

\rightarrow

Ethanamine)

(vii)

Nitromethane into dimethylamine

\text{CH}_3\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{CH}_3\text{NH}_2 \xrightarrow{\text{CHCl}_3 + \text{KOH}} \text{CH}_3\text{NC} \xrightarrow{\text{H}_2/\text{Pt}} \text{CH}_3\text{NHCH}_3

(Nitromethane

\rightarrow

Methanamine

\rightarrow

Methyl isocyanide

\rightarrow

Dimethylamine) Alternatively, using alkylation:

\text{CH}_3\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{CH}_3\text{NH}_2 \xrightarrow{2\text{CH}_3\text{Cl}} (\text{CH}_3)_3\text{N}^+\text{Cl}^- \text{ (mixture forms, less ideal)}

A better controlled alkylation:

\text{CH}_3\text{NH}_2 + \text{CH}_3\text{Cl} \rightarrow (\text{CH}_3)_2\text{NH}_2^+\text{Cl}^- \xrightarrow{\text{NaOH}} (\text{CH}_3)_2\text{NH}

(viii)

Propanoic acid into ethanoic acid

\text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{CH}_3\text{CH}_2\text{CONH}_2 \xrightarrow{\text{Br}_2 + \text{NaOH}} \text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{KMnO}_4/\text{H}^+} \text{CH}_3\text{COOH}

(Propanoic acid

\rightarrow

Propanamide

\rightarrow

Ethanamine

\rightarrow

Ethanol

\rightarrow

Ethanoic acid)

Q6Exercises

Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved.

Solution

A method for the identification and distinction of primary, secondary, and tertiary amines is the Hinsberg's Test. This test uses benzenesulphonyl chloride (

\text{C}_6\text{H}_5\text{SO}_2\text{Cl}

) as the reagent, which is known as Hinsberg's reagent.

Principle: Primary, secondary, and tertiary amines react differently with Hinsberg's reagent.

1. Primary Amine ( $1^\circ$ ) A primary amine reacts with benzenesulphonyl chloride to form an N-alkylbenzenesulphonamide. The hydrogen attached to the nitrogen in this sulphonamide is acidic because of the strong electron-withdrawing sulphonyl group. Hence, it is soluble in an aqueous alkali solution (like KOH or NaOH).

Chemical Equation:

\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + \text{H}-\text{NHR} \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{NHR} + \text{HCl}

(N-alkylbenzenesulphonamide)

\text{C}_6\text{H}_5\text{SO}_2\text{NHR} + \text{KOH} \rightarrow [\text{C}_6\text{H}_5\text{SO}_2\text{NR}]^-\text{K}^+ + \text{H}_2\text{O}

(Soluble potassium salt)

Observation: Forms a precipitate which dissolves in alkali.

2. Secondary Amine ( $2^\circ$ ) A secondary amine reacts with benzenesulphonyl chloride to form an N,N-dialkylbenzenesulphonamide. This sulphonamide does not have any hydrogen atom attached to the nitrogen. Therefore, it is not acidic and is insoluble in alkali.

Chemical Equation:

\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + \text{H}-\text{NR}_2 \rightarrow \text{C}_6\text{H}_5\text{SO}_2\text{NR}_2 + \text{HCl}

(N,N-dialkylbenzenesulphonamide)

\text{C}_6\text{H}_5\text{SO}_2\text{NR}_2 + \text{KOH} \rightarrow \text{No reaction (insoluble)}

Observation: Forms a precipitate which is insoluble in alkali.

3. Tertiary Amine ( $3^\circ$ ) A tertiary amine does not have a hydrogen atom attached to the nitrogen atom. Therefore, it does not react with benzenesulphonyl chloride.

Chemical Equation:

\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + \text{R}_3\text{N} \rightarrow \text{No reaction}

Observation: No reaction occurs. The amine remains insoluble and can be separated.

Summary of the Test:

Primary amine: Reacts with Hinsberg's reagent to form a product that is soluble in alkali.
Secondary amine: Reacts with Hinsberg's reagent to form a product that is insoluble in alkali.
Tertiary amine: Does not react with Hinsberg's reagent.

Q7Exercises

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.

Solution

(i)

Carbylamine reaction (Isocyanide test): This reaction is a test for primary amines (both aliphatic and aromatic). When a primary amine is heated with chloroform and an ethanolic solution of potassium hydroxide, it forms an isocyanide or carbylamine. These compounds have a very unpleasant, foul smell. Secondary and tertiary amines do not give this test. General Reaction:

\text{R}-\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{R}-\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}

(Amine)

\rightarrow

(Isocyanide)

(ii)

Diazotisation: Diazotisation is the process of converting a primary aromatic amine into a diazonium salt. This is achieved by treating the amine, dissolved in a cold aqueous mineral acid (like HCl), with a solution of sodium nitrite (

\text{NaNO}_2

) at a low temperature (273-278 K or 0-5 °C). General Reaction:

\text{Ar}-\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \text{Ar}-\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

(Aromatic amine)

\rightarrow

(Arenediazonium chloride) The resulting diazonium salts are important intermediates in the synthesis of many aromatic compounds.

(iii)

Hofmann's bromamide reaction: This is a method for the preparation of primary amines from amides. In this degradation reaction, an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide. The resulting amine contains one carbon atom less than the parent amide. This reaction is useful for stepping down a carbon series. General Reaction:

\text{R}-\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} \rightarrow \text{R}-\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}

(Amide)

\rightarrow

(Primary amine)

(iv)

Coupling reaction: This is a reaction of arenediazonium salts in which the diazo group (

\text{-N}_2^+

) is retained. The diazonium salt acts as an electrophile and reacts with electron-rich aromatic compounds like phenols and anilines to form brightly coloured azo compounds (

\text{Ar}-\text{N=N}-\text{Ar}'

). This reaction is an electrophilic substitution. Example:

\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-, \text{pH 9-10}} \text{C}_6\text{H}_5-\text{N=N}-\text{C}_6\text{H}_4-\text{OH} \text{ (p-Hydroxyazobenzene - Orange dye)}

(v)

Ammonolysis: Ammonolysis is the process of cleavage of a C-X bond in an alkyl or benzyl halide by an ammonia molecule (

\text{NH}_3

). It is a nucleophilic substitution reaction used to prepare amines. The reaction is typically carried out by heating an alkyl halide with an ethanolic solution of ammonia in a sealed tube. A disadvantage is that the primary amine formed can further react with the alkyl halide, leading to a mixture of primary, secondary, tertiary amines, and quaternary ammonium salts. General Reaction:

\text{R}-\text{X} + \text{NH}_3 \rightarrow \text{R}-\text{NH}_3^+\text{X}^- \xrightarrow{\text{NaOH}} \text{R}-\text{NH}_2

Further reaction:

\text{RNH}_2 \xrightarrow{\text{RX}} \text{R}_2\text{NH} \xrightarrow{\text{RX}} \text{R}_3\text{N} \xrightarrow{\text{RX}} \text{R}_4\text{N}^+\text{X}^-

(vi)

Acetylation: Acetylation is the introduction of an acetyl group (

\text{CH}_3\text{CO-}

) into a molecule. In the context of amines, it is the reaction of primary or secondary amines with acylating agents like acid chlorides (e.g., acetyl chloride) or acid anhydrides (e.g., acetic anhydride) to form amides. This reaction is a nucleophilic substitution. It is often used to protect the amino group in aromatic amines during electrophilic substitution reactions. General Reaction:

\text{R}-\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{R}-\text{NHCOCH}_3 + \text{HCl}

(Amine)

\rightarrow

(N-alkylacetamide)

(vii)

Gabriel phthalimide synthesis: This is a method used for the preparation of pure primary amines. It involves three main steps: 1) Phthalimide is treated with ethanolic potassium hydroxide to form potassium phthalimide. 2) The potassium salt is then heated with an alkyl halide to form N-alkylphthalimide. 3) Alkaline hydrolysis of the N-alkylphthalimide yields a pure primary amine and regenerates the phthalate salt. This method cannot be used to prepare aromatic primary amines. General Reaction:

Phthalimide $\xrightarrow{\text{KOH}}$ Potassium phthalimide
Potassium phthalimide $\xrightarrow{\text{R-X}}$ N-Alkylphthalimide
N-Alkylphthalimide $\xrightarrow{\text{NaOH(aq)}} \text{R}-\text{NH}_2 + \text{Sodium phthalate}$

Q8Exercises

Accomplish the following conversions:

(i)

Nitrobenzene to benzoic acid

(ii)

Benzene to $m$ -bromophenol

(iii)

Benzoic acid to aniline

(iv)

Aniline to 2,4,6-tribromofluorobenzene

(v)

Benzyl chloride to 2-phenylethanamine

(vi)

Chlorobenzene to $p$ -chloroaniline

(vii)

Aniline to $p$ -bromoaniline

(viii)

Benzamide to toluene

(ix)

Aniline to benzyl alcohol.

Solution

(i)

Nitrobenzene to benzoic acid

\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn/HCl}} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273\text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCN/KCN}} \text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{C}_6\text{H}_5\text{COOH}

(Nitrobenzene

\rightarrow

Aniline

\rightarrow

Benzenediazonium chloride

\rightarrow

Benzonitrile

\rightarrow

Benzoic acid)

(ii)

Benzene to $m$ -bromophenol

\text{C}_6\text{H}_6 \xrightarrow{\text{conc. } \text{HNO}_3 + \text{conc. } \text{H}_2\text{SO}_4} \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Br}_2, \text{FeBr}_3} m-\text{BrC}_6\text{H}_4\text{NO}_2 \xrightarrow{\text{Sn/HCl}} m-\text{BrC}_6\text{H}_4\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273\text{K}} m-\text{BrC}_6\text{H}_4\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_2\text{O}, \Delta} m-\text{BrC}_6\text{H}_4\text{OH}

(Benzene

\rightarrow

Nitrobenzene

\rightarrow

m-Bromonitrobenzene

\rightarrow

m-Bromoaniline

\rightarrow

m-Bromobenzenediazonium chloride

\rightarrow

m-Bromophenol)

(iii)

Benzoic acid to aniline

\text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{SOCl}_2} \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{NH}_3} \text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2, \text{KOH}} \text{C}_6\text{H}_5\text{NH}_2

(Benzoic acid

\rightarrow

Benzoyl chloride

\rightarrow

Benzamide

\rightarrow

Aniline)

(iv)

Aniline to 2,4,6-tribromofluorobenzene

\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{3\text{Br}_2(\text{aq})} 2,4,6-\text{Br}_3\text{C}_6\text{H}_2\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273\text{K}} 2,4,6-\text{Br}_3\text{C}_6\text{H}_2\text{N}_2^+\text{Cl}^- \xrightarrow{\text{HBF}_4} 2,4,6-\text{Br}_3\text{C}_6\text{H}_2\text{N}_2^+\text{BF}_4^- \xrightarrow{\Delta} 2,4,6-\text{Br}_3\text{C}_6\text{H}_2\text{F}

(Aniline

\rightarrow

2,4,6-Tribromoaniline

\rightarrow

Diazonium salt

\rightarrow

Diazonium fluoroborate

\rightarrow

2,4,6-Tribromofluorobenzene)

(v)

Benzyl chloride to 2-phenylethanamine

\text{C}_6\text{H}_5\text{CH}_2\text{Cl} \xrightarrow{\text{KCN (ethanolic)}} \text{C}_6\text{H}_5\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{H}_2/\text{Ni}} \text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2

(Benzyl chloride

\rightarrow

Phenylacetonitrile

\rightarrow

2-Phenylethanamine)

(vi)

Chlorobenzene to $p$ -chloroaniline

\text{C}_6\text{H}_5\text{Cl} \xrightarrow{\text{conc. } \text{HNO}_3 + \text{conc. } \text{H}_2\text{SO}_4} p-\text{ClC}_6\text{H}_4\text{NO}_2 + o-\text{ClC}_6\text{H}_4\text{NO}_2 \xrightarrow{\text{Fractional Crystallisation}} p-\text{ClC}_6\text{H}_4\text{NO}_2 \xrightarrow{\text{Sn/HCl}} p-\text{ClC}_6\text{H}_4\text{NH}_2

(Chlorobenzene

\rightarrow

p-Chloronitrobenzene

\rightarrow

p-Chloroaniline)

(vii)

Aniline to $p$ -bromoaniline First, protect the amino group by acetylation to control the reaction and avoid polysubstitution.

\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{(\text{CH}_3\text{CO})_2\text{O}, \text{Pyridine}} \text{C}_6\text{H}_5\text{NHCOCH}_3 \xrightarrow{\text{Br}_2, \text{CH}_3\text{COOH}} p-\text{BrC}_6\text{H}_4\text{NHCOCH}_3 \xrightarrow{\text{H}^+ \text{ or } \text{OH}^-, \Delta} p-\text{BrC}_6\text{H}_4\text{NH}_2

(Aniline

\rightarrow

Acetanilide

\rightarrow

p-Bromoacetanilide

\rightarrow

p-Bromoaniline)

(viii)

Benzamide to toluene

\text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2, \text{KOH}} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273\text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{H}_3\text{PO}_2 + \text{H}_2\text{O}} \text{C}_6\text{H}_6 \xrightarrow{\text{CH}_3\text{Cl}, \text{Anhyd. } \text{AlCl}_3} \text{C}_6\text{H}_5\text{CH}_3

(Benzamide

\rightarrow

Aniline

\rightarrow

Benzenediazonium chloride

\rightarrow

Benzene

\rightarrow

Toluene)

(ix)

Aniline to benzyl alcohol This conversion is not straightforward as it requires adding a -CH2- group. A possible route is:

\text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273\text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCN/KCN}} \text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{H}_3\text{O}^+, \Delta} \text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{LiAlH}_4} \text{C}_6\text{H}_5\text{CH}_2\text{OH}

(Aniline

\rightarrow

Benzenediazonium chloride

\rightarrow

Benzonitrile

\rightarrow

Benzoic acid

\rightarrow

Benzyl alcohol)

Q9Exercises

Give the structures of A, B and C in the following reactions:

(i)

$\text{CH}_3\text{CH}_2\text{I} \xrightarrow{\text{NaCN}} \text{A} \xrightarrow{\text{OH}^-/\text{Partial hydrolysis}} \text{B} \xrightarrow{\text{NaOH}+\text{Br}_2} \text{C}$

(ii)

$\text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{CuCN}} \text{A} \xrightarrow{\text{H}_2\text{O}/\text{H}^+} \text{B} \xrightarrow{\text{NH}_3, \Delta} \text{C}$

(iii)

$\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{KCN}} \text{A} \xrightarrow{\text{LiAlH}_4} \text{B} \xrightarrow{0^\circ\text{C}}{\text{HNO}_2} \text{C}$

(iv)

$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{A} \xrightarrow[273 \text{K}]{\text{NaNO}_2+\text{HCl}} \text{B} \xrightarrow[\Delta]{\text{H}_2\text{O}/\text{H}^+} \text{C}$

(v)

$\text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3, \Delta} \text{A} \xrightarrow{\text{NaOBr}} \text{B} \xrightarrow{\text{NaNO}_2/\text{HCl}} \text{C}$

(vi)

$\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{A} \xrightarrow[273 \text{K}]{\text{HNO}_2} \text{B} \xrightarrow{\text{C}_6\text{H}_5\text{OH}} \text{C}$

Solution

(i)

A: $\text{CH}_3\text{CH}_2\text{I} + \text{NaCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN}$ (Propanenitrile)
B: $\text{CH}_3\text{CH}_2\text{CN} + \text{H}_2\text{O} \xrightarrow{\text{OH}^-, \text{partial}} \text{CH}_3\text{CH}_2\text{CONH}_2$ (Propanamide)
C: $\text{CH}_3\text{CH}_2\text{CONH}_2 + \text{NaOH} + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2$ (Ethanamine)

A:

\text{CH}_3\text{CH}_2\text{CN}

B:

\text{CH}_3\text{CH}_2\text{CONH}_2

C:

\text{CH}_3\text{CH}_2\text{NH}_2

(ii)

A: $\text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{CuCN} \rightarrow \text{C}_6\text{H}_5\text{CN}$ (Benzonitrile)
B: $\text{C}_6\text{H}_5\text{CN} + 2\text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{COOH}$ (Benzoic acid)
C: $\text{C}_6\text{H}_5\text{COOH} + \text{NH}_3 \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CONH}_2$ (Benzamide)

A:

\text{C}_6\text{H}_5\text{CN}

B:

\text{C}_6\text{H}_5\text{COOH}

C:

\text{C}_6\text{H}_5\text{CONH}_2

(iii)

A: $\text{CH}_3\text{CH}_2\text{Br} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN}$ (Propanenitrile)
B: $\text{CH}_3\text{CH}_2\text{CN} \xrightarrow{\text{LiAlH}_4} \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ (Propan-1-amine)
C: $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 + \text{HNO}_2 \xrightarrow{0^\circ\text{C}} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ (Propan-1-ol)

A:

\text{CH}_3\text{CH}_2\text{CN}

B:

\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2

C:

\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}

(iv)

A: $\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2$ (Aniline)
B: $\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \xrightarrow{273\text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ (Benzenediazonium chloride)
C: $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{OH}$ (Phenol)

A:

\text{C}_6\text{H}_5\text{NH}_2

B:

\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-

C:

\text{C}_6\text{H}_5\text{OH}

(v)

A: $\text{CH}_3\text{COOH} + \text{NH}_3 \xrightarrow{\Delta} \text{CH}_3\text{CONH}_2$ (Ethanamide)
B: $\text{CH}_3\text{CONH}_2 + \text{NaOBr} \rightarrow \text{CH}_3\text{NH}_2$ (Methanamine) (Note: NaOBr is equivalent to Br2+NaOH)
C: $\text{CH}_3\text{NH}_2 + \text{NaNO}_2/\text{HCl} \rightarrow \text{CH}_3\text{OH}$ (Methanol)

A:

\text{CH}_3\text{CONH}_2

B:

\text{CH}_3\text{NH}_2

C:

\text{CH}_3\text{OH}

(vi)

A: $\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe/HCl}} \text{C}_6\text{H}_5\text{NH}_2$ (Aniline)
B: $\text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 \xrightarrow{273\text{K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ (Benzenediazonium chloride)
C: $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \rightarrow \text{p-HOC}_6\text{H}_4-\text{N=N}-\text{C}_6\text{H}_5$ (p-Hydroxyazobenzene)

A:

\text{C}_6\text{H}_5\text{NH}_2

B:

\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-

C:

p

-Hydroxyazobenzene

Q10Exercises

An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound 'B' which on heating with $\text{Br}_2$ and KOH forms a compound 'C' of molecular formula $\text{C}_6\text{H}_7\text{N}$ . Write the structures and IUPAC names of compounds A, B and C.

Solution

Let's deduce the structures by working backward from compound C.

Compound C: The molecular formula is $\text{C}_6\text{H}_7\text{N}$ . This corresponds to Aniline ( $\text{C}_6\text{H}_5\text{NH}_2$ ).
- Structure of C: $\text{C}_6\text{H}_5\text{NH}_2$
- IUPAC Name of C: Benzenamine or Aniline
Formation of C from B: Compound C (Aniline) is formed by heating compound B with $\text{Br}_2$ and KOH. This is the Hofmann bromamide degradation reaction, which converts an amide into a primary amine with one less carbon atom. Therefore, compound B must be an amide with one more carbon atom than aniline. The amide corresponding to aniline is Benzamide ( $\text{C}_6\text{H}_5\text{CONH}_2$ ).
- Structure of B: $\text{C}_6\text{H}_5\text{CONH}_2$
- IUPAC Name of B: Benzamide
Formation of B from A: Compound B (Benzamide) is formed when compound 'A' is treated with aqueous ammonia and heated. Amides are typically formed from carboxylic acids (or their derivatives) upon reaction with ammonia followed by heating. Therefore, compound A must be the carboxylic acid corresponding to benzamide, which is Benzoic acid ( $\text{C}_6\text{H}_5\text{COOH}$ ). The reaction first forms an ammonium salt, which upon heating dehydrates to form the amide.
- Structure of A: $\text{C}_6\text{H}_5\text{COOH}$
- IUPAC Name of A: Benzoic acid

Summary of Reactions:

A $\rightarrow$ B: $\text{C}_6\text{H}_5\text{COOH} \text{ (A)} + \text{NH}_3(\text{aq}) \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CONH}_2 \text{ (B)} + \text{H}_2\text{O}$
B $\rightarrow$ C: $\text{C}_6\text{H}_5\text{CONH}_2 \text{ (B)} + \text{Br}_2 + 4\text{KOH} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{NH}_2 \text{ (C)} + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}$

Q11Exercises

Complete the following reactions:

(i)

$\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + \text{alc. KOH} \rightarrow$

(ii)

$\text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \rightarrow$

(iii)

$\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{SO}_4 \text{(conc.)} \rightarrow$

(iv)

$\text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_2\text{H}_5\text{OH} \rightarrow$

(v)

$\text{C}_6\text{H}_5\text{NH}_2 + \text{Br}_2(\text{aq}) \rightarrow$

(vi)

$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow$

(vii)

$\text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{(i) HBF}_4} \xrightarrow{\text{(ii) NaNO}_2/\text{Cu}, \Delta}$

Solution

(i)

\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \rightarrow \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}

(Phenyl isocyanide)

(ii)

\text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{H}_3\text{PO}_2 + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_6 + \text{N}_2 + \text{H}_3\text{PO}_3 + \text{HCl}

(Benzene)

(iii)

\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{SO}_4 \text{(conc.)} \rightarrow \text{C}_6\text{H}_5\text{NH}_3^+\text{HSO}_4^- \xrightarrow{453-473\text{K}} p-\text{H}_2\text{NC}_6\text{H}_4\text{SO}_3\text{H}

(Anilinium hydrogensulphate, which rearranges to Sulphanilic acid)

(iv)

\text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_2\text{H}_5\text{OH} \rightarrow \text{C}_6\text{H}_6 + \text{CH}_3\text{CHO} + \text{N}_2 + \text{HCl}

(Benzene, Ethanal)

(v)

\text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2(\text{aq}) \rightarrow 2,4,6-\text{Br}_3\text{C}_6\text{H}_2\text{NH}_2 + 3\text{HBr}

(2,4,6-Tribromoaniline - white precipitate)

(vi)

\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{CH}_3\text{COOH}

(N-Phenylacetamide or Acetanilide)

(vii)

\text{C}_6\text{H}_5\text{N}_2\text{Cl} \xrightarrow{\text{(i) HBF}_4} \text{C}_6\text{H}_5\text{N}_2^+\text{BF}_4^- \xrightarrow{\text{(ii) NaNO}_2/\text{Cu}, \Delta} \text{C}_6\text{H}_5\text{NO}_2 + \text{N}_2 + \text{NaBF}_4

(Nitrobenzene)

Q13Exercises

Write the reactions of (i) aromatic and (ii) aliphatic primary amines with nitrous acid.

Solution

Nitrous acid (

\text{HNO}_2

) is an unstable acid and is prepared in situ by reacting sodium nitrite (

\text{NaNO}_2

) with a mineral acid like HCl.

(i) Reaction of Aromatic Primary Amines with Nitrous Acid: Aromatic primary amines, such as aniline, react with nitrous acid at low temperatures (273-278 K or 0-5 °C) to form a relatively stable arenediazonium salt. This reaction is called diazotisation.

Chemical Equation:

\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}

(Aniline)

\rightarrow

(Benzenediazonium chloride)

These diazonium salts are stable in cold solution and are very useful synthetic intermediates.

(ii) Reaction of Aliphatic Primary Amines with Nitrous Acid: Aliphatic primary amines also react with nitrous acid to form alkyldiazonium salts. However, these salts are highly unstable, even at low temperatures. They immediately decompose to form a mixture of products, including an alcohol, with the liberation of nitrogen gas. The reaction is not synthetically useful for preparing alcohols due to rearrangements and side products, but the quantitative evolution of nitrogen gas is used in the Van Slyke method for the estimation of amino acids.

Chemical Equation:

\text{R}-\text{NH}_2 + \text{HNO}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}} [\text{R}-\text{N}_2^+\text{Cl}^-] \text{ (Unstable)} \xrightarrow{\text{H}_2\text{O}} \text{R}-\text{OH} + \text{N}_2 \uparrow + \text{HCl}

(Aliphatic amine)

\rightarrow

(Alcohol)

Q14Exercises

Give plausible explanation for each of the following:

(i)

Why are amines less acidic than alcohols of comparable molecular masses?

(ii)

Why do primary amines have higher boiling point than tertiary amines?

(iii)

Why are aliphatic amines stronger bases than aromatic amines?

Solution

(i)

Amines are less acidic than alcohols: Acidity is the ability to donate a proton (H+). When an amine loses a proton, it forms an amide ion (

\text{R}-\text{NH}^-

). When an alcohol loses a proton, it forms an alkoxide ion (

\text{R}-\text{O}^-

).

\text{R}-\text{NH}_2 \rightleftharpoons \text{R}-\text{NH}^- + \text{H}^+

\text{R}-\text{OH} \rightleftharpoons \text{R}-\text{O}^- + \text{H}^+

Oxygen is more electronegative than nitrogen. Therefore, the O-H bond in alcohol is more polar than the N-H bond in amine. This makes the release of

\text{H}^+

easier from an alcohol. Furthermore, the negative charge on the resulting alkoxide ion is on the more electronegative oxygen atom, making it more stable than the amide ion where the negative charge is on the less electronegative nitrogen atom. Since a more stable conjugate base corresponds to a stronger acid, alcohols are more acidic than amines.

(ii)

Primary amines have higher boiling points than tertiary amines: Boiling points are influenced by the strength of intermolecular forces. Primary amines (

\text{RNH}_2

) have two hydrogen atoms bonded to the nitrogen atom, allowing them to form extensive intermolecular hydrogen bonds with other amine molecules. Tertiary amines (

\text{R}_3\text{N}

) have no hydrogen atoms directly bonded to the nitrogen atom, so they cannot form intermolecular hydrogen bonds with each other. They only exhibit weaker van der Waals forces and dipole-dipole interactions. Because more energy is required to break the stronger hydrogen bonds in primary amines compared to the weaker forces in tertiary amines, primary amines have significantly higher boiling points than tertiary amines of comparable molecular mass.

(iii)

Aliphatic amines are stronger bases than aromatic amines: The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom for donation to a proton.

In aliphatic amines (e.g., ethanamine, $\text{C}_2\text{H}_5\text{NH}_2$ ): The alkyl group (e.g., ethyl) is an electron-releasing group (+I effect). It pushes electron density towards the nitrogen atom, increasing the electron density on nitrogen and making the lone pair more available for protonation. This increases the basic strength.
In aromatic amines (e.g., aniline, $\text{C}_6\text{H}_5\text{NH}_2$ ): The lone pair of electrons on the nitrogen atom is in conjugation with the benzene ring. It gets delocalized over the ring through resonance. This delocalization makes the lone pair less available for donation to a proton. Consequently, aromatic amines are much weaker bases than aliphatic amines.