Class 12 - Chemistry - Amines

Question 13.1.

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

(i) (CH3)2CHNH2 (ii) CH3 (CH2) 2NH2 (iii) CH3NHCH (CH3)2

(iv) (CH3)3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2)2NCH3

(vii) m–BrC6H4NH2?

Answer:

(i) 1-Methylethanamine (10 amine)

(ii) Propan-1-amine (10 amine)

(iii) N−Methyl-2-methylethanamine (20 amine)

(iv) 2-Methylpropan-2-amine (10 amine)

(v) N−Methylbenzamine or N-methylaniline (20 amine)

(vi) N-Ethyl-N-methylethanamine (30 amine)

(vii) 3-Bromobenzenamine or 3-bromoaniline (10 amine)

 

 

Question 13.2.

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.?

 Answer:

  • The best test for distinguishing methyl amine and dimethylamine is the Carbylamines test.

Carbylamine Test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide

form foul-smelling isocyanides or carbylamines.

In this case, Methylamine (which is an aliphatic primary amine) gives a positive carbylamine test while dimethylamine won’t.

Reaction:

                                       ∆

CH3-NH2+CHCl3 + 3KOH --> CH3-NC-3KCl+ 3H2

Methylamine (10)             Methylisocyanide (foul smell)

                                       ∆

(CH3)2-NH2+CHCl3 + 3KOH --> No reaction

 

  • Hinsberg’s reagent (benzenesulphonyl chloride, C6H5SO2Cl);It can be used to distinguish secondary and tertiary amines.

Hinsberg Test: Secondary amines react with Hinsberg’s reagent to form a product that is insoluble in an alkali.

For example, N, N−diethylamine reacts with Hinsberg’s reagent to form N,

 

N−diethylbenzenesulphonamide, which is insoluble in an alkali, Tertiary amines, however, do not react with Hinsberg’s reagent.

Reaction:-

 Class_12_Chemistry_Amines__N-N_DiethylBenzeneSulphonide

  • Ethylamine and aniline can be distinguished using the azo-dye test.
  • A dye is obtained when aromatic amines react with HNO2 (NaNO2 + dilute HCl) at 0-5°C, followed by a reaction with the alkaline solution of 2-naphthol.
  • The dye is usually yellow, red, or orange in colour.
  • Aliphatic amines give a brisk effervescence due (to the evolution of N2 gas) under similar conditions.

   Class_12_Chemistry_Amines__Ethylamine_&_Aniline_Test

  • Aniline and benzylamine can be distinguished with the help of nitrous acid.

 

 

Nitrous acid test: Benzylamine reacts with nitrous acid to form unstable diazonium salt, which in turn gives alcohol with the evolution of nitrogen gas,

while aniline reacts with nitrous acid to form a stable diazonium salt without releasing nitrogen gas.

Reaction:

Class_12_Chemistry_Amines__Nitrous_Acid_Test

  • Carbylamine test can be used to distinguish between Aniline and N-methylaniline.

Carbylamine Test: Primary amines, on heating with chloroform and ethanolic potassium hydroxide, form foul smelling isocyanides or carbylamines.

Aniline, being an aromatic primary amine, gives positive carbylamine test. However, N-methylaniline, being a secondary amine does not.

Reaction:

                                       ∆

C6H5-NH2+CHCl3 + 3KOH --> C6H5-NC-3KCl+ 3H2

Benzylamine (10)             Benzylisocyanide (foul smell)

                                       ∆

C6H5-NHCH3+CHCl3 + 3KOH --> No reaction

N-Methylaniline

 

Question 13.3.

Account for the following:

(i) pKb 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?

Answer:

  • pKb of aniline is more than that of methylamine.

       pkb value is the negative logarithm of the basicity constant (Kb) .

          i.e., pKb = -log Kb

Evidently, smaller the value of pKb, stronger is the base (strong tendency to donate electrons)

⇒ the structure of methyl amine is CH3NH2

⇒ the structure of aniline is:

Class_12_Chemistry_Amines_Aniline_Structure

Aniline (C6H5NH2) shows resonance:

Class_12_Chemistry_Amines__Resonance_Structure_of_Aniline

As a result of resonance, the lone pair of electrons on the nitrogen atom gets delocalized over benzene ring.

As a result, electron density on the nitrogen decreases and thus is less easily available to donate electrons making it less basic.

In contrast, in methyl amine (CH3NH2), delocalization of the lone pair of electrons on the nitrogen atom by resonance is not possible.

Furthermore, CH3 being an electron- releasing group, due to which +I effect of CH3

increases the electron density on the N- atom and thus is more easily available to donate electrons making it more basic.

Class_12_Chemistry_Amines__I_Effect

(+I effect) (Pushing electrons)

Therefore, aniline is weaker base than methylamine and hence its pKb value is higher than that of methylamine.

  • Ethylamine is soluble in water whereas aniline is not.

The structure of ethylamine is CH3CH2NH2

⇒ the structure of aniline is:

Class_12_Chemistry_Amines_Aniline_Structure

Ethylamine form H-bonds with water. It dissolves in water due to intermolecular H-bonding as shown below:

 Class_12_Chemistry_Amines_Ethylamine_Hydrogen_Bonds

Thus, ethylamine is soluble in water.

However, in aniline, due to the larger hydrophobic part, i.e., hydrocarbon part (C6H5 group),

which tends to retard the formation of H-bonds. The extent of H-bonding decreases and hence aniline is insoluble in water.

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

The structure of methylamine is CH3NH2

Methylamine is more basic than water due to the presence of CH3 (electron releasing group) and +I effect (pushing electrons).

Being more basic, methylamine accepts a proton from water liberating OH- ions.

Class_12_Chemistry_Amines_Structure_Of_MethylAmine

Dissociation of ferric chloride in water to give Fe3+ and Cl-

FeCl3→ Fe3+ + 3Cl-

Now, the liberated OH- ions combine with Fe3+ ions present in H2O to form brown precipitate of hydrated ferric oxide.

2Fe3+ + 6OH- --> 2Fe (OH) 3 or Fe2O3.3H2O

                         Hydrated Ferric Oxide (brown ppt)

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

The below diagram shows the position of ortho (o), para (p) and meta (m) derivatives of amino group:

 Class_12_Chemistry_Amines_Derivatives_Of_Amino_Group

 

Electrophilic addition reaction of amines: In addition to the reaction of the amino group (NH2 group),

aromatic amines also undergo typical electrophilic substitution reactions of the aromatic ring.

In all these reactions, the NH2 group strongly activates the aromatic ring through delocalization

of the lone pair of electrons of the N-atom over the aromatic ring.

 Class_12_Chemistry_Amines_Resonance_Structures_1

 

As a result, electron density increases more at ortho and para positions.

Therefore NH2 group directs the incoming group to ortho and para positions, i.e., NH2 are an o-, p-directing group.

But it has been observed that on nitration of aniline gives a substantial amount of m-nitroanilne.

Explanation: Nitration is usually carried out in acidic medium in the presence of concentrated HNO3 and concentrated H2SO4.

As a result, most of the aniline is converted into anilinium ion (gets protonated) and since - +NH3 is m-directing group,

therefore, an unexpected large amount of m-nitroaniline is obtained.

 

 

Nitration of aniline gives mainly p-nitroaniline

Class_12_Chemistry_Amines_p_&_o_NitroAniline

Nitration of anilinium ions gives m-nitroanilne (due to protonation)

 

Hence, nitration of aniline gives a mixture of p-nitroaniline and m-nitroaniline in approx. 1:1 ratio.

 Class_12_Chemistry_Amines_m-NitroAniline

  • Aniline does not undergo Friedel-Crafts reaction.

        Friedel- Crafts reaction: When any benzene or its derivative is treated with alkyl halide (R-X, X=Cl) or

acetyl chloride (CH3-COCl) in the presence of anhydrous aluminium chloride (AlCl3) to form alkyl or acetyl substituted benzene

or its derivative, this reaction is called Friedel-Crafts reaction.

 

 

⇒ Friedel-Crafts alkylation: When any benzene or its derivative is treated with alkyl halides (R-X, X=Cl, Br)

in the presence of anhydrous aluminium chloride (AlCl3) to form alkyl substituted benzene or its derivatives,

this reaction is called Friedel-Crafts alkylation. For example:

Class_12_Chemistry_Amines_Methyl_Benzene

Friedel-Crafts acylation: When any benzene or its derivative is treated with acetyl chloride (R-COCl)

in the presence of anhydrous aluminium chloride (AlCl3) to form acetyl substituted benzene or its derivatives,

this reaction is called Friedel-Crafts acylation. For example:

Class_12_Chemistry_Amines_Acetyl_Benzene

Aniline does not undergo Friedel-Crafts reaction.

Explanation: Aniline is a Lewis base (electron-pair acceptor) while AlCl3 is Lewis acid (electron-pair donor).

They combine with each to form a salt.

Class_12_Chemistry_Amines_Lewis_Base_&_Lewis_Acid

Due to presence of positive charge on nitrogen (N) atom in the salt, the group N+H2AlCl3-

acts as a strong electron withdrawing group (strong deactivating group).

As a result, it reduces the electron density in the benzene ring and hence aniline does not undergo Friedel-Crafts (alkylation or acetylation) reaction

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

Diazonium salts: Diazonium salts have the general formula

(R/Ar—N2+Cl-) where R stands for the alkyl group and Ar stands for the aryl group; the structure is given below:

Class_12_Chemistry_Amines_Arenediazonium_Chloride

Formation of Diazonium salts:

Class_12_Chemistry_Amines_Diazonium_Salts_1

Diazonium salt is obtained by treating aromatic amine (aniline) dissolved in dil. HCl with HNO2 at 273-278K (0° -5° C)

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

Explanation:

Aromatic amine form arenediazonium salts, which are stable for a short time in solution at low temperature (273-278K).

The diazonium salts of aromatic amines are more stable due to the dispersal of the positive charge on the benzene ring (resonance) as shown below:

Class_12_Chemistry_Amines_Resonance_Structure

The aliphatic amines, on the other hand, form highly unstable alkane diazonium salt (R—N2+Cl-)

they rapidly decompose even at low temperature (<272-278K) forming carbocation and nitrogen gas.

Class_12_Chemistry_Amines_Carbocation

Hence, diazonium salts of aromatic amine are much more stable than aliphatic diazonium salts.

 Class_12_Chemistry_Amines_Diazonium_Salts

Note: Carbocation is an ion in which carbon atom consists of positive charge

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines

Gabriel phthalimide synthesis is a very convenient method for the preparation of pure aliphatic amines ( especially primary amines)

Phthalimide on treatment with ethanolic KOH gives potassium phthalimide which on heating with a suitable alkyl halide gives N-substituted phthalimides.

These upon subsequent hydrolysis with dil.HCl under pressure or with alkali give primary amines.

Step 1: Phthalimide is treated with KOH to form potassium phthalimide

 Class_12_Chemistry_Amines__Gaberiel_Phthalimide_Sysnthesis

Step 2: Potassium phthalimide is treated with suitable alkyl halide to form N-substituted phthalimides.

Class_12_Chemistry_Amines__Gaberiel_Phthalimide_Sysnthesis_1

Step 3: N-substituted phthalimides undergoes hydrolysis in the presence of dilute HCl or with alkali (NaOH) to give primary amines.

Class_12_Chemistry_Amines__Gaberiel_Phthalimide_Sysnthesis_2

Therefore, Gabriel phthalimide synthesis results in the formation of primary(1° amine) only.

Secondary or tertiary amines are not formed through this synthesis.

Hence, Gabriel phthalimide synthesis preferred for the formation of primary amines only.

 

Question 13.4.

Arrange the following:

(i) In decreasing order of the pKb values:

C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

(ii) In increasing order of basic strength:

C6H5NH2, C6H5N (CH3)2, (C2H5)2NH and CH3NH2

(iii) In increasing order of basic strength:

(a) Aniline, p-nitroaniline and p-toluidine

(b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2.

(iv) In decreasing order of basic strength in gas phase:

C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3

(v) In increasing order of boiling point:

C2H5OH, (CH3)2NH, C2H5NH2

(vi) In increasing order of solubility in water:

C6H5NH2, (C2H5)2NH, C2H5NH2

Answer:

(i) In decreasing order of the pKb values:

C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2

pkb value is the negative logarithm of the basicity constant (Kb)

i.e., pKb = -log Kb

Evidently, smaller the value of pKb, stronger is the base (strong tendency to donate electrons)

Aliphatic amines(R-NH2) are more basic (tendency to donate electrons) than aromatic amines (C6H5NH2) because of the following reasons:

⇒ in aliphatic amines, alkyl groups are present. Alkyl groups are electron releasing groups;

Class_12_Chemistry_Amines_Basic_Character_of_Amines

hence they increase the electron density of N-atom and thus is easily available to donate electrons.

This property makes aliphatic amines more basic.

⇒ In aromatic amines, aryl group is present. Aromatic amine shows resonance:

 Class_12_Chemistry_Amines_Resonance_in_Aromatic_Amines

As a result of resonance, the lone pair of electrons on the nitrogen atom gets delocalized over benzene ring.

As a result, electron density on the nitrogen decreases and thus is less easily available to donate electrons making it less basic.

Hence aliphatic amines(R-NH2) are more basic than aromatic amines (C6H5NH2)

Now, in C2H5NH2, one ethyl group (alkyl) is present and in (C2H5)2NH2 two ethyl groups are present.

As we know that more alkyl groups are present, more basic will be the amine.

Hence, (C2H5)2NH2 is more basic than the C2H5NH2

Now in C6H5NH2, due to delocalisation of lone pair of electrons of the N-atom over the benzene ring, makes it less basic than (C2H5)2NH2 and C2H5NH2

Now, in C6H5NHCH3, due to presence of CH3 group, makes it more basic than C6H5NH2 but less basic than

(C2H5)2NH2 and C2H5NH2 due to the presence of aromatic ring which is responsible for the delocalisation of lone

pair of electrons of N-atom over the benzene ring.

Combining all these facts, the relative basic strength of these four amines decrease in the order:

(C2H5)2NH2 > C2H5NH2 > C6H5NHCH3 > C6H5NH2

Since, a stronger base has a lower pkb value, therefore, pKb values decrease in the reverse order:

C6H5NH2 > C6H5NHCH3 > C2H5NH2 > (C2H5)2NH2

(ii) In increasing order of basic strength:

C6H5NH2, C6H5N (CH3)2, (C2H5)2NH and CH3NH2

In (C2H5)2NH2, two ethyl groups are present and in CH3NH2 one methyl group is present.

As we know that more alkyl groups are present, more basic will be the amine. Hence, (C2H5)2NH2 is more basic than the CH3NH2

Now in C6H5NH2, due to delocalisation of lone pair of electrons of the N-atom over the benzene ring (decreases the electron density of N-atom)

makes it less basic than (C2H5)2NH2 and CH3NH2

 

 

Now, in C6H5N (CH3)2 due to presence of two CH3 groups (increases the electron density of N-atom) makes it more basic than C6H5NH2

but less basic than (C2H5)2NH2 and CH3NH2 due to the presence of aromatic ring which is responsible

for the delocalisation of lone pair of electrons of N-atom over the benzene ring (decreases the electron density of N-atom)

Combining all these facts, the relative basic strength of these four amines increases in the order:

C6H5NH2 < C6H5NHCH3 < CH3NH2 < (C2H5)2NH

 (iii) In increasing order of basic strength:

  • Aniline, p-nitroaniline and p-toluidine

(a) Electron-donating groups such as –CH3, -OCH3, -NH2, etc. increase the basicity and electron-withdrawing groups

such as –NO2, -CN, -SO3H, -COOH, -X(halogen), etc. decrease the basicity of amines.

Explanation: Electron-donating groups releases electrons, stabilizes the conjugate acid (cation) and thus increases the basic strength.

Electron-withdrawing groups withdraws electrons, destabilizes the conjugate acid (cation) and thus decreases the basic strength.

 

 

In p-nitroaniline, NO2 group is present. As we know that NO2 group is an electron

withdrawing group which decreases the basic strength of amine.

In p-toluidine, CH3 group is present and as we know that CH3 group is an electron donating (releasing) group which increases the basic strength of amine.

Hence, p-toluidine is more basic than p-nitroaniline

Now in C6H5NH2 (aniline), due to delocalisation of lone pair of electrons of the N-atom

over the benzene ring (decreases the electron density of N-atom)

makes it less basic than p-toluidine but more basic than p-nitroaniline (NO2 group is present which decreases the density of aniline)

Combining all these facts, the relative basic strength of these three amines increases in the order:

p-nitroaniline< aniline < p-toluidine

  • C6H5NH2, C6H5NHCH3, C6H5CH2NH2.

In C6H5NH2 and C6H5NHCH3, the N-atom is directly attached to the aromatic ring.

Hence, they both show resonance in which delocalisation of lone pair of electrons N-atom takes place.

As a result the electron density of N-atom decreases. Hence both are weaker bases.

However in C6H5NHCH3, CH3 group (electron withdrawing) is present which increases the overall density of electrons.

Hence, C6H5NHCH3 is more basic than C6H5NH2

In C6H5CH2NH2, the N-atom is not directly attached to the aromatic ring. As a result, it does not show resonance.

There is no effect on the electron density of lone pair of electrons of N-atom.

Hence, it is more basic than C6H5NH2 and C6H5NHCH3

Combining all these facts, the relative basic strength of these three amines increases in the order:

C6H5NH2 < C6H5NHCH3 < C6H5CH2NH2

(iv) In decreasing order of basic strength in gas phase:

C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3

Amines in gas phase or in non-aqueous solvents, there is no solvation effect (getting influenced by the solvent).

It means that the stabilization of conjugate acid formed due to the formation of hydrogen bonding is absent.

Class_12_Chemistry_Amines_Strength_of_Amines

Hence, the basic strength of amines depends only on the +I effect of the alkyl groups.

Alkyl groups are electron releasing groups,

they release electron to the nitrogen in amine and increase the overall electron density of electrons and thus is easily available to donate electron.

This property makes it more basic. As s result, more alkyl groups are attached, the higher the +I effect.

Hence, the higher the +I effect, stronger is the base (high tendency to accept electrons)

In (C2H5)3N, 3 alkyl groups are present, hence it is more basic. In (C2H5)2NH, 2 alkyl groups are present,

hence it is less basic than (C2H5)3N. In C2H5NH2, only one alkyl group is present, hence it is less basic than (C2H5)3N and (C2H5)2NH.

In NH3, no alkyl group is present, so there is no +I effect. Hence it is less basic among all the amines.

Combining all these facts, the relative basic strength of these four amines decreases in the order:

(C2H5)3N > (C2H5)2NH > C2H5NH2 > NH3

(v) In increasing order of boiling point:

C2H5OH, (CH3)2NH, C2H5NH2

As we know that boiling point of compounds depend upon the formation of H-bonding.

Amines have higher boiling points than hydrocarbons of simple molecular masses.

This is due to the reason that amines being polar, form intermolecular H-bonding

(except tertiary amine which do not have hydrogen atom linked to N-atom, i.e., R3N)

Further, since the electronegativity (tendency to attract a shared pair of electrons) of

nitrogen in amine is lower (3.0) than that of oxygen (3.5) in alcohol,

therefore, amines form weaker H-bonds than electronegative oxygen atom.

Class_12_Chemistry_Amines_InterMolecular_H-Bonding_Strong

Intermolecular H-bonding(strong)

Hence, C2H5OH has higher boiling point than (CH3)2NH and C2H5NH2.

Because in C2H5OH, the electronegativity of O-atom is higher

than H-atom which makes strong intermolecular H-bonding whereas in (CH3)2NH and C2H5NH2,

the electronegativity of N-atom is higher than H-atom but lower than O-atom in C2H5OH which makes weak intermolecular H-bonding.

Further, since, the extent of H-bonding depends upon the number of H-atoms on the N-atom.

More the no. of H-atoms linked to nitrogen, the higher the boiling point.

Since in (CH3)2NH have one hydrogen atom and in C2H5NH2

have two H-atoms linked to nitrogen, therefore,

C2H5NH2 has higher boiling point than (CH3)2NH.

Combining all these facts, the boiling point of the given three compounds increases in the order:

(CH3)2NH < C2H5NH2 < C2H5OH

(vi) In increasing order of solubility in water:

C6H5NH2, (C2H5)2NH, C2H5NH2

All the three classes of aliphatic amines form H-bonds with water. As the size of alkyl group increases,

the solubility decreases due to a corresponding increase in the hydrophobic part (hydrocarbon part) of the molecule.

On the other hand, aromatic amines are insoluble in water due to presence of larger hydrocarbon part (C6H5-group)

which tends to retard the formation of H-bonds.

Class_12_Chemistry_Amines_Intermolecular_H-Bonding_With_Water

Intermolecular H-bonding with water

Now, among the given compounds, C6H5NH2 is insoluble in water due to presence of C6H5-group (hydrocarbon part).

In (C2H5)2NH, two alkyl groups are present and in C2H5NH2 only one alkyl group is present,

hence C2H5NH2 is more soluble in water than (C2H5)2NH.

Combining all these facts, solubility of the given three compounds increases in the order:

C6H5NH2 < (C2H5)2NH < C2H5NH2

 

Question 13.5.

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?

Answer:

  • Ethanoic acid into Methanamine

Class_12_Chemistry_Amines_Ethanoic_acid_into_Methanamine

To convert ethanoic acid into methanamine (CH3NH2), we need ethanamide (amide group-RCONH2) and

we can get ethanamide from ethanoyl chloride (acid chloride group- RCOCl).

Step-1: Convert ethanoic acid into ethanoyl chloride

Ethanoic acid (CH3COOH) reacts with thionyl chloride (SOCl2) or PCl5 or PCl3 to form ethanoyl chloride (CH3COCl) by the replacement of OH group by Cl atom.

Step-2 Convert ethanoyl chloride into ethanamide

Ethanoyl chloride (CH3COCl) reacts with ammonia (in excess) to form ethanamide (CH3CONH2) by removal of NH4Cl

Step-3: Convert ethanamide into methanamine (Hoffman Bromamide reaction)

Ethanamide (CH3CONH2) is treated with an aqueous or ethanolic solution of potassium hydroxide (KOH) and bromine (Br2);

it gives ethanamine (CH3NH2-final product)

Note: Hoffman Bromamide Reaction is a reaction in which a primary amide is treated with an aqueous KOH or NaOH and bromine;

it gives a primary amine which has one carbon atom less than the original amide.

  • Hexanenitrile into 1-aminopentane

 Class_12_Chemistry_Amines_Hexanitrile_Into_1-AminoPentane

 The structure of Hexanenitrile is CH3CH2CH2CH2CH2CN or C5H11CN.

To convert Hexanenitrile into 1-aminopentane, first we need hexanamide (amide group-RCONH2)

and we can get hexanamide from hexanoyl chloride (acid chloride group – RCOCl) and we can get hexanoyl chloride from hexanoic acid.

Step 1: Convert Hexanenitrile into Hexanoic acid

Hexanenitrile (C5H11CN) undergoes hydrolysis to form Hexanoic acid (C5H11COOH)

Step 2: Convert Hexanoic acid into hexanoyl chloride

Hexanoic acid (C5H11COOH) reacts with thionyl chloride (SOCl2) or PCl5 or PCl3 to form hexanoyl chloride (C5H11COCl)

by the replacement of OH group by Cl atom.

Step 3: Convert Hexanoyl chloride into hexanamide

Hexanoyl chloride (C5H11COCl) reacts with excess ammonia to form hexanamide (C5H11CONH2) by the removal of NH4Cl.

Step 4: Convert hexanamide into 1- amino pentane by Hoffman Bromamide reaction)

Hexanamide (C5H11CONH2) is treated with an aqueous or ethanolic solution of potassium hydroxide (KOH) and bromine (Br2),

it gives 1-aminopentane (final product) which has one carbon atom less than the hexanamide.

  • Methanol to ethanoic acid

Class_12_Chemistry_Amines_Methanol_To_EthanoicAcid

The structure of methanol is CH3OH. To convert methanol to ethanoic acid (the number of carbon atoms is increasing

from one carbon atom to two carbon atoms) we need ethanenitrile and we can get ethanenitrile from methyl chloride.

Note: If the number of carbon atoms is increasing, we need a nitrile group and if the number of atoms is decreasing,

we need an amide group (Hoffman Bromamide reaction)

Step 1: Convert methanol into methyl chloride

Methanol (CH3OH) reacts with thionyl chloride (SOCl2) or PCl5 or PCl3 to form methyl chloride (CH3Cl) by the replacement of OH group by Cl atom.

Step 2: Convert methyl chloride into ethanenitrile

Methyl chloride (CH3Cl) reacts with ethanolic NaCN/ KCN to form ethanenitrile (CH3CN) by the removal of NaCl / KCl

Step 3: Convert ethanenitrile into ethanoic acid

Ethanenitrile (CH3CN) undergoes hydrolysis to form ethanoic acid (final product) which has one carbon atom more than the ethanenitrile

  • Ethanamine into Methanamine

Class_12_Chemistry_Amines_Ethanamine_Into_Methanamine

The structure of ethanamine is CH3CH2NH2. To convert ethanamine into methanamine (the number of carbon atoms is decreasing

from two carbon atoms to one carbon atom), first we need ethanamide (amide group- RCONH2) and

we can get acetamide from acetic acid. By oxidation of ethanol, we can get ethanoic acid.

Step 1: Convert Ethanamine into ethanol with the help of diazonium salt

Ethanamine (CH3CH2NH2) reacts with NaNO2 and HCl to give Diazonium salt (R--N2+Cl-) which undergoes hydrolysis to form ethanol.

Note: The diazonium salts or diazonium compounds are the class of organic compounds with

general formula R−N2+X where X is an organic or inorganic anion (for example, Cl–, Br–, BF4, etc.) and R is an alkyl or aryl group.

Hence, they have two nitrogen atoms with one being charged.

Example - Benzenediazonium chloride (C6H5N2+Cl)

Class_12_Chemistry_Amines_Benzene_Diazonium_Chloride_1

Step 2: Convert ethanol to ethanoic acid by oxidation

Ethanol (CH3CH2OH) undergoes oxidation in the presence of strong oxidizing agent KMNO4 to form ethanoic acid (CH3COOH)

Step 3: Convert ethanoic acid into ethanamide (amide group)

Ethanoic acid (CH3COOH) is treated with ammonia (in excess) to form ethanamide (CH3CONH2)

Step 4: Convert ethanamide into methanamine by Hoffman Bromamide Reaction

Ethanamide (CH3CONH2) is treated with alcoholic NaOH or KOH in the presence of bromine; it gives methanamine

(CH3NH2 - final product) which has one carbon atom less than the ethanamide.

  • Ethanoic acid into propanoic acid

Class_12_Chemistry_Amines_EthanoicAcid_Into_Propanoic_Acid

To convert ethanoic acid into propanoic acid (the number of carbon atoms is increasing from two carbon atoms to three carbon atoms),

we need propionitrile (CH3CH2CN) and we can get propionitrile from ethyl chloride.

To get ethyl chloride, we need ethanol which can be formed by the reduction of ethanoic acid.

Step 1: Convert ethanoic acid into ethanol by reduction

Ethanoic acid (CH3COOH) undergoes reduction in the presence of lithium aluminium hydride (LiAlH4) to form ethanol (CH3CH2OH)

Step 2: Convert ethanol into ethyl chloride

Ethanol (CH3CH2OH) reacts with PCl5 to form ethyl chloride (CH3CH2Cl) by the replacement of OH group by Cl atom.

Step 3: Convert ethyl chloride into ethyl cyanide/ propionitrile

Ethyl chloride (CH3CH2Cl) reacts with ethanolic NaCN / KCN to give ethyl cyanide/ propionitrile (CH3CH2CN)

Step 4: Convert ethyl cyanide/ propionitrile into propanoic acid

Ethyl cyanide (CH3CH2CN) undergoes hydrolysis to form a propanoic acid (CH3CH2COOH-final product)

which has one carbon atom more than the propionitrile.

  • Methanamine into Ethanamine

Class_12_Chemistry_Amines_Methanamine_Into_Ethanamine

The structure of methanamine is CH3NH2 and the structure of ethanamine is CH3CH2NH2.

To convert methanamine into ethanamine (the number of carbon atoms are increasing from one carbon atom two carbon atoms)

so we need ethanenitrile which we can get from ethyl chloride. We can obtain ethyl chloride

from alcohol which can be obtained from diazonium salt (R--N2+Cl-)

Step 1: Convert methyl amine to methanol

Methyl amine (CH3NH2) is first treated with HNO2 and HCl, which gives a fresh diazonium salt (R--N2+Cl-)

and then diazonium salt undergoes hydrolysis to form methanol.

Step 2: Convert methanol to methyl chloride

Methanol is treated with PCl5 or thionyl chloride, gives ethanenitrile (CH3Cl) by the replacement of OH atom by Cl atom.

Step 3: Convert methyl chloride to ethanenitrile

Methyl chloride (CH3Cl) is treated with ethanolic NaCN or KCN, gives ethanenitrile (CH3CN)

Step 4: Convert ethanenitrile to ethanamine

Ethanenitrile (CH3CN) undergoes a reduction in the presence of sodium and ethyl alcohol to form ethanamine (CH3CH2NH2 – final product)

which has one carbon atom more than the ethanenitrile.

  • Nitromethane into dimethylamine

Class_12_Chemistry_Amines_Dimethylamines

The structure of Nitromethane is CH3NO2 and structure of dimethylamine (CH3NHCH3),

as dimethylamine is a secondary amine so to obtain this, methyl isocyanide (CH3NC)

is required that can be obtained from methanamine through carbylamine reaction.

Step 1: Convert nitromethane to methanamine

Nitromethane (CH3NO2) undergoes reduction in the presence of Sn and HCl to form methanamine (CH3NH2)

Step 2: Convert methanamine to methyl isocyanide by Carbylamine reaction

Methanamine (CH3NH2) is heated with alcoholic potassium hydroxide and chloroform, the methyl isocyanide (CH3NC) is formed.

Note: Carbylamine reaction is given only by primary amines

Primary amines when heated with chloroform and alcoholic potassium hydroxide give isocyanides (carbylamines)

having very unpleasant smell, which can be easily detected.

  R-NH2                     +   CHCl3      + 3KOH (alc.) -->     R-NC + 3KCl + H2O

Primary Amine        Chloroform      Potassium Hydroxide       Carbylamine

Class_12_Chemistry_Amines_Reaction_1

Step 3: Convert methyl isocyanide to diethylamine

Methyl isocyanide (CH3NC) undergoes reduction in the presence of sodium and ethyl alcohol to form diethylamine (CH3NHCH3-final product)

(viii) Propanoic acid into ethanoic acid

Class_12_Chemistry_Amines_Propanoic_Acid

The structure of propanoic acid is CH3CH2COOH and the structure of ethanoic acid is CH3COOH.

To convert propanoic acid to ethanoic acid (the number of carbon atoms are decreasing), we need propionamide (amide group-RCONH2).

Then propionamide undergoes Hoffman Bromamide reaction to form methylamine.

Then methylamine formed can be converted to ethanol to form ethanoic acid by oxidation.

Step 1: Convert propanoic acid to propionamide

Propanoic acid (CH3CH2COOH) reacts with ammonia (in excess) to form propionamide (CH3CH2CONH2)

Step 2: Convert propionamide to ethyl amine by Hoffman Bromamide reaction

Propionamide (CH3CH2CONH2) reacts with potassium hydroxide and bromine to form ethylamine (CH3CH2NH2)

which has one carbon atom less than the propionamide (Hoffman Bromamide reaction)

Step 3: Convert ethyl amine to methanol by forming diazonium salt

Ethyl amine (CH3CH2NH2) reacts with NaNO2 and HCl to form diazonium salt (R--N2+Cl-)

then diazonium salt undergoes hydrolysis to from ethanol (CH3CH2OH)

Step 4: Convert ethanol to ethanoic acid by oxidation

Ethanol (CH3CH2OH) undergoes oxidation in the presence of strong oxidizing agent KMNO4 to form ethanoic acid (CH3COOH-final product)

 

 

Question 13.6.

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

 

Answer:

Primary, secondary and tertiary amines can be identified and distinguished by Hinsberg’s test.

In this test, the amines are allowed to react with Hinsberg’s reagent, benzenesulphonyl chloride (C6H5SO2Cl).

The three types of amines react differently with Hinsberg’s reagent. Therefore, they can be easily identified using Hinsberg’s reagent.

Primary amines react with benzenesulphonyl chloride to form N-alkylbenzenesulphonyl amide which is soluble in alkali.

Class_12_Chemistry_Amines_N-Propyl_Benzene_Sulphonamide

Due to the presence of a strong electron-withdrawing sulphonyl group in the sulphonamide,

the H−atom attached to nitrogen can be easily released as proton. So, it is acidic and dissolves in alkali.

Secondary amines react with Hinsberg’s reagent to give a sulphonamide which is insoluble in alkali.

Class_12_Chemistry_Amines_N,N_Dimethyl_Benzene_Sulphonamide

There is no H−atom attached to the N-atom in the sulphonamide. Therefore, it is not acidic and insoluble in alkali.

On the other hand, tertiary amines do not react with Hinsberg’s reagent at all.

 

 

Question 13.7.

Write short notes on the following:

  • Carbylamine reaction
  • Diazotisation
  • Hofmann’s bromamide reaction
  • Coupling reaction
  • Ammonolysis
  • Acetylation
  • Gabriel phthalimide synthesis

Answer:

  • Carbylamine reaction

Carbylamine reaction is used as a test for the identification of primary amines.

When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed.

These carbylamines have very unpleasant odours. Secondary and tertiary amines do not respond to this test.

Class_12_Chemistry_Amines_Carbyl_Amine_Reaction

  • Diazotisation

Aromatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl)

at low temperatures (273-278 K) to form diazonium salts. This conversion of aromatic primary amines into diazonium salts is known as diazotization.

For example, on treatment with NaNO2 and HCl at 273−278 K, aniline produces benzenediazonium chloride, with NaCl and H2O as by-products.

C6H5 + NaNO2 + HCl --> C6H5N2Cl + NaCl + 2H2O

  • Hofmann’s bromamide reaction

When an amide is treated with bromine in an aqueous or ethanolic solution of sodium hydroxide,

a primary amine with one carbon atom less than the original amide is produced.

This degradation reaction is known as Hoffmann bromamide reaction.

This reaction involves the migration of an alkyl or aryl group from the carbonyl carbon atom of the amide to the nitrogen atom.

Class_12_Chemistry_Amines_HoffMann_Bromide_Reaction

  • Coupling reaction

The reaction of joining two aromatic rings through the −N=N−bond is known as coupling reaction.

Arenediazonium salts such as benzene diazonium salts react with phenol or aromatic amines to form coloured azo compounds.

It can be observed that, the para-positions of phenol and aniline are coupled with the diazonium salt.

This reaction proceeds through electrophilic substitution.

 Class_12_Chemistry_Amines_Coupling_Reactions

  • Ammonolysis

When an alkyl or benzyl halide is allowed to react with an ethanolic solution of ammonia,

it undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (−NH2) group.

This process of cleavage of the carbon-halogen bond is known as ammonolysis.

Class_12_Chemistry_Amines_Substituted_Ammonium_Salt1

When this substituted ammonium salt is treated with a strong base such as sodium hydroxide, amine is obtained.

Class_12_Chemistry_Amines_Subsituted_Ammonium_Salt

Though primary amine is produced as the major product, this process produces a

mixture of primary, secondary and tertiary amines, and also a quaternary ammonium salt as shown.

Class_12_Chemistry_Amines_Quaternary_Ammonium_Salt

  • Acetylation

Acetylation (or ethanoylation) is the process of introducing an acetyl group into a molecule.

Class_12_Chemistry_Amines_Acetyl_Molecule

Aliphatic and aromatic primary and secondary amines undergo acetylation reaction by nucleophilic substitution

when treated with acid chlorides, anhydrides or esters.

This reaction involves the replacement of the hydrogen atom of −NH2 or > NH group by the acetyl group,

which in turn leads to the production of amides.

To shift the equilibrium to the right hand side, the HCl formed during the reaction is removed as soon as it is formed.

This reaction is carried out in the presence of a base (such as pyridine) which is stronger than the amine.

Class_12_Chemistry_Amines_N-PhenylEthanamide_1

When amines react with benzoyl chloride, the reaction is also known as benzoylation. For example,

Class_12_Chemistry_Amines_N-Ethyl_Benzamide

  • Gabriel phthalimide synthesis

Gabriel phthalimide synthesis is a very useful method for the preparation of aliphatic primary amines.

It involves the treatment of phthalimide with ethanolic potassium hydroxide to form potassium salt of phthalimide.

This salt is further heated with alkyl halide, followed by alkaline hydrolysis to yield the corresponding primary amine.

 Class_12_Chemistry_Amines_Sodium_Salt_of_Pthalic_Acid_1

 

Question 13.8.

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.

Answer:

  • Nitrobenzene to benzoic acid

First nitrobenzene is reduced to aniline by reduction with hydrogen gas with palladium catalyst followed by nitration and substitution

and final hydrolysis by acid by a proton yields benzoic acid.

Class_12_Chemistry_Amines_Benzoic_Acid_As_Product

  • Benzene to m-bromophenol

Benzene by nitration with conc. hydrochloric acid and nitrous acid gives nitrobenzene which on reaction with bromine liquid gives m-bromobenzene.

 Class_12_Chemistry_Amines_m_bromophenol

Upon reduction with tin and acid followed by heating with diluted hydrochloric acid gives m-bromophenol.

  • Benzoic acid to aniline

Benzoic acid, reacted with sulphonyl chloride undergoes reaction, followed by ammine and bromine liquid in presence of sodium hydroxide gives aniline.

This reaction is called as Hoffman bromamide degradation.

Class_12_Chemistry_Amines_Aniline_As_Product

  • Aniline to 2, 4, 6-tribromofluorobenzene

Aniline reaction with bromine water, followed by sodium nitrite gives 2, 4, 6- tribromodiazoniumchloride,

which is very reactive, gives 2, 4, 6-tribrmofluorobenzene upon treatment with hydrofluoroburic acid.

Class_12_Chemistry_Amines_TriBromoFluroBenzene

  • Benzyl chloride to 2-phenylethanamine

Benzyl chloride reacted with alcoholic NaCN and reduction gives 3-phenylethananmine.

Class_12_Chemistry_Amines_3_Phenyl_Ethanamine

  • Chlorobenzene to p-chloroaniline

Chlorobenzene upon nitration with nitronium ion and para product obtained undergoes reduction to give p-chloroaniline.

Class_12_Chemistry_Amines_pChloroAniline

  • Aniline to p-bromoaniline

As aniline is a very activating group, it is first reacted with anhydride to make it less activating,

which on reaction with bromine in acetic acid, followed by acid hydrolysis gives p-bromoaniline.

Class_12_Chemistry_Amines_p-BromoAniline

  • Benzamide to toluene

Benzamide undergoes Hoffman bromamide degradation to give aniline upon treatment with sodium nitrite gives benzenediazonium chloride,

reacts with phosphoric acid , followed by Friedel crafts reaction with methyl chloride in solvent gives toluene.

Class_12_Chemistry_Amines_Toulene

 

(ix) Aniline to benzyl alcohol.

Aniline reacts with sodium nitrite following by substitution reaction with KCN, by acid hydrolysis gives benzoic acid

which upon treatment with reducing agent gives benzyl alcohol.

 Class_12_Chemistry_Amines_Benzoyl_Alcohol

 

Question 13.9.

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

 Class_12_Chemistry_Amines_Structures

 

Answer:

  • Ethyl iodide reacts with NaCN gives substitution reactions to give propanitrile upon partial hydrolysis gives B,
  • upon reaction with sodium hydroxide gives C.

 Class_12_Chemistry_Amines_Ethanamine

  • Benzenediazoniumchloride gives nucleophilic substitution reactions gives A,
  • upon hydrolysis the CN ion is replaced by OH ion which is less better leaving group gives B upon heating with ammonia gives C.

Class_12_Chemistry_Amines_Benzamide_Product

  • Ethyl bromide gives nucleophilic substitution reactions gives B, upon reduction gives B
  • followed by reacting with nitrous acid, i.e. oxidation gives propanol.

Class_12_Chemistry_Propanol

  • Nitrobenzene upon reduction with iron/acid gives A, reacting with sodium nitrite gives benzenediazonium chloride,
  • followed by complete Hydrolysis gives phenol.

Class_12_Chemistry_Amines_Phenol

  • Acetic acid upon heating with ammonia gives A, which is an amide.
  • This amide reacts with NaOBr which extracts the carbonyl carbon giving B, followed by reacting with sodium nitrite gives methanol.

Class_12_Chemistry_Amines_Methanol

  • Nitrobenzene upon reduction with iron/acid mixture gives A,
  • followed by oxidation with nitrous acid gives B and reacting with phenol undergoes addition reaction to give C.

 Class_12_Chemistry_Amines_p-Hydroxyazobenzene

 

Question 13.10.

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’

which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N.

Write the structures and IUPAC names of compounds A, B and C.

Answer:

It is given that compound ‘C’ having the molecular formula, C6H7N is formed by heating compound ‘B’ with Br2 and KOH.

This is a Hoffmann bromamide degradation reaction.

Therefore, compound ‘B’ is an amide and compound ‘C’ is an amine.

The only amine having the molecular formula, C6H7N is aniline, (C6H5NH2).

Class_12_Chemistry_Amines_Aniline

Therefore, compound ‘B’ (from which ’C’ is formed) must be benzamide, (C6H5CONH2).

Class_12_Chemistry_Amines_Benzamide

Further, benzamide is formed by heating compound ‘A’ with aqueous ammonia.

Therefore, compound ‘A’ must be benzoic acid.

Class_12_Chemistry_Amines_Benzoic_Acid

The given reactions can be explained with the help of the following equations:

 Class_12_Chemistry_Amines_Reaction

 

Question 13.11.

Complete the following reactions:

(i) C6H5NH2 + CHCl3 + alc.KOH →

(ii) C6H5 N2Cl + H3PO2 + H2O  -->

(iii) C6H5NH2 + H2SO4 (conc) →

(iv) C6H5 N2Cl + C2H5OH→

(v) C6H5NH2 + Br2 (aq) →

(vi) C6H5NH2 + (CH3CO) 2 O→

(vii) C6H5N2Cl --> (In presences of HBF4) and (NaNO2/Cu.∆)

Answer:

                                       (Carbylamine reaction)

(i) C6H5NH2 + CHCl3 + alc.KOH → 3H2O + 3KCl + C6H5-NC

  Aniline                                                          Phenyl-isocyanide

  • C6H5N2Cl +     H3PO2 + H2O  --> C6H6 + N2 + H3PO3 + HCl

           Benzediazonium Chloride                 Benzene

                 

  • C6H5NH2 + H2SO4 (conc) → C6H5NH3HSO4

Aniline                            Anilinium hydrogen sulphate

  • C6H5 N2Cl    +             C2H5OH→ C6H6      +   CH3CHO + N2 + HCl

            Benzediazonium chloride    Ethanol Benzene    Ethanal  

Class_12_Chemistry_Amines_TriBromoAniline_1

Class_12_Chemistry_Amines_TriBromoAniline

 

 (vii) C6H5N2Cl à (In presences of HBF4) and (NaNO2/Cu.∆)

        C6H5N2Cl            -->        C6H5NO2 + N2 + NaBF4

    Benzediazonium chloride  Nitrobenzene

Question 13.12.

Why aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis?

Answer:

Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines.

It involves nucleophilic substitution (SN2) of alkyl halides by the anion formed by the phthalimide.

Class_12_Chemistry_Amines_Sodium_Salt_of_Pthalic_Acid

But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Class_12_Chemistry_Amines_Aryl_Halides

Hence, aromatic primary amines cannot be prepared by this process.

 

 

Question 13.13.

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

Answer:

  • Aromatic amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl) at 273 − 278 K
  • to form stable aromatic diazonium salts i.e., NaCl and H2O.

Class_12_Chemistry_Amines_BenzenediazoniumChloride

  • Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO2 and a mineral acid such as HCl)
  • to form unstable aliphatic diazonium salts, which further produce alcohol and HCl with the evolution of N2

 Class_12_Chemistry_Amines_Aliphatic_Primary_Alcohol

 

Question 13.14.

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?

Answer:

  • Amines undergo protonation to give amide ion.

R-NH2 --> R – N-H + H+

               Amide ion

Similarly, alcohol loses a proton to give alkoxide ion.

R-OH -->  R- O-        + H +

Alcohol   Alkoxide ion

In an amide ion, the negative charge is on the N-atom whereas in alkoxide ion, the negative charge is on the O-atom.

Since O is more electronegative than N, O can accommodate the negative charge more easily than N.

As a result, the amide ion is less stable than the alkoxide ion. Hence, amines are less acidic than alcohols of comparable molecular masses.

  • In a molecule of tertiary amine, there are no H−atoms whereas in primary amines, two hydrogen atoms are present.
  • Due to the presence of H−atoms, primary amines undergo extensive intermolecular H−bonding.
  • As a result, extra energy is required to separate the molecules of primary amines.
  • Hence, primary amines have higher boiling points than tertiary amines.

Class_12_Chemistry_Amines_Primary_&_Teritary_Amines

 

  • Due to the −R effect of the benzene ring, the electrons on the N- atom are less available in case of aromatic amines.
  • Therefore, the electrons on the N-atom in aromatic amines cannot be donated easily.
  • This explains why aliphatic amines are stronger bases than aromatic amines.

Share this with your friends  

Download PDF


You can check our 5-step learning process


.