Class 12 Chemistry Amines NCERT Solutions: Download Chapter 13 Questions with Solutions PDF

A 13.1 (i) Primary, because the nitrogen atom is attached to the only 1 carbon atom.
(ii) Tertiary, because the Nitrogen atom is attached to the 3 carbon atoms.
(iii) Primary, because the nitrogen atom is attached to the only 1 carbon atom.
(iv) Secondary, because the nitrogen atom is attached to the only 1 carbon.

A 13.2 (i) & (ii) There are total 8 geometrical isomers of the given compound.

(iii) a) Pairs 1,2,6,7 Exhibit Position isomerism; means the change in position of the substituent.

1. Pairs 1,3 and 1,4 and 2,3 and 2,4 exhibit chain isomerism i.e. in this type of isomerism the different structures can be produced by changing the chain of the
2. Pairs 5,6 and 5,7 exhibit metamerism; e. different group on either side of the central atom.
3. All Primary amines exhibit functional isomers. All secondary amines share functional isomerism and same for The functional isomerism means same functional group.

A 13.3 Benzene into aniline

When Benzene is treated with HNO₃/H₂SO₄ it forms nitrobenzene. When Nitrobenzene reduced with Sn/HCL it forms Aniline because Sn/HCl is a reducing Mixture.

When Benzene is reacted with nitrating mixture it forms nitrobenzene. When it Reduced H₂/Pd in ethanol or Sn/HCl, it forms Aniline. When Aniline reacts 2 times with CH₃Cl It forms N, N- dimethylaniline.

When 1,4-dichlorobutane reacts with NaCN it forms Di cyanide compound, After Hydrogenation it forms the Hexane 1,6-Diamine.

• (i) C₂H₅NH₂, C₆H₅NH₂, NH₃, C₆H₅CH₂NH₂ and (C₂H₅)₂NH
• (ii) C₂H₅NH₂, (C₂H₅ )₂NH, (C₂H₅ )₃N, C₆H₅NH₂

• (iii) CH₃NH₂, (CH₃ )₂NH, (CH₃)₃N, C₆H₅NH₂, C₆H₅CH₂ NH₂ .

A 13.4 (i) Alkyl group contribute inductive effect which increases the basic strength of

NH₃ 2H ₅NH ₂<(C ₂H ₅) ₂NH.

Then C₆H₅NH₂ is having –I effect that reduces strength. And C₆H₅CH₂NH₂ increases the basic strength but not as much as C₂H₅ group.

Hence final order will be C₆H₅NH₂ 3 6H ₅CH ₂NH ₂ 2H ₅NH ₂<(C ₂H ₅) ₂NH.

• (ii) By taking into consideration –R effect and steric hindrance of groups we can arrange them in the order

C₆H₅NH₂< C₂H₅NH₂<(C2H5)₃N<(C₂H₅)₂NH.

Because (C₂H₅)₃N has a lot of steric hindrances that reduces the basic strength.

• (iii) In C₆H₅NH₂ , N is directly attached to the ring that causes delocalization of electrons of the benzene ring. Whereas in case of C₆H₅CH₂NH₂ it is not directly connected to benzene ring Hence it has more basic strength.

Due to –I effect of (CH₃)₃ ^– a group it has more basic strength than C₆H₅CH₂NH₂. Hence final order will be

C₆H₅NH₂< C₆H₅CH₂NH₂<(CH₃)₃N< CH₃NH₂< (CH₃)₂NH

A 13.5 (i) CH₃CH₂CH₂NH₂ + HCl → CH₃CH₂CH₂N⁺H₃Cl^-

The final product is (N-propyl ammonium chloride.)

(ii) (C₂H₅)₃N + HCl → (C₂H₅)₃N⁺HCl^-

The final product is (Tri ethyl ammonium chloride)

A 13.6

On excessive alkylation with methyl iodide aniline gets converted into N,N,N-Trimethylanilinium iodide. After reacting it with sodium carbonate it get converted into N,N,N-Trimethylanilinium carbonate.

A 13.7 When aniline is treated with benzoyl chloride in the presence of base it gets converted into N-Phenylbenzamide.

A 13.8 The different isomers of the molecular formula: C₃H₉N are given in the table. However only 1° amines will liberate nitrogen gas on the treatment with h=the nitrous acid are given as follows:

Note: Only primary amines liberate nitrogen gas on treatment with nitrous acid.

A 13.9 When 3-methylaniline treated with NaNO₂ + HCl it gets converted into chlorine complex.

When that complex reacted with HBF₄ It gets converted into Barium Fluoride complex. This complex reacts with NaNO₂ in presence of copper to give 3-Nitrotoluene.

When aniline reacts with Br₂ water it gets converted into 2,4,6 tribromobenzamine. When this further reacted with NaNO₂/HCl it forms Chloride complex. This complex forms 1,3,5 tribromobenzene after treating with H₃PO₂ in presence of water.

(i) (CH₃ ) ₂CHNH₂ (ii) CH₃ (CH₂ ) ₂NH₂ (iii) CH₃NHCH(CH₃ ) ₂ (iv) (CH₃ ) ₃CNH₂ (v) C₆H₅NHCH₃ (vi) (CH₃CH₂ ) ₂NCH₃ (vii) m–BrC₆H₄NH₂

_{A 13.1}

1 - Methylethanamine

The root name is based on the longest chain with the -NH₂ attached. The chain is numbered so as to give the amine unit the lowest possible number. The longest chain is ethane chain which is further suffixed with ‘amine’.

2 - Propan-1-amine

The longest chain here is propane. The naming is such that amine unit should get a lowest possible number. Propane-1-amine can also be written as 1-propylamine.

3 - N−Methyl-2-methylethanamine

The chain is numbered so as to give the amine unit the lowest possible number. The other alkyl group is treated as a substituent, with N as the locant. The N locant is listed before numerical locants. The longest chain is ethane which has a Methyl substituent group.

4 - 2-Methylpropan-2-amine

Propane is the longest chain and is joined to the amine group at the 2^nd position. Another substituent methyl group is also at the same position.

5 - N−Methylbenzamine or N-methylaniline

Aniline is considered as the principle group. The other alkyl group is treated as a substituent, with N as the locant.

6 - N-Ethyl-N-methylethanamine

This is a tertiary group of amine. The root name is based on the longest chain with an amine having the lowest possible number. The alkyl groups are treated as substituents with N as the locant.

7 - 3-Bromobenzenamine or 3-bromoaniline

Benzene is considered as the principle group. The substituent is bromine which is numbered 3.

A 13.2 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 wont.

Hinsberg’s reagent (benzenesulphonyl chloride, C₆H₅SO₂Cl). 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.

Azo-dye test can be used to distinguish ethylamine and aniline.

Azo-dye test: A dye is obtained when aromatic amines react with HNO₂ (NaNO₂ + dil.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.

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.

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.

A 13.3 Pk_b value is the negative logarithm of the basicity constant (K_b) .i.e., pK_b = -logK_b

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

⇒ The structure of methyl amine is CH₃NH₂

⇒ The structure of aniline is:

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.

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

_⇒ The structure of ethylamine is CH₃CH₂NH₂

⇒ The structure of aniline is:

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

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.

⇒ 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.

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.

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.

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 is 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 anilne gives mainly p-nitroaniline

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

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 annhydrous 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 annhydrous aluminium chloride (AlCl₃) to form alkyl substituted benzene or its derivatives, this reaction is called Friedel-Crafts alkylation. For example:

⇒ Friedel-Crafts acylation: When any benzene or its derivative is treated with acetyl chloride (R-COCl) in the presence of annhydrous aluminium chloride (AlCl₃) to form acetyl substituted benzene or its derivatives, this reaction is called Friedel-Crafts acylation. For example:

Due to presence of positive charge on nitrogen (N) atom in the salt, the group N ⁺ H ₂ AlCl ^- 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: Diazonium salts have the general formula (R/Ar—N ⁺Cl^-) where R stands for the alkyl group and Ar stands for the aryl group. The structure is given below:

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:

The aliphatic amines, on the other hand, form highly unstable alkane diazonium salt (R—N ⁺Cl^-). They rapidly decompose even at low temperature(<272-278K) forming carbocation and nitrogen gas.

Hence, diazonium salts of aromatic amine are much more stable than aliphatic diazonium salts. Note: Carbocation is an ion in which carbon atom consists of positive charge

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 hallide gives N-substituted phthalimides. These upon susbsequent hydrolysis with dil.HCl under pressure or with alkali gives primary amines.

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

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

Overall reaction:

∴ 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.

• (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) C₆H₅NH_2, 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 .

A 13,4 Pk_b value is the negative logarithm of the basicity constant (K_b) .i.e., pK_b = -logK_b

Evidently, smaller the value of pK_b , 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, hence they increase the elecron density of N-atom and thus is easily available to donate electrons. This poperty makes aliphatic amines more basic.

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 C₆H₅NHCH_3, 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 pk_b value, therefore, pK_b values decrease in the reverse order:

C₆H₅NH₂ > C₆H₅NHCH₃ > C₂H₅NH₂ > (C₂H₅)₂NH₂

 

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

 

• 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

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 delocalization 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

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 are absent.

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 C₂H₅NH_2, 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

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.

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

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.

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

A 13.5

A 13.6 Primary amine: A primary (1°) amine is an amine that has the following general structural formula.

R= alkyl or aryl group

Secondary amine: A secondary (2°) amine is an amine that has the following general structural formula.

R¹ and R²= alkyl or aryl group

Tertiary amine: A tertiary amine is an amine that has the following structure

R¹, R² and R³ are alkyl or aryl groups

Identification of Primary, Secondary and Tertiary amines

Primary, secondary and tertiary amines can be identified by the following test:

Hinsberg’s test: This is an excellent test for the identification of primary, secondary and tertiary amines. In this test, the amine is shaken with benzenesulphonyl chloride ( Hinsberg’s reagent) in the presence of an excess of aqueous KOH solution when

• A primary amine gives a clear solution which on acidification gives an N-alkylbenzene sulphonamide which is soluble in

Due to the presence of strong electron withdrawing sulphonyl group in the sulphonamide, the H-atom attached to nitrogen can be easily released as a proton. So it is acidic and dissolves in alkali.

• A secondary amine reacts with Hinsberg's reagent to give a sulphonamide which is soluble in There is no H-atom attached to the N-atom in the sulphonamide Therefore it is not acidic and soluble in alkali.
• A Tertiary amine does not react with Hinsberg’s reagent at all

A 13.7 (i)  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.

For example,

(ii)  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.

(iii)  Hoffmann 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.

For example,

(iv)  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.(v)  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.

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

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.

(vi)  Acetylation

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

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

(vii)  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.

A 13.8 

A 13.9 (i) Ethyl iodide reacts with NaCN gives a substitution reactions to give propanitrile upon partial hydrolysis gives B , upon reaction with sodium hydroxide gives C.

(ii) 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.

(iii) Ethylbromide gives nucleophilic substitution reactions gives B ,upon reduction gives B followed by reacting with nitrous acid, i.e oxidation gives propanol.

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

(v) Acetic acid upon heating with ammmonia 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.

(vi) 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.

A 13.10

It is given that compound 'C' having the molecular formula,C₆H₇N is formed by heating compound 'B' with Br₂and KOH. This is a Hoffmann bromamide degradation reaction (in which isocyanide compound is formed). Therefore, compound 'B' is an amide and compound 'C' is an amine. The only amine having the molecular formula, C₆H₇N is aniline. Therefore, compound 'B' (from which 'C' is formed) must be benzamide.

Further, benzamide is formed by heating compound 'A' with aqueous ammonia. Therefore, compound 'A' must be benzoic acid. The given reactions can be explained with the help of the following equations:

• C₆H₅NH₂ + CH₃Cl + KOH →
• C₆H₅N₂Cl + H₃PO₂ + H₂O →
• C₆H₅NH₂ + H₂SO₄ (conc) →
• C₆H₅N₂Cl + C₂H₅OH à
• C₆H₅NH₂ + Br₂ (aq) à
• C₆H₅NH₂ + (CH₃CO)₂O à
• C₆H₅N₂Cl

A 13.11 C₆H₅NH₂ + CHCl₃ + KOH

It is a carbylamine reaction in which a isocyanide compound is formed along with side products of potassium chloride.Basically the name of reaction is given is due to formation of a foul smelling compound called as isocyanide.

2-      C₆H₅N₂Cl + H₃PO₂ + H₂O

Benzenediazonium chloride is a very reactive compound which oxidises hypophosphorous acid to hypophosphoric acid and the reactant is reduced to benzene.

3-      C₆H₅NH₂ + H₂SO₄(conc.)

Aniline undergoes sulphonation to anillium hydrogensulphate.

4-      C₆H₅N₂Cl + C₂H₅OH

Aniline is very activating group which undergoes reaction to give ortho and para product. But in acidic medium aniline accquires positive charge on N atom which is metadirecting group and hence the product.

5-      C₆H₅NH₂ + Br₂(aq)

It is clubbing reaction of a very reactive compound undergoing addition to give benzene.

6-      C₆H₅NH₂ + (CH₃CO)₂O

As mentioned aniline is avery activating group, so to reduce its activity it is reacted with anhydride to give n-phenylethanamine.

7-      C₆H₅N₂Cl

Hydrofluroburic acid reacts qith benzenediazonium chloride along with sodium nitrite to give a reduced product as Nitrobenzene.

A 13.12 Gabriel phthalimide synthesis is used for the preparation of aliphatic primary amines. It involves nucleophilic substitution (S_N2) of alkyl halides by the anion formed by the phthalimide.But aryl halides do not undergo nucleophilic substitution with the anion formed by the phthalimide.

Therefore aromatic primary amines cannot be formed by gabriel phthalimide process.

A 13.13 (i) Aromatic amines react with nitrous acid (prepared in situ fromNaNO₂ and a mineral acid such as HCl) at 273 - 278 K to form stable aromatic diazonium salts i.e., NaCl and water. This reaction is widely used for preparation of variety of compounds.

(ii) 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 is very reactive, which further produce alcohol and HCl with the evolution of nitrogen gas.

A 13.14 Aromatic amines react with nitrous acid (prepared in situ from NaNO₂and a mineral acid such as HCl) at 273 - 278 K to form stable aromatic diazonium salts e., NaCl and water.

• Aliphatic primary amines react with nitrous acid (prepared in situ from NaNO₂and a mineral acid such as HCl) to form unstable aliphatic diazonium salts, which further producealcohola and acid called as hydrochloric acid and evolution of nitrogen
• Aliphatic amines are stronger bases than aromatic amines due to following reasons:
  • The lone pair of electrons of the nitrogen atom of aromatic amines is involved in

conjugation with the π−bond pairs of the ring as follows

• Anilinium ion obtained by accepting a proton is less stabilized by resonance. Due to the −R effect of the benzene ring, the electrons on the N- atom are less available in the case of aromatic amines. Therefore, the electrons on the N-atom in aromatic amines cannot be donated

So, aniline is a weaker base than alkyl amines, in which +1 effect increases the electron density on the nitrogen atom.