Amines

(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.

(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.

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

 

(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

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

 

(b) 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 influenc

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.

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 the 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.