Chapter 13: Amines

Ans:

In the nitrating mixture of conc. $\mathrm{H}_{2} \mathrm{SO}_{4}+$ conc. $\mathrm{HNO}_{3}, \mathrm{HNO}_{3}$ act as base to generate the nitronium ion which is an initiation species for the nitration reaction.

$\mathrm{HNO}_{3}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \underset{\text { Electrophile }}{\mathrm{NO}_{2}^{+}}+\underset{4}{\mathrm{HSO}_{4}^{-}}+\mathrm{H}_{2} \mathrm{O}$

Ans:

As $-\mathrm{NH}_{2}$ is a strong activating group, the aniline will readily undergo electrophilic substitution reaction, and it is difficult to cease reaction at the mono substitution stage. So, direct acetylation of aniline is not possible as it undergoes oxidation. 

Therefore, the activating group $-\mathrm{NH}_{2}$ is protected by an acetylation process and then followed by the nitration reaction to obtain a monosubstituted nitrated product. 

The acetylated complex formed utilises the lone pair of nitrogen and is less available for donation, this helps to carry out the nitration reaction easily. 

Ans: When Benzyl amine is treated with nitrous acid, firstly BDC is formed which is unstable and decomposes to Benzyl alcohol. 

$\underset{\text { Benzyl amine }}{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{NH}_{2}}+\mathrm{HONO} \stackrel{\mathrm{HCl}}{\longrightarrow} \underset{\text { Benzyl alcohol }}{\longrightarrow} \underset{6}{\mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}}+\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O}$

Ans: On treatment of nitriles with strong reducing agents such as lithium aluminium hydride in ether or sodium metal in alcohol, primary amines are obtained. 

$\underset{\text { Alkyl nitrile }}{R-\cdots} \mathrm{CN}+4[\mathrm{H}] \frac{\mathrm{Na} / \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}{\text { or } \mathrm{Li} \mathrm{Ah}_{4} / \mathrm{ether}} \longrightarrow R-\mathrm{Cl}_{2} \mathrm{NH}_{2}$

Ans:

Ans: Hinsberg's reagent is also known as benzenesulphonyl chloride ($\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$). When it reacts with primary and secondary amines, it produces sulphonamides. Allowing secondary and tertiary amines to react with Hinsberg's reagent allows them to be distinguished (benzenesulphonyl chloride $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$). Secondary amines react with Hinsberg's reagent to form an alkali-insoluble product. N, N-diethylamine, for example, reacts with Hinsberg's reagent to form N,N-diethylbenzenesulphonamide, which is insoluble in alkalis. Tertiary amines, on the other hand, are unaffected by Hinsberg's reagent.

Ans: The reaction of aniline with nitrous acid at 273-278 K produces benzene diazonium chloride. The reaction of sodium nitrite with hydrochloric acid produces nitrous acid in the reaction mixture. Diazotisation is the process of converting primary aromatic amines into diazonium salts. Because of its instability, the diazonium salt is generally not stored or used immediately after preparation.

The crystalline solid benzene diazonium chloride is colorless.

It is easily soluble in water and stable at room temperature, but it reacts with water when warmed.

In the dry state, it decomposes easily.

Ans: As $-\mathrm{NH}_{2}$ is a strong activating group, the aniline will readily undergo electrophilic substitution reaction, and it is difficult to cease reaction at the mono substitution stage. 

Therefore, the activating group $-\mathrm{NH}_{2}$ is protected by an acetylation process.

The acetylated complex formed utilizes the lone pair of nitrogen and are less available for donation, this helps to carry out the nitration reaction easily. 

Ans: As the electronegativity of oxygen is more than the electronegativity of a nitrogen atom, the $\mathrm{O}-\mathrm{H}$ bond is more polar than the $\mathrm{N}-\mathrm{H}$ bond, therefore $\mathrm{MeOH}$ is stronger acid than $\mathrm{MeNH}_{2}$ or $\mathrm{MeNH}_{2} 2$ is stronger base than $\mathrm{MeOH}$.

Ans: Amides are the byproducts of the acylation reaction. The reaction is carried out in the presence of a stronger base than the amine, such as pyridine, which removes the formed HCl and shifts the equilibrium to the right.