Test: Preparation of Amines(22 Dec) - JEE MCQ

The correct answer is option C
CH_3​CH₂​CONH₂​ (A)⟶ CH₃​ − CH₂ ​− NH₂​ (B)⟶ ​CH_3​ − CH₂ ​− OH
In the above sequence A & B respectively are Br₂​/KOH and HNO₂​
The first step is Hoffmann bromamide degradation reaction in which an amide (propanamide) is converted to an amine  (ethylamine) containing one carbon atom less. The reagent A is bromine in presence of KOH. In the second step, aliphatic primary amine (ethyl amine) reacts with nitrous acid (reagent B) to form aliphatic primary alcohol (ethyl alcohol).
 

Fe + HCl is preferred due to the following reasons:
- Scrap iron i.e Fe is cheap and commercially easily available
- FeCl₂ formed will get hydrolysed and release HCl, so, HCl which is required in the reaction will be produced by itself
- Hence only small amount of HCl is required to initiate the reaction.

Aniline is prepared commercially by the catalytic hydrogenation of nitrobenzene or by the action of ammonia on chlorobenzene. The reduction of nitrobenzene can also be carried out with iron borings in aqueous acid. A primary aromatic amine, aniline is a weak base and forms salts with mineral acids.

compound is most basic due to localized lone pair of electron on nitrogen atom while other compounds have delocalized lone pair of electron.