Exams › JEE Main › Chemistry

Reaction of propanamide with Br2/KOH (aq) produces:

1. (1) Propanenitrile
2. (2) Ethylnitrile
3. (3) Ethylamine
4. (4) Propylamine

Correct answer: (3) Ethylamine

Solution

Propanamide undergoes a reaction known as the Hofmann degradation when treated with bromine and aqueous potassium hydroxide, resulting in the formation of ethylamine through the cleavage of the carbon chain and the release of nitrogen.

Related JEE Main Chemistry questions

• In Kjeldahl’s method, the nitrogen present in the sample is determined in the form of:
• What product is formed when benzenediazonium chloride (C6H5N2+Cl−) is treated with CuCl?
• When o-methoxybromobenzene is first reacted with sodamide and then treated with ammonia, which compound is obtained as the final product?
• An organic compound A, when treated with NH3, forms B. On heating, B gives C, and C on treatment with Br2 in the presence of KOH yields ethylamine. The compound A is:
• An organic nitrogen-containing compound gives an oily nitrosoamine when treated with aqueous nitrous acid at low temperature. Which compound is it?
• Which of the following statements are true? (i) In the Sandmeyer reaction, groups such as Cl, Br and CN are introduced into a benzene ring in the presence of Cu+ ions. (ii) In the Gattermann reaction, substitution into the benzene ring occurs in the presence of copper powder and HCl. (iii) The yield obtained in the Gattermann reaction is better than that in the Sandmeyer reaction.

⚔️ Practice JEE Main Chemistry free + battle 1v1 →