Question

The correct statements of the following are: A) Aniline forms a stable benzene diazonium chloride at 285K.
B) N - Phenylethanamide is less reactive towards nitration than aniline.
C) p - CH\(_3\)C\(_6\)H\(_4\)COCl is Hinsberg reagent.

• A. A \& B only
• B. A \& C only
• C. B only
• D. C only

Hint:

The Hinsberg test involves the reaction of amines with benzoyl chloride (p - CH\(_3\)C\(_6\)H\(_4\)COCl) to distinguish between primary, secondary, and tertiary amines based on the solubility of the products.

Answer 1

The Correct Option is C

Let's analyze each statement: A) Aniline forms a stable benzene diazonium chloride at 285K. Benzene diazonium chloride is stable at low temperatures (0-5°C or 273-278K). At 285K, it would decompose. So statement A is incorrect. B) N-Phenylethanamide is less reactive towards nitration than aniline. N-Phenylethanamide (acetanilide) is less reactive towards electrophilic substitution reactions like nitration compared to aniline. This is because the lone pair of electrons on the nitrogen atom in acetanilide is delocalized over the carbonyl group, making it less available for donation to the benzene ring. Thus, statement B is correct. C) p-CH$_3$C$_6$H$_4$COCl is Hinsberg reagent. Hinsberg reagent is benzenesulfonyl chloride (C$_6$H$_5$SO$_2$Cl). p-CH$_3$C$_6$H$_4$COCl is not Hinsberg reagent. Thus, statement C is incorrect. Therefore, the only correct statement is B. Final Answer: (3) B only