Question

Given below are two statements :
Statement I : Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine.
Statement II : Nitration of aniline with $HNO_3/H_2SO_4$ at 288 K produces m-nitroaniline in higher amount than o-nitroaniline (pH adjusted).
In the light of the above statements, choose the correct answer from the options given below :

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Statement I describes the Hoffmann bromamide degradation reaction. Statement II describes the unusual nitration behavior of aniline.
Step 2: Detailed Explanation:
Statement I: The Hoffmann bromamide reaction converts an amide ($RCONH_2$) to a primary amine ($RNH_2$) with one less carbon. Benzamide ($C_6H_5CONH_2$) reacts to form Aniline ($C_6H_5NH_2$), not benzylamine ($C_6H_5CH_2NH_2$). Thus, Statement I is False.
Statement II: In strongly acidic nitrating mixtures, aniline is protonated to the anilinium ion ($-NH_3^+$), which is a strongly deactivating meta-directing group. This leads to 47% m-nitroaniline, whereas o-nitroaniline is only 2%. Thus, Statement II is True.
Step 3: Final Answer:
Statement I is false but Statement II is true.