Before You Begin

Question #44914

Difficulty: 8/10 | 🟢 Free

The Question

An amount of $12.1 \text{ g}$ of benzamide ( $\ce{C6H5CONH2}$ ) undergoes Hofmann bromamide degradation with a $75\%$ yield to produce an aromatic amine, $\mathbf{A}$ . The obtained sample of $\mathbf{A}$ is divided into two equal portions, Portion I and Portion II.

Portion I is treated with excess bromine water at $298 \text{ K}$ to yield compound $\mathbf{B}$ quantitatively.

Portion II is first protected using acetic anhydride in the presence of pyridine, then treated with $\ce{Br2}$ in acetic acid at $298 \text{ K}$ , and finally subjected to acidic hydrolysis to yield the major mono-substituted product $\mathbf{C}$ quantitatively.

Assuming $100\%$ yield for all steps involving the conversion of $\mathbf{A}$ to $\mathbf{B}$ and $\mathbf{C}$ , what is the total mass of bromine atoms incorporated into the final isolated products $\mathbf{B}$ and $\mathbf{C}$ combined? (Molar mass: Benzamide $= 121 \text{ g/mol}$ , $\ce{Br} = 80 \text{ g/mol}$ )

Options

A.
$12.0 \text{ g}$
B.
$6.0 \text{ g}$
C.
$16.0 \text{ g}$
D.
$18.0 \text{ g}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Electrophilic Substitution Control|Hoffmann Bromamide Rearrangement|Ammonolysis Complexity

Why This Question Is Tricky

Common Mistakes

This problem tests the student's mastery of Electrophilic Aromatic Substitution (EAS) control and the Hofmann bromamide degradation.

The core 'trap' lies in the differential reactivity of aniline versus its protected counterpart, acetanilide.

Students frequently overlook the 'polysubstitution trap': the unprotonated $-\ce{NH2}$ group exerts such a powerful $+M$ (mesomeric) effect that reaction with aqueous bromine instantaneously yields 2,4,6-tribromoaniline, making mono-substitution impossible under these conditions.

To achieve mono-substitution, the amine must be protected via acetylation.

The acetyl group diminishes the ring-activating power of nitrogen through cross-conjugation with the carbonyl oxygen and introduces steric bulk, directing the incoming electrophile primarily to the para position.

The distractors mathematically account for students who incorrectly assume either mono-bromination or tri-bromination occurs in both pathways, or those who fail to apply the initial reaction yield.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Amines - Class 12
Concept	Electrophilic Substitution Control
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free