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Question #44923
The Question
A chemist attempts to synthesize a haloarene using the following reaction sequence:
\begin{align*} \text{Step 1: } & \ce{C6H5NH2} \xrightarrow[\text{Pyridine}]{\ce{PhSO2Cl} \text{ (1.0 eq)}} \textbf{P} \ \text{Step 2: } & \textbf{P} \xrightarrow{\ce{Br2}, \ce{FeBr3}} \textbf{Q} \text{ (Major)} \ \text{Step 3: } & \textbf{Q} \xrightarrow[\Delta]{\ce{H3O+}} \textbf{R} \ \text{Step 4: } & \textbf{R} \text{ (2.0 eq)} \xrightarrow[\text{273 K}]{\ce{NaNO2} \text{ (1.0 eq)}, \ce{HCl} \text{ (1.0 eq)}} \textbf{S} \text{ (Major)} \end{align*}
Identify the major organic product \textbf{S}.
Options
- A.
\ce{1,3-bis(4-bromophenyl)triaz-1-ene}
- B.
\ce{4-bromo-2-((4-bromophenyl)diazenyl)aniline}
- C.
\ce{4-bromobenzenediazonium chloride}
- D.
\ce{4,4'-dibromoazobenzene}
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
Sandmeyer vs Gattermann|Hinsberg Reagent Selectivity|Ammonolysis Complexity
Why This Question Is Tricky
Common Mistakes
This problem tests the student's understanding of stoichiometric constraints and competitive pathways in diazotization.
While students memorize as the reagent for diazonium salt formation, they often overlook the necessity of excess acid.
With limiting acid, the unreacted primary amine acts as a nucleophile, leading to N-coupling (diazoamino formation) rather than a stable diazonium salt.
Distractor B traps students who know about azo coupling but forget that primary amines undergo N-coupling before C-coupling.
Distractor C traps those who blindly apply Sandmeyer conditions without checking stoichiometry.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Amines - Class 12 |
| Concept | Sandmeyer vs Gattermann |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |