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Question #44924
The Question
Benzene undergoes electrophilic aromatic substitution with \ce{Cl_2 / FeCl_3} to give \textbf{A}, which upon nitration with a mixture of conc. \ce{HNO_3} and conc. \ce{H_2SO_4} yields \textbf{P} as the major isomer. \textbf{P} is then subjected to two different reaction pathways:
\textbf{Pathway 1:}
- \ce{Sn / HCl}, then \ce{NaOH(aq)} \textbf{Q}
- \ce{NaNO_2 / HCl}, \textbf{R}
- \ce{CuCN / KCN}, \textbf{S}
- \ce{CH_3MgBr}, dry ether, then \ce{H_3O^+}, \textbf{T}
\textbf{Pathway 2:}
- \ce{LiAlH_4} (excess), dry ether, , then \ce{H_2O} \textbf{U}
Identify the major products \textbf{T} and \textbf{U}.
Options
- A.
\textbf{T} = 1-(4-chlorophenyl)ethan-1-one, \textbf{U} = Azobenzene
- B.
\textbf{T} = 1-(4-chlorophenyl)ethan-1-one, \textbf{U} = 4,4'-dichloroazobenzene
- C.
\textbf{T} = 1-(4-chlorophenyl)ethan-1-amine, \textbf{U} = 4-chloroaniline
- D.
\textbf{T} = 4-chlorobenzaldehyde, \textbf{U} = 4,4'-dichlorohydrazobenzene
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|Reduction Selectivity (Sn/HCl vs LiAlH4)|Aniline Resonance Pitfalls
Why This Question Is Tricky
Common Mistakes
The question tests chemoselectivity in reduction reactions, contrasting the non-selective nature of LiAlH4 with the controlled reduction by Sn/HCl.
A common trap is assuming LiAlH4 reduces nitroarenes to anilines (it actually forms azo compounds) or failing to account for the hydrodehalogenation of aryl halides by excess LiAlH4.
Students who miss the reductive cleavage of the C-Cl bond will incorrectly choose 4,4'-dichloroazobenzene.
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 |