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Question #44482

Difficulty: 8/10 | 🟢 Free

The Question

Compound \textbf{P} (1-chloro-4-nitrobenzene) is subjected to the following sequence of reactions: \begin{enumerate} \item \ce{CH3Cl} (1 equiv.), Anhydrous \ce{AlCl3}, $\Delta$ $\rightarrow$ \textbf{Q} \item \ce{CH3ONa} (1 equiv.), \ce{CH3OH}, $\Delta$ $\rightarrow$ \textbf{R} \item \ce{Sn / HCl}, $\Delta$ , followed by neutralization with \ce{NaOH(aq)} $\rightarrow$ \textbf{S} \item \ce{NaNO2 / HCl}, $273-278\text{ K}$ , followed by \ce{CuCN / KCN}, $\Delta$ $\rightarrow$ \textbf{T} \end{enumerate} Identify the major organic product \textbf{T}.

Options

A.
4-methoxybenzonitrile
B.
4-methoxy-3-methylbenzonitrile
C.
4-chloro-3-methylbenzonitrile
D.
4-chlorobenzonitrile

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Friedel-Crafts Alkylation Limitations|Nucleophilic Aromatic Substitution (SNAr)|C-X Bond Polarization and Polarizability

Why This Question Is Tricky

Common Mistakes

This problem tests the contrasting electronic requirements of EAS and SNAr.

The primary trap (Option B) catches students who mechanically apply Friedel-Crafts alkylation, failing to recognize that nitro-deactivated rings are completely inert to F-C reactions (nitrobenzene is often used as an inert solvent for them).

Conversely, the very same electron-withdrawing nature of the nitro group is what activates the ring for SNAr in the subsequent step by stabilizing the carbanion intermediate.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Hydrocarbons - Class 11 Part 2
Concept	Friedel-Crafts Alkylation Limitations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free