JEE Chemistry 2023 Question with Solution

Given: Four proposed orders of reactivity are shown for SN2, SN1, electrophilic substitution, and nucleophilic substitution.

Find: Which of the statements (A), (B), (C), and (D) are correct.

For SN2, steric hindrance controls the rate. The less hindered substrate reacts faster, so the stated order with increasing substitution is correct.

For SN1, the reaction proceeds through a carbocation intermediate. A benzylic carbocation is highly stabilized by resonance with the aromatic ring, so the given order is correct.

For electrophilic aromatic substitution, electron-donating groups activate the ring, while electron-withdrawing groups deactivate it. Also, Cl is weakly deactivating, whereas NO2 is strongly deactivating. Hence, the given order is correct.

For nucleophilic aromatic substitution, strong electron-withdrawing groups at ortho or para positions stabilize the intermediate and increase reactivity. NO2 enhances this reactivity much more than Br, so the given order is correct.

Therefore, (A), (B), (C), and (D) are all correct. The correct option is D.

Given: The question compares reactivity orders for four different mechanisms.

Find: Check each statement separately.

1. (A) SN2 Reaction:

• The nucleophile attacks from the backside.
• A transition state is formed in which the nucleophile and leaving group are partially bonded to carbon.
• Steric hindrance plays the decisive role.
• The reactivity order is Primary (1°) > Secondary (2°) > Tertiary (3°).
• Therefore, the given order is correct.

2. (B) SN1 Reaction:

• The reaction proceeds through a carbocation intermediate.
• The rate-determining step is carbocation formation.
• Benzylic carbocations are highly stable because of resonance.
• Therefore, the given order is correct.

3. (C) Electrophilic Substitution:

• Electron-donating groups increase electron density on the aromatic ring and activate it toward electrophiles.
• Electron-withdrawing groups reduce electron density and deactivate the ring.
• Cl is weakly deactivating, whereas NO2 is strongly deactivating.
• Therefore, the given order is correct.

4. (D) Nucleophilic Substitution:

• In nucleophilic aromatic substitution, electron-withdrawing groups at ortho and para positions to the leaving group stabilize the intermediate.
• NO2 is strongly electron-withdrawing compared with Br.
• Therefore, the given order is correct.

Conclusion: All four statements are correct, so the correct option is D.