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Which of the following correctly represents the order of basic strength? (A) Cyclohexadienyl anion < tert-alkyl-substituted cyclohexadienyl anion. (B) Alkoxide adjacent to a conjugated diene < more-substituted simple alkoxide. (C) Dimethylamino group bearing two nitro substituents is more basic than the same group bearing three nitro substituents. (D) Para-aminophenol with a tert-butyl group is more basic than ortho-aminophenol with a tert-butyl group.

1. (A) Cyclohexadienyl anion < tert-alkyl substituted cyclohexadienyl anion
2. (B) Conjugated diene alkoxide < more substituted alkoxide
3. (C) NMe2 with two NO2 > NMe2 with three NO2
4. (D) Aminophenol with tert-butyl at para > aminophenol with tert-butyl at ortho

Correct answer: (C) NMe2 with two NO2 > NMe2 with three NO2

Solution

(A): Tert-alkyl groups donate electrons, destabilising the carbanion (raising its energy), making it a stronger base. So tert-alkyl substituted > unsubstituted; the order given (less < more) is correct, BUT the comparison of anion stabilities for carbanions vs stabilised systems requires care — the statement is broadly correct in direction. (B): A conjugated diene delocalises the negative charge on the adjacent alkoxide (resonance stabilisation) -> more stable anion -> weaker base. A more-substituted alkoxide without conjugation is destabilised by alkyl donation -> stronger base. So conjugated diene alkoxide < more-substituted alkoxide is correct in direction. (C): Each additional NO2 group withdraws electron density from N by resonance/induction, reducing basicity. Two NO2 -> more basic than three NO2. TRUE and unambiguous. (D): The position (para vs ortho) affects steric and electronic factors differently; the statement as worded is ambiguous. Among these, (A), (B), and (C) are correct in principle; (C) is the most unambiguous and standard textbook statement.

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