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Asked by Kalyani Last Modified
Answer 
Buddhadeb Chandra
Tutor, The ultimate solution to the fear of Organic Chemistry, PhD from DRDO in Chemical Science.
Option (B) is the right answer.
Explanation: Before solving such types of problem we should have the concept of conjugate acid-base theory. Here in question given three conjugate base of their acid 1) CH3-CCH, 2) HCCH, 3) H3C-CH2- respectively. We know that unstable conjugate base provides stable acids and vice versa. If we order the decreasing stability of 1), 2) and 3) then we get 1>2>3, one get one extra substituent than 2 and 3 is a charge entity. According to this 2 provides more stable conjugate base than 1 i,e; B>A. But for 3 it contain one negative charge even after absorption of one proton. Therefor here conjugate acid base theory is not applicable. Hybridization play an important role in stabilization of particular entity. For A) and B) negative charge on sp hybridised molecular orbital (MO) whereas for C) it is on sp3 hybridized. Again sp MO provides extra stability to nagative charge because of its spherical size than dumble size sp3 orbital. Due to spherical size it remain more closely to nucleous and stabilized by delocalizing the charge around the nucleous. Due to dumble size sp3 MO remain far away from the nucleous. Thats why probability of delocalization is less. For this reason C) is highly unstable.
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