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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Priyanshu kumar Best Answer
Option A is correct as after deprotonation carbanion is most stabilised in case of A. In 2nd carbanion is not stabilized as no resonance is possible as after delocalisation bridge head carbon will be sp2 hybridised which is not possible according to Bredt's rule 3rd and 4th is also not more acidic than A as carbanion is destabilized by +R effect of methyl and ethoxy group
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