Theoretically , I think the stability of carbanion formation increasing in the following order A
thank u
but:
no there are not this answer (C>B>A) in the following answers
you must Choose one of the options below:
1-C>A>B
2-B>A>C
3-A>B>C
4-B>C>A
????????
I think the right answer is B>C>A. Probably the question is about the conjugated base stability. In B the base generated is aromatic, in C stabilized by hyperconjugation and in A is anti aromatic. The more stable base will have the more fast reaction.
I think it's B>C>A .. as the delocalized electrons in B gives it more stability and faster reaction rate ... but A has the slowest rate because there are so much negative charge in a small ring ,this will make the system relatively unstable ....All this makes C in the middle of the order
The B> C>A is more logical as B satisfies the requirement of aromaticity once anion is generated and should be more stable. Then, C because its benzyllic anion and will be comparatively stable than C, though there may not be any assistance from benzene resonance for its stability. A will be least stable as, A is highly strained due to 3 membered ring and anion will further destabilize it as its anion in the ring will increase anti-aromatic character. Hence, in my opinion, their stability will decide rate of reactivity as B> C> A.
Although i think that driving a computational study will give a better answer since there are a lot of parameters need to be taken care. Before discussing about the carbanions, we should regard the steric hindrance of the reactants. With this perspective, the reactivity order should be B> A>C. Because the slowest step will determine the rate of reaction and as everybody knows this step is much slower then the reactions after formation of the carbanion.
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