It is known that Methyl has a greater +I effect than Ethyl, but here, in this case, the condition is satisfied in the para benzene carbocations whereas, in the meta forms, it is the opposite. Shouldn't meta methyl benzene carbocation be more stable than meta ethyl benzene carbocation? Please, Can anyone explain?
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$\begingroup$ I would like to point out that the species you are talking about are not benzene carbocations but phenylic carbocations. $\endgroup$– Nisarg BhavsarMay 25, 2021 at 5:49
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$\begingroup$ And yes, no resonance or hyperconjugation effects are observed from the meta position and thus the higher inductive effect of ethyl group should make it a stabler structure. $\endgroup$– Nisarg BhavsarMay 25, 2021 at 5:52
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$\begingroup$ Sorry for the mistake. $\endgroup$– Guddu LarganMay 25, 2021 at 6:06
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$\begingroup$ Does this mean methyl has a greater +M effect than ethyl. How? $\endgroup$– Guddu LarganMay 25, 2021 at 6:07
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$\begingroup$ Ethyl and methyl don't have an M effect they have a H effect. $\endgroup$– Nisarg BhavsarMay 25, 2021 at 9:36
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1 Answer
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We know that effect of neither hyperconjugation nor resonance is observed if the groups are attached in meta position. Hence, stability of structures I and IV will be decided on the basis of inductive effect only. Alkyl groups exert +I effect, and ethyl group exerts greater +I than methyl, hence stabilizing the electron deficient benzylic carbocation more. So, IV>I.
Now note that both of these structures must be less stable than II and III, because these are stabilized by both hyperconjugation and inductive effect, and hyperconjugation is a more stabilizing phenomenon than inductive effect. Since II has more $\alpha$-H than III(3 and 2, respectively), there would be more hyperconjugative structures for II, hence more stability. Hence, finally, the order is I<IV<III<II. This is the first option.
