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1: 1‐(2‐methylbutan‐2‐yl)cyclopentan‐1‐ylium; 2: 1‐tert‐butylcyclopentan‐1‐ylium

(1) should be more stable as the $\ce{CH3}$ group is electron donating. But apparently (2) is more stable. Is there another phenomenon taking place here?

andselisk
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user638473
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    As formulated one should expect 2 to be more stable. Point is eventually why is the ethyl fragment not stabilising more than a methyl, as usually found and as resumed here https://chemistry.stackexchange.com/questions/47008/why-do-larger-alkyl-groups-show-a-greater-inductive-i-effect – Alchimista Mar 04 '19 at 16:15
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    This is getting absurd. I quite wonder how detached from reality these textbook exercises would get. For all practical purposes there no difference here. You can argue about "inductive effect" vs "hyperconjugation" however long you wish... – Mithoron Mar 04 '19 at 17:43
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    On the same line of @Mithoron though the example might be of interest in a proper contest (e.g. computational approach) it seems to stretched simple rules to the limit. – Alchimista Mar 05 '19 at 08:36
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    So does that mean these structures simply cannot be compared? – user638473 Mar 05 '19 at 11:58
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    No - you YES can. But they should be rather the same and whatever difference should be experimentally seen than explained, not the vice versa. It is the weight of subtle hyperconjugation that might differentiate between. – Alchimista Mar 05 '19 at 16:06
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    @Alchimista can you not argue using the Boyer's strain?, The pentagonal geometry increases the s character on the carbocation...which further develops it's way asking for the $e^-$ density from 3 hydrogens using hyperconjugation (methyl) or 2 (for ethyl) –  Aug 13 '20 at 08:33
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    @Alchimista Indeed the s char is high, I checked and found it to be 35% s and remaining 64% –  Aug 13 '20 at 08:41

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