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I am a class 8th student preparing for chemistry olympiad. I am having slight confusion regarding how to decide the reactivity towards electrophilic addition reactions. What should we give more priority to? To the stability of carbocations formed, or the electron density on the substrate?

a: 1,2-dimethyl-1-cyclohexene; b: 1-methylcyclohex-1-ene; c: cyclohexene; d: methylenecyclohexane

In this question if we finally add an electrophile and check the stability of cation using hyperconjugation effect. b = d > a > c seems to be the answer, but the solution says a > b due to the electron density because of the hyperconjugation effect on both sides of the π-bond which increases the electron density.

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    $\begingroup$ I am still slightly confused by this question. It would be great so see the actual wording and what is asked. Could you cite the source of it, please. I'd guess it's part of a preparation kit of some kind. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 27, 2023 at 18:58

2 Answers 2

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The reaction rate depends on the energy difference between the state of the substrate and the transition state. The energy of the transition state is close to that of the carbocation but not quite so there can be subtle surprises.

My first blush predictions are: a, d, b, c

Reasoning: a. extra methyl stabilizes both ground and transition states but should have greater effect on transition state. Also entropy favored 2 sites d. Same intermediate as b; the transition state has lower ring interactions. Exo double bond is of higher energy [I think; I did not cheat and look it up]. Models might help here. b. tertiary cation. c. secondary cation higher energy.

Both ground state and transition state must be considered. The classic example of this is benzene that introduced the new concept of resonance stabilizing the ground state.

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    $\begingroup$ The reputable source here is the original data giving the rates etc. of these compounds under SN1 conditions., not the answer on an answer sheet to a MC question. $\endgroup$
    – jimchmst
    Commented Aug 24, 2023 at 22:10
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I agree with the poster above. A, D, B, C seems to be the correct order. The logic behind this answer is that A gets more inductive donation from the methyl groups across the double bond. For D, the double bond is less substituted, so more unstable and more prone to reactivity. For B, inductive donation from the methyl group. For C, it does not receive any inductive donation at all, and it's quite substituted, so it is stable as is.

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  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Aug 26, 2023 at 1:18
  • $\begingroup$ It is the responsibility of the framer of the test question to support the answer with data and citations. The real question about considering ground state or transition state or intermediate energies is much deeper than the "corr $\endgroup$
    – jimchmst
    Commented Aug 27, 2023 at 6:13
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    $\begingroup$ @jimchmst I guess you wanted to write some more. $\endgroup$
    – Martin - マーチン ♦
    Commented Aug 27, 2023 at 18:59

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