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Which of the following statements is correct about methyl group?

  1. It stabilizes both carbocation and free radical equally.
  2. It stabilizes a free radical more than a carbocation.
  3. It stabilizes a carbocation more than a free radical.
  4. None of these.

I thought the answer would be (1). Keeping in mind these structures: The methyl group can stabilize the carbocation and the free radical group attached to it equally through hyperconjugation. However, the answer given is (3). How is this possible? Why is the carbocation more stable than the free radical? Are the structures I drew wrong?

Gaurang Tandon
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dr.drizzy
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  • Can you give an example where stabilization by radical formation is seen. – Avyansh Katiyar Feb 27 '18 at 12:59
  • @AvatarShiny You'll have to scroll a bit and you will see free radical stabilization due to hyperconjugation. https://hubpages.com/education/Thorough-Chemistry-of-Hyperconjugation – dr.drizzy Feb 27 '18 at 13:03
  • related https://chemistry.stackexchange.com/questions/64875/why-are-radical-intermediates-are-more-stable-on-tertiary-carbons – Mithoron Feb 27 '18 at 20:31
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    @raajsuriya You can't draw a resonance structure involving all three hydrogen atoms on an adjacent methyl. They can't all be oriented properly for hyperconjugation at the same time. Let's just say that looking briefly at the linked web page, I'm not impressed. – Zhe Feb 27 '18 at 22:52
  • @Zhe The link was not important I googled that quickly for Avtar Shiny. Do you know why carbocations are stabilized more than free radicals through hyperconjugation? Or is this something I should memorize? Thank you for your time – dr.drizzy Feb 28 '18 at 05:48
  • First, why do you think that's true. Second, what do you mean by "stabilized more?" – Zhe Feb 28 '18 at 13:46
  • @Zhe I have edited my question. Maybe this will be clearer. – dr.drizzy Feb 28 '18 at 20:28
  • Well, it certainly can't be equal since their different. But I don't understand how you plan to compare stability of a radical against stability of a cation. They're different and have different modes of reactivity... – Zhe Feb 28 '18 at 20:55
  • I only came across this question, I'm just as confused as well. @Zhe – dr.drizzy Feb 28 '18 at 21:13
  • I think the question is not so difficult (at leats if we accept some ideas). Suppose those structures don't have hyperconjugation, and then, they have. So, you need a difference of energy for each process and compare it. Possibly, it will be enough to say that carbocation atracts more electron density, so, it will be more stabilized.. –  Mar 01 '18 at 16:34
  • Related: https://chemistry.stackexchange.com/a/31729/31775 – Apoorv Potnis Mar 08 '18 at 09:55

1 Answers1

1

A carbonation is deficient in 2 electrons whereas a free radical is only 1 electron deficient. Thus any type of stabilisation working on a carbonation will be much more effective as it is more electron deficient. Hence, hyperconjugation or inductive electron donation stabilises a carbocation more than it does a free radical.

Gaurang Tandon
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Dipan Das
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