When we add one equivalent of $\ce{Br2}$ then which double bond will get brominated?


The double bond in the molecule with methyl group will be more stable as it will have more hyper conjugating structures. Thus the other double bond should get brominated.

But my textbook says the double bond with methyl group will get brominated. Can someone explain why?


2 Answers 2


You misunderstood what hyperconjugation is. It occurs when the electrons in a $\ce{C-H}$ $\sigma$-bond are shared with an adjacent atom's empty or partially filled $\mathrm{p}$-orbital, giving the species an extended molecular orbital and thereby stabilizing it. This means that hyperconjugation doesn't really come into play until bromine is added across one of the double bonds. Let's consider the two possible intermediates:



The first one is a tertiary carbocation, and the second is a secondary carbocation. The first intermediate is more stable due to the hyperconjugative effects of the methyl group, and will require less energy to form than the second. This means the reaction will occur much faster at the more substituted double bond, and is why that is the major product.

  • 1
    $\begingroup$ Thanks, ringo. I thought when br2 is added to a double bond then cyclobromium ring is form and br- ions comes from the opposite and the step becomes concerted without formation of carbocation. $\endgroup$
    – nikola
    Jul 15, 2016 at 5:45
  • $\begingroup$ @nikola It is formed, but the tertiary carbocation is the most important resonance structure. $\endgroup$
    – ringo
    Jul 15, 2016 at 5:47
  • 2
    $\begingroup$ You might also add that the substituted alkene will be more nucleophilic due to the increased electron density from hyperconjugation with the $\pi ^*$ orbital. $\endgroup$
    – bon
    Jul 15, 2016 at 9:04

Hyper conjugation supports transition state structures. More hyperconjugation = lower energy of TSS = faster reaction = this moiety is more prone to react.


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