Bromination of 4‐methylcyclohex‐1‐ene

My problem is with the methyl group. If it wasn't there, I had no problem. When the halonium bridge is made does the methyl group shift to give a more stable transition state or not?

  • $\begingroup$ I would believe that Br is attached mostly on the Carbon 4, and Br or OH on the Carbon 3. $\endgroup$
    – Maurice
    Commented Dec 27, 2019 at 13:15
  • $\begingroup$ Why? I would expect a mixture close to 1:1 $\endgroup$
    – Waylander
    Commented Dec 27, 2019 at 13:19

2 Answers 2


In order for a methyl shift to occur, there would need to be an empty orbital resulting from a carbocation intermediate for the methyl group to occupy.

Halohydrin mechanism for trans addition of Br and OH

Note that in the bromonium intermediate shown above, neither carbon in the three-membered ring has an entire empty orbital available. Hence methyl shift doesn't occur and the mechanism continues as expected, resulting in halohydrin formation.

  • 1
    $\begingroup$ That's not quite true; alkyl shifts can occur with simultaneous loss of a leaving group, cf. pinacol/semipinacol-type rearrangements. The C–LG σ* orbital would play the same role as the empty orbital which you are speaking of. $\endgroup$ Commented Dec 27, 2019 at 18:58
  • $\begingroup$ @orthocresol Purely out of curiosity, would a methyl-shifted product be more favorable than the above scheme? I’m not far enough in my organic chemistry curricula to know the correct answer. $\endgroup$
    – elucidium
    Commented Dec 27, 2019 at 19:09
  • 2
    $\begingroup$ The methyl would have to shift over two carbons. Or more precisely, you would need a hydride shift followed by a methyl shift. I find that rather unlikely, so I agree with Waylander (comment on another answer) that there is no selectivity in this reaction. $\endgroup$ Commented Dec 27, 2019 at 19:18

methyl groups will not shift in a non-classical carbocation mechanism. This reaction will first have a $TS$ where the $\ce{Br2}$ molecule makes a 3 membered ring with the $\pi%$ bond and the water molecule will attach itself to $\ce{C3}$, as that is a more stable electron-deficient site due to the inductive effect of the methyl group.

  • 1
    $\begingroup$ Your answer would be fine for cyclohexene but you omit regiochemistry. An inductive effect of methyl is not sufficient. A Reaxys search gave no hits but 7 for direct epoxidation with no sign of facial selectivity. See chemistry.stackexchange.com/questions/86341/… for a guideline on how to think about such a problem. $\endgroup$
    – user55119
    Commented Dec 27, 2019 at 18:35

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