If we have an alkene reacting with bromine, then I know that the the reaction happens by a cyclic intermediate. But if we add alcoholic KOH to the products formed, then what will be the major products formed?

I don't understand the meaning of trans addition in case of only one π and σ bonds where the bond formed after addition is free to rotate as it is σ bond. If bromine adds in an anti manner, then can it not convert to the cis form?

I am not sure if using the terms like 'trans' and 'cis' in case of σ bond is correct. I am not sure whether it's a molecule of bromine or HBr will be eliminated after addition of alcoholic KOH.

  • $\begingroup$ HBr gets elimianted. Depending on the molecule you may get 1 or 2 equivalents of HBr lost so diene formation is possible. $\endgroup$ – Waylander Mar 20 at 8:18
  • $\begingroup$ @Waylander I am afraid. It's not very clear to me. $\endgroup$ – Pan Mar 20 at 8:33
  • $\begingroup$ @Waylander can you please explain why will HBr be eliminated and not Br2 ? $\endgroup$ – Pan Mar 22 at 2:41
  • $\begingroup$ The mechanism of alkene formation by KOH is removal of a proton by OH- to give H2O and leaving a carbanion. The carbanion very quickly eliminates Br-, KBr is formed in solution but overall the components of HBr are lost from the moelcule. $\endgroup$ – Waylander Mar 22 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.