6
$\begingroup$

In regard to the mechanism of halogen addition, consider the situation when we add a halogen (say, bromine) to an unsymmetrical alkene about the double bond (say, propene) in the presence of $\ce{CCl4}$.

I knew the mechanism would be something of this sort:

A cyclic bromonium ion would be formed with the two carbons of the double bond. The second bromide ion (acting as a nucleophile) attacks on the more substitued carbon from the side opposite to the bromonium ion to cause ring opening and formation of a vic-dibromide.

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.

I have two questions in this regard:

  1. Which of the two is the correct mechanism?
  2. If the former is indeed the correct mechanism, then would the nucleophile attack the least substituted carbon? (seeing as how the partial positive charge on the more substituted carbon is nuetralised to a greater extent due to $+I$ effect of attached methyl/ethyl groups).
$\endgroup$
1
  • $\begingroup$ If only ethene forms a cyclic bromonium ion, then there would be no stereospecificity in the bromination of the 2-butenes. Secondly, the active brominating agent is Br3+ Br-. $\endgroup$
    – user55119
    Apr 8, 2021 at 17:52

1 Answer 1

1
$\begingroup$

The first mechanism is more likely unless specific conditions of the experiment dictate otherwise. In the presence of $\ce{CCl4}$, the addition of $\ce{Br2}$ takes place via the formation of a cyclic halonium ion as intermediate as follows:

enter image description here

$\endgroup$
1
  • 1
    $\begingroup$ That's a bromonium cation and fishing hook arrows are for radical mechanisms. $\endgroup$
    – Mithoron
    Aug 28, 2023 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.