Draw the stepwise mechanism of hydrohalogenation of ethylenecyclopentane?

1 Answer
Truong-Son N.
Sep 12, 2015

The mechanism for hydrobromination, a specific version of hydrohalogenation, looks like this:

The steps involved are:

  • The pi bond on ethylenecyclopropane (or 2-cyclopentyl-1-ethene) donates its electrons to grab a proton from #"HBr"#, and so #"H"# must break its bond with #"Br"#, giving a weak #"Br"^(-)# nucleophile
  • Intramolecular ring expansion, faster than the #"Br"^(-)# can backside-attack and giving a more stable ring, occurs first. Then...
  • since it is the better nucleophile in solution relative to #"H"_2"O"#, #"Br"^(-)# donates its electrons to bond to the cationic carbon in the intermediate.

You also get both an R and S stereoisomers due to the chiral center you achieve with the methylene, isopropyl, bromide, and hydride substituents being all different. The left one is R, and the right one is S.

The mechanism for the other most-often-taught hydrohalogenation method, hydrochlorination, is identical for this case, other than using #"Cl"# instead.

Related questions
Impact of this question
2744 views around the world
You can reuse this answer
Creative Commons License