6
$\begingroup$

terminal alkene on ethyl bond of cyclopentane reacts with H+ and HOCH3 to form a cyclohexane with 1 methyl group and 1 Och3 substituent Here is my problem. The mechanism seems pretty straightforward, but I am having trouble converting the cyclopentane to a cyclohexane. During a synthesis problem, does the starting material add the extra C onto its chain during carbocation rearrangement or a methyl/hydride shift?

As for the problem at hand, I believe the first step is for the pi bond to attack the $\ce{H+}$, and that will leave me with a carbocation at the more substituted carbon. From there, I know $\ce{HOCH3}$ will attack, but somehow, the molecule needs to rearrange its format in order for the $\ce{HOCH3}$ to come in. What needs to happen here?

$\endgroup$
  • 2
    $\begingroup$ I figure there's an alkyl shift after the initial protonation. $\endgroup$ – orthocresol Jul 27 '15 at 6:20
4
$\begingroup$

Your first step is correct. The reactive exomethylene group attacks the proton leaving a secondary carbocation (remember Markovnikovs rule). Now the five-membered ring can rearrange into a more stable six-membered ring by releasing ring strain. This is a fast intramolecular step. Finally methanol attacks the rearranged carbocation releasing a proton which is used catalytically in the next reaction.

Remember though that you can often not be completely sure about a mechanism before it is verified in practice (i.e. in the lab through an experiment). Sometimes true mechanism can be quite different from what appears to be most reasonable on paper!

$\endgroup$

Your Answer

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

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