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?

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    $\begingroup$ I figure there's an alkyl shift after the initial protonation. $\endgroup$ Jul 27, 2015 at 6:20

1 Answer 1


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!


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