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Cyclopentyl to cyclohexyl carbocation rearrangement
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+ Biochemistry
+ Synthesis
+ Carbocation
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Kaori Kitao
Cyclopentyl to cyclohexyl carbocation rearrangement
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!
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!
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!
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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