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How to convert methylcyclopentanol to cyclopentanone?

I tried eliminating the alcohol group as $\ce{H2O},$ but that gives Saytzeff product. If only I can get the Hofmann product (methylene cyclopentane), I would subject it to ozonolysis and get cyclopentanone.

I don't know exactly what to use as the reagent to get that Hofmann product. Could there be other methods for the conversion?

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    $\begingroup$ Just to make things clear, do you mean 1-methylcyclopentanol is the precursor and methylidenecyclopentane is the Hofmann product? $\endgroup$ – andselisk May 20 at 19:44
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If you leave it to E1, you will not get the selectivity you need.

The difference between the methyl group and the methylene unit is steric bulk. You can select for deprotonation of the methyl for elimination via a bulkier base in an E2 reaction.

In addition, you will need to convert the substrate so that it has a good leaving group. For an alcohol, this is typically accomplished by first converting to a tosylate or a mesylate.

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  • $\begingroup$ Thank u so much... So that means I will have to treat it first with TsCl and then with tertiary butoxide (a bulky base) to get the Hoffman product and carry out the further reaction. $\endgroup$ – Yashwini May 21 at 7:11
  • $\begingroup$ If you are dealing with 1-methylcyclopentanol, this reactive tosylate may form the chloride in situ. Why not treat the alcohol with HCl to form the chloride initially and then proceed? $\endgroup$ – user55119 May 21 at 16:49
  • $\begingroup$ @user55119 I strongly doubt that that tosylate will react as is without heating (Please provide a reference if you have reason to believe my intuition is wrong. It might be.). Sn1 reactions can be messy, and I have a strong preference not doing that when possible. $\endgroup$ – Zhe May 21 at 20:15
  • $\begingroup$ From my own experience. Used to put reaction in a freezer to form tosylate and avoid formation of chloride in more complex tertiary alcohols. It doesn't matter much in this case if it is a mixture of tosylate and chloride. $\endgroup$ – user55119 May 21 at 20:42
  • $\begingroup$ Ah OK. Still good to know though. Thanks for the feedback. @user55119 $\endgroup$ – Zhe May 21 at 20:45

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