I came across the following problem in my book:

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Now my confusion is how will cyclohexene be obtained? I read that primary alcohols dehydrate by E2 mechanism because of formation of a primary carbocation. Further I read in Solomons organic chemistry that rearrangement in Primary alcohols occur due to the reversibility of the reaction:

enter image description here

So I tried to figure out the mechanism:

enter image description here

So I am unable to find how cyclohexene will be obtained. For ring expansion to take place there must be positive charge on the alpha carbon but in E2 mechanism there is no possibility of carbocation formation


1 Answer 1


Concentrated sulphuric acid is a highly acidic medium and thus it very exclusively prefers the $E_1$ mechanism over $E_2$ mechanism and thus a carbocation is surely formed in this conditions.

The example you provide proceeds through the following mechanism:

Dehydration in acidic conditions

Now the mechanism you provide, other than following the wrong pathway(very minor pathway) also has a major mistake. During the protonation of 1-methylenecyclopentane the tertiary carbocation is not more stable than the primary carbocation. The tertiary carbocation is $\mathrm{sp^2}$ hybridized and thus prefers a bond angle of 120° which cannot be provided by the cyclopentane ring as it is not flexible enough.

It prefers the very less angle strain in case of $\mathrm{sp^3}$ hybridized orbitlas but with less positive inductive effect over higher angle strain but more positive inductive effect. That's the reason the below equilibrium is highly backward favoured.

carbocation rearrangement

  • 6
    $\begingroup$ Can you cite a reference or evidence for the final point? I would have always said the tertiary cation was more stable, but perhaps I've been wrong in this assumption? $\endgroup$
    – PCK
    Commented May 18, 2021 at 8:24
  • 2
    $\begingroup$ I agree. I would like to see a reference for this $\endgroup$
    – Waylander
    Commented May 18, 2021 at 13:04
  • 1
    $\begingroup$ (-1) for the last step. A reference would be appreciated since this seems very unlikely. $\endgroup$ Commented May 21, 2021 at 12:58

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