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How do you get from 2,3-dimethylcyclopentanol to 1,2-dimethylcyclopent-1-ene?

I know the -OH group leaves as water (being protonated by acid) forming a carbocation. Then the weak nucleophile (water) attacks the beta-hydrogen to form what I think is 2,3-dimethylcyclopent-1-ene. But why is it 1,2-dimethylcyclopent-1-ene?

I know Zaitsev's rule wants the more substituted alkene, but forming the 1,2-dimethylcyclopent-1-ene would leave the carbocation hanging there, wouldn't it (the carbocation would still be unsatisfied)?

acid-catalyzed dehydration of 2,3-dimethylcyclopentanol 2,3-dimethylcyclopent-1-ene

Update: enter image description here

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    $\begingroup$ Seems like OH protonation and water loss would lead to a secondary carbocation centered at C1 (I'll use the carbon numbering of the alcohol reactant). A hydride shift could move the C2 hydrogen to C1, thus creating a tertiary (more stable) carbocation at C2. The C3 hydrogen is then lost to solvent, leading to 1,2-dimethylcyclopent-1-ene. Your mechanism leading to 1,2-dimethylcyclopent-1-ene is also plausible. The key question is, what is the rate of the C2 hydride shift relative to the rate of simple deprotonation at C2 of the initial secondary carbocation? $\endgroup$
    – Curt F.
    Commented Apr 20, 2015 at 5:02
  • $\begingroup$ Oh! Maybe the hydride shift is more reasonable! It satisfies the carbocation and creates a tertiary and more stable carbocation. And then further allows for the formation of the more substituted alkene. I guess I should first consider a hydride shift then move on to a methyl shift. $\endgroup$
    – liya77
    Commented Apr 20, 2015 at 5:08
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    $\begingroup$ @liya77 You should consider a hydride shift before a methyl shift for 2 reasons. 1. The hydride group has better migratory aptitude and 2. The carbocation formed on doing a hydride shift will be more stable than that obtained through a methyl shift in most cases. $\endgroup$
    – Binary Geek
    Commented Apr 20, 2015 at 6:50

1 Answer 1

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See this:

enter image description here ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

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    $\begingroup$ This answer "looks OK". But maybe you should be working on the quality of that image. Maybe a larger resolution. $\endgroup$
    – M.A.R.
    Commented Apr 20, 2015 at 11:25
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    $\begingroup$ @ADG There is (free) chemical drawing software available you know (e.g. Chemdraw), both online and offline. Just use Google. $\endgroup$
    – Jori
    Commented Apr 20, 2015 at 11:44
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    $\begingroup$ @ADG I think Jori and MAR are just trying to be helpful. There really is no need to make such passive aggressive comments. Using free chemical drawing software helps clear any ambiguities in drawings, has overall better readability and presentation, is often scalable, and is better for storage purposes. $\endgroup$
    – user3735
    Commented Apr 20, 2015 at 18:29
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    $\begingroup$ Let's address other people's comments in a constructive manner. You are off to a rocky start already, so really reflect on what you've written before you submit it. $\endgroup$
    – jonsca
    Commented Apr 20, 2015 at 23:14
  • $\begingroup$ @jonsca see edit. $\endgroup$
    – RE60K
    Commented Apr 22, 2015 at 5:43

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