# Mechanism for acid-catalyzed dehydration of 2,3-dimethylcyclopentanol

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)?

Update:

• 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? – Curt F. Apr 20 '15 at 5:02
• 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. – liya77 Apr 20 '15 at 5:08
• @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. – Binary Geek Apr 20 '15 at 6:50

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