Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: briteyellowness on August 15, 2005, 06:55:10 PM
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question: given that the dehydration of pinacol (2,3-dimethyl-2,3-butanediol) with strong acid and heat is neither an oxidation nor a reduction, which of the following best describes the product?
answer: ketone
i have no idea why? isn't the carbon tertiary and forming a ketone would mean having five bonds with it, which isn't possible? the only thing i might be able to see forming is maybe an epoxide?
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This reaction comes about from what is known as the pinacol rearrangement (at least, this is what my organic prof. refered to it as). The acid will protonate a hydroxide group to form water. The water will leave to form a carbocation. The carbocation will rearrange via a 1,2-methyl shift to form a secondary carbocation connected to the other hydroxide. This rearrangement is favored because of resonance-stabilization; the lone pair on the oxygen can donate electron density to the empty p-orbital of the carbocation, and in general, resonance-stabilized secondary carbocations are more stable than tertiary carbocations with no resonance-stabilization. Subsequent deprotonation of the oxygen results in the formation of 3,3-dimethyl-2-butanone (pinacolone[?]).
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Yep, Yggdrasil has it right. Mitch has kindly drawn out the mechanism (http://www.chemicalforums.com/index.php?page=name#Pinacol%20Rearrangement) on our mechanism page.
However, I wouldn't call the stabilization of the carbocation a resonance stabilization. It's really a "neighboring group" effect. Technically resonance structures must have the same number of sigma bonds.