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Topic: Zaitsev v. Hoffmann products  (Read 7311 times)

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Offline Burningkrome

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Zaitsev v. Hoffmann products
« on: April 25, 2012, 08:33:35 AM »

So...studying the Hoffmann and Zaitsev elimination reaction. Briefly (see image), I completely understand that a large, sterically hindered molecule like t-butoxide creates the Hoffmann product. BUT, if a NON-sterically hindered based is used (that can reach the crowded hydrogen in the image) ... do you achieve predominantly the Zaitsev product, or a racemic mixtiure of Zaitsev and Hoffmann?

If not...why not? :-)


Offline Babcock_Hall

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Re: Zaitsev v. Hoffmann products
« Reply #1 on: April 26, 2012, 01:05:06 PM »
A racemic mixture is a mixture that has equal parts of two enantiomers.  It does not have anything to do with this problem.

Offline bonfire09

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Re: Zaitsev v. Hoffmann products
« Reply #2 on: April 29, 2012, 11:27:29 AM »
hoffman and zaitsev reactions are opposing ideas. hoffman products are generally ones that have double bonds near the least substituted carbon. Zaitsev products have double bonds where the carbons are the most susbstituted. Though hoffman products are generally much less formed than zaitsevs but they still occur

Offline orgopete

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Re: Zaitsev v. Hoffmann products
« Reply #3 on: April 29, 2012, 08:50:39 PM »
I think of Zaitsev elimination reactions a SN1-like. They occur from elimination from electron donation of the most electron rich carbon, that is the most substituted. I think of Hoffman elimination products as the result of deprotonation of the most acidic hydrogen, this will be the least substituted.

The base, solvent, and leaving groups can all affect these elements. A very good leaving group, like iodide will stretch the C-I bond and draw electron density toward it. This will favor Zaitsev, as will a polar solvent. A poor leaving group or one with a higher pKa, will not draw electrons in the same manner. A fluoride or quaternary ammonium salt are more basic, poorer leaving groups, pull electrons from neighboring groups to a lesser degree, and are dependent upon deprotonation initiating the reaction. The least substituted carbon is the most acidic hydrogens. This will give you a Hoffman product.
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