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Topic: Zaitsev's rule  (Read 4504 times)

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Offline tezha

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Zaitsev's rule
« on: August 10, 2011, 02:01:56 PM »
A textbook states that in the following reaction the less substituted alkene is the main product formed, which does not really obey Zaitsev's rule. Do I understand it correctly that the reason for that is that there are two methyl groups on the first carbon which can be both attacked by the base so the chance of forming a less substituted product is twice as big? Or is there another reason?


Offline opsomath

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Re: Zaitsev's rule
« Reply #1 on: August 10, 2011, 02:07:34 PM »
This is a classic Hofmann elimination, which is known to produce the less substituted alkene in the vast majority of cases. The reason is that the leaving group is very bulky, and the less substituted carbon is better able to assume the correct orientation for E2 elimination, that is, the anti conformation with the proton getting removed trans to the nitrogen leaving group.

Offline fledarmus

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Re: Zaitsev's rule
« Reply #2 on: August 10, 2011, 02:07:59 PM »
Look at a model of your intermediate, with all the hydrogens in place. What orientation would the hydrogen that is being eliminated have in relation to the empty orbital on the cation?

The driving that you mention, the number of possible ways of forming the first product as opposed to the second, is also telling, but it is much larger than you indicate. There are actually 6 different protons that can be eliminated to form the major product, and only one to form the minor product.

Offline tezha

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Re: Zaitsev's rule
« Reply #3 on: August 10, 2011, 02:38:59 PM »
Right, thanks, I totally forgot about this hoffman thing!

Offline orgopete

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Re: Zaitsev's rule
« Reply #4 on: August 12, 2011, 04:34:41 PM »
I shall give an alternate explanation that I think should be considered. Trimethylamine, the leaving group, is a much stronger base than bromide or iodide alternatives. The effect is the reaction will be less SN1-like. If the rate limiting step requires removal of the more acidic hydrogens (CH3>CH2>CH), that would give a Hoffmann product. You may think of this as somewhere between an E2 and an E1cb elimination.

If a reaction were completely concerted, then the more acidic the hydrogens and the larger the possible collisions, the greater the amount of Hoffmann product one should expect. The Zaitzev product is the result of removal of the least acidic hydrogens. The more SN1-like a reaction is, the more the product will favor a Zaitzev product. In this case, the neighboring intramolecular electrons are donated. They correspond with the more substituted carbon. The participation of those electrons will increase the acidity of the more substituted hydrogens.



This is only a suggestion. Some reactions are dominated by stereoelectronic effects. If the electrons that are anti, e.g., by restricted rotation, will give the predominant product.
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