July 02, 2022, 02:24:12 PM
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Topic: Hoffmann elimination of an amine  (Read 254 times)

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Offline jestquim

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Hoffmann elimination of an amine
« on: February 02, 2022, 09:32:01 AM »
The Hoffmann elimination is a reaction that allows an alkene to be obtained from an amine. Mechanistically, it is a three-step reaction. In the first stage, the compound undergoes exhaustive methylation with methyl iodide in order to transform the amino group into a good leaving group. Then, in a subsequent step, the compound is treated with acouso-silver oxide, which allows the introduction of a base (necessary for elimination E2). Finally, by heating the reaction mixture, the alkene is formed.

I fully understand the mechanism and each of its stages. However, I do not know how to predict the product of a Hoffmann elimination. For example, I need to know what products are obtained after subjecting species (A) to a Hoffmann elimination:

Offline rolnor

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Re: Hoffmann elimination of an amine
« Reply #1 on: February 02, 2022, 04:33:06 PM »
I think you get a mixture

Offline natalie5933

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Re: Hoffmann elimination of an amine
« Reply #2 on: February 03, 2022, 03:08:55 AM »
Well you will get a mixture of products. You will have a Zaitsev product and Hofmann product. The more substituted alkene is the Zaitsev's product however this is a minor product in such reactions and Hofmann product (the less substituted alkene) predominated even though more substituted alkenes are generally more stable. The reason behind this is the bulky leaving group. All this is nicely explained with diagrams here - https://www.masterorganicchemistry.com/2017/10/18/the-hofmann-elimination/

For the molecule you have posted here the major product will be 4-methyl-hex-2-ene as the alkene.

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