E2 stereochemistry

Problem # 531

E2 elimination reactions require anti-coplanar geometry. (note: some textbooks call this anti-periplanar).

Let's work through an E2 reaction, and rotate the molecule eblow into an anti-coplanar geometry to predict the product of this E2 reaction.

Solution

To predict the stereochemistry of the alkene product from an E2 reaction, we have to rotate the leaving group (Br) and the beta-proton into anti-coplanar geometry. It's easiest to put the H and Br in the plane of the page.

After we do this, we see that both the ethyl and phenyl groups are wedges, and the two methyls are dashes. So after the E2 reaction, the resulting alkene will have its two methyl groups cis to each other.

Problem Details

MendelSet practice problem # 531 submitted by Matt on July 2, 2011.

Subject: Organic Chemistry

Subject Topic(s): Alkene Formation: Elimination, Dehydration, and Zaitsev's Rule, Stereochemistry, Chirality, and Optical Activity

Question Type: Concept (explain why this is so..)

Keyword(s): E2 stereochemistry

Problem # 319

For a molecule to undergo an E2 reaction, the leaving group and the beta-proton must be in an anti-coplanar conformation (one atom straight up, the other straight down). Based on this, which compound undergoes E2 reaction with KOtBu faster? Why?

Solution

The methyl group is bulky and so is most stable in the equatorial position.

The 1,4 cis compound's most stable chair form has its leaving group (Br) and beta-hydrogen in an anti-coplanar (anti-periplanar in some textbooks) conformation.

So most of the time, 1,4 cis is in a chair conformation that is able to undergo an E2 reaction.

The opposite is true for the 1,4 trans compound. Most of the time it is in a chair conformation that is unable to undergo an E2 reaction.

Therefore, the 1,4 cis compound will undergo an E2 elimination reaction faster than the 1,4 trans compound.