I was given a problem of reaction between (2R,3S)-3,4-dimethylpentan-2-ol and $\ce{POCl3}$ and pyridine. I know that $\ce{POCl3}$ will be attacked with hydroxyl group and the hydrogen on hydroxyl will be cleaved with pyridine so the hydroxyl group becomes good leaving group.

The reaction mechanism proceeds on E2 mechanism no matter whether the alcohol is primary, secondary, or tertiary. The problem is E2 mechanism can proceed if the stereochemistry of antiperiplanarity is fulfilled. When I drawn the stereochemistry of the alcohol, it turned that methyl on alpha-carbon and beta-carbon made the beta-hydrogen on these carbons cis to the hydroxyl group (the methyl on alpha carbon is wedge so the bonded hydrogen and hydroxyl group are dash), thus the reaction can not proceed.

Is that correct? If it is correct, will the reaction mechanism proceed on other mechanism?


2 Answers 2


I am uncertain whether or not you are assuming that the E2 product of a "cis elimination" is the (E)-alkene. Alternatively, you were given the geometry of the alkene and were asked to provide a mechanism for its formation. In either case, let's go through the steps.

You are correct that the alcohol 1 reacts with phosphorus oxychloride to form derivative 2a in addition to pyridine hydrochloride. Recall that there is free rotation about the C2-C3 bond in 2a, which is shown as its Newman projection in 2b in the anti-periplanar conformation ready for E2-elimination. Clearly, this elimination leads to (Z)-3,4-dimethylpent-2-ene. (Z)-4. If indeed it is the stereoisomer (E)-4 that is the target, then one can invoke SN2 displacement on 2a by chloride to invert the configuation at C2 leading to chloride 3a. Now E2-elimination of 3b leads to (E)-4.

enter image description here


The reaction will take place via E2 mechanism as the C2-C3 $\sigma$ bond will undergo rotation, giving the product. The concentration of the minor product will also be comparatively more as compared to the (2R,3R) or (2S,3S) form, but it will not form the major product. Stereochemistry plays an important role in these reactions when the C-C bond cannot rotate about it's axis and hence no reaction occurs. An example is given below:

Image edited

In the first reaction, bond rotation occurs and E2 elimination takes place, whereas since the second compound is cyclic, no bond rotation can occur and the compound undergoes an E2 elimination, giving the less substituted product as the major product.

  • 2
    $\begingroup$ You drew the 2S,3R enantiomer, not that it matters. E2 anti elimination of the phosphorus derivative would give the Z-alkene. Elimination in the second case probably gives more of the less substituted alkene. $\endgroup$
    – user55119
    Nov 20, 2019 at 3:11
  • $\begingroup$ Ah yes my bad. Edited with a better image now $\endgroup$ Nov 20, 2019 at 5:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.