I have this reaction with a dr = 96:4 but I can't explain the product theoretically. I suppose it is some orbital interaction? enter image description here

My hyporthesis is that there is a $\pi$-$\sigma^{*}$ interaction between the alcohol-group ($\sigma^*$) that draws electrons out of the $\pi$ orbital of the double bond? I cannot formulate it really. And why the ratio is so pronounced.


No, you’re overthinking this. It’s just sterics.

The iodine will form a iodonium ion (three-membered ring with positive charge on iodine) where the double bond was. Since iodine is huge, it will come from the less sterically hindered side which is from behind the monitor plain.

The carboxylic acid then attacks the iodonium rom above closing a five-membered lactone ring (kinetically favoured against a six-membered one). The side chain carbon retains iodine pointing backwards (you can twist along the $\ce{C-C}$ bond to push the carbon into the monitor plain and iodine backwards) while the lactone oxygen coming from above means that the methyl substituent on the ring must be pointing behind the plain, again.

Also, stereochemistry on the hydroxide should be retained since it does not take part in the reaction.

A crappy mechanism
Scheme 1: A not-well drawn mechanism for the reaction.

  • $\begingroup$ Couldn't the hydroxide react with adjacent carbon to give the epoxide intermediate? $\endgroup$
    – wuschi
    Oct 8 '15 at 11:45
  • $\begingroup$ @wuschi It could, and that could explain the $4~\%$ of the diastereomeric ratio. If the epoxide is formed, iodine will still point backwards, but the carboxylate will also attack from behin the monitor plain meaning both methyl and hydroxyl group will point forwards. Unless of course OP (or source) actually observed $96~\%$ of the other diastereomer in which case the argument needs to be turned around. $\endgroup$
    – Jan
    Oct 8 '15 at 11:48
  • $\begingroup$ You said, the iodine comes from the less sterically hindered side, i.e. from behind the monitor. But why would this side be sterically favoured? The two carbons of the double bond are both sp2, they form a plane, so no side is inherently more sterically hindered (and the hydroxy group can be turned away). Am I missing something? $\endgroup$
    – ste
    Oct 8 '15 at 13:18
  • 1
    $\begingroup$ @ste That alcoholic carbon next to the double bond has two further non-hydrogen residues. No matter if you draw it the way I did or have the hydroxyl group in the plain, if the hydrogen is not in the plain it will point backwards thus preferring backward. You can try to turn away the hydroxyl group, but you won’t succeed. $\endgroup$
    – Jan
    Oct 8 '15 at 13:22

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