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When a terminal alkenyl borane is succumbed to halogenation where the halogen and the base are added in two separate steps, the geometry of the double bond is changed. When they are added simultaneously, the double bond geometry is retained. Why is it so? enter image description here

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  • $\begingroup$ You've actually drawn a boronic acid instead of a borane. I suspect that the the reaction doesn't proceed with a boronic acid, since these are markedly less electrophilic. "The failure to utilize the third group in organoboranes prepared from terminal olefins restricts the yield of iodide from a particular terminal olefin to a maximum of 66.7%." (JACS 1968, 90, 5038). And indeed, I can't find anything on Reaxys for this particular transformation. $\endgroup$
    – orthocresol
    Dec 13, 2016 at 3:27
  • $\begingroup$ It was taken from Smith's Organic Synthesis. $\endgroup$
    – EJC
    Dec 14, 2016 at 16:41
  • $\begingroup$ Thank you for the tipoff, and you are right, I am wrong. The reaction proceeds with boronic acids. In fact, with the borane, additional complications occur because of migration of B-C bonds in the iodonium ion. March's Advanced Organic Chemistry 7th ed. p705 mentions the reaction and gives the reference: JOC 1989, 54, 6075. I'll update the answer tomorrow - gotta head to bed now... $\endgroup$
    – orthocresol
    Dec 14, 2016 at 18:01
  • $\begingroup$ You could add the mechanism for the side reaction: alkyl migration. $\endgroup$
    – EJC
    Dec 14, 2016 at 19:23

1 Answer 1

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I remember learning that these two reactions proceed via different mechanisms. I'm going to use an alkenylborane to illustrate them.

Organoboranes are electron-deficient and electrophilic; when hydroxide is added together with iodide, the first step is the formation of an anionic ate complex. This is now nucleophilic via the $\ce{B-C}$ $\sigma$ bond, and can therefore react with the electrophilic iodine:

Mechanism 1

This mechanism is somewhat analogous to the hydroboration-oxidation $\ce{H2O2}$ step, where the carbon "nucleophile" is intramolecularly transferred from boron to an electrophilic oxygen atom. This step is an intermolecular transfer of carbon from boron to iodine instead.

Incidentally, I remember that with alkylboranes the migration occurs with inversion of stereochemistry, but I cannot seem to find a reference. Given the premise of the question, I would hypothesise that alkenylboranes react with retention of stereochemistry (which would not be surprising since stereochemistry at $\mathrm{sp^2}$ and $\mathrm{sp^3}$ centres, and the stereoelectronic requirements for inversion/retention, are entirely different).


As for the other reaction, hydroxide isn't added first, so iodine (electrophilic) has to react with the double bond (nucleophilic). The subsequent steps are best described by a diagram instead.

Mechanism 2

Again I am a little frustrated at not being able to find a reference, but I am fairly sure this was how I learnt it. I literally looked up all reactions of $\ce{BR3}$ with iodine on Reaxys but couldn't find anything on it, so...

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