In the hydration reaction that is addition of H2O , in the presence of (H.B.O/T.H.F/H2O2), where no rearrangement takes place, Why does H is placed on more hindered C atom and OH on less hindered C atom ?

  • $\begingroup$ H is smaller and less cumbersome than OH. $\endgroup$ – Maurice Feb 10 at 9:25
  • 2
    $\begingroup$ And what chemical exactly is "HBO"? $\endgroup$ – orthocresol Feb 10 at 10:20

The reaction Hydroboration-Oxidation follows anti-Markovnikov rule because of 2 factors:

1)Electronic consideration: When the borane ($\ce{BH_3}$) approaches the $\pi$ bond of the alkene, the attack of the $\pi$ bond to the empty p orbital of the boron triggers a hyride shift (the mechanism of this is similar to that of pericyclic reactions). There is a partial positive charge developed on the vinylic position of the alkene that triggers the hydride shift. reference picture for 1

  1. Steric considerations: In the first step of the mechanism of hydroboration, we add $\ce{BH_2}$ and $\ce{H}$ simultaneously. As the $\ce{BH_2}$ is bigger in size than $\ce{H}$, the transition state of this step would be less in energy and also less crowded if the $\ce{BH_2}$ is positioned at the less steric position. reference picture for 2

Both factors suggest that the mechanism must follow anti-markovnikov addition. After the attach of $\ce{BH_2}$, the further mechanism will replace the $\ce{BH_2}$ with $\ce{OH}$. Here I showed why the $\ce{H}$ attaches to the more crowded site and follows anti-Markovnikov addition.

For the further part of the mechanism, you can view the page 414 of the David Klein 2012 edition.

All the reference pictures are taken from the David Klein Organic Chemistry 2012 edition book.


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