8
$\begingroup$

Let us say that you have 1-propene reacting with $\ce{BH3-THF}$ and then later with hydrogen peroxide and sodium hydroxide. I eventually get 1-propanol. The way it was explained to was that after reacting with $\ce{BH3-THF}$ we have $\ce{CH3CH2CH2BH2}$ and after the second set of reagents we get our product.

But, I never understood the mechanism, the explanation seems forced. $\ce{BH2}$ goes to the less hindered carbon which explains why its at the terminal. And then for some reason, $\ce{BH2}$ gets replaced by $\ce{OH}$.

$\endgroup$
10
$\begingroup$

I had to write the complete mechanism for you so that you can understand it better. See the image below. I have explained it in the simplest form. Some of the points I would like to discuss before you look at the mechanism are:

  1. In BH3, Boron is less electronegative than H (i.e. hydrogen) , hence in the syn addition I have shown, hydrogen ends up being attached to a more substituted carbon atom.

  2. For every bonding orbital we have an anti-bonding orbital which lies exactly opposite to the bonding orbital and each bonding-antibonding orbital pair can contain a maximum of 2 electrons in them.

Ok now lets head towards the mechanism which I wrote exclusively for you:

hydroboration-pg 1

hydroboration-pg 2

hydroboration-pg 3

If you have any further doubts in the mechanism then you can ask in the comment section below my answer. I hope it helps.

$\endgroup$
  • 1
    $\begingroup$ B(OH)3 or H3BO3 ionises as H+ & [B(OH)4]- in alkaline aqueous media, so the ionization step seems suspect. It is also a very weak monoprotic acid. Now if you changed the final byproduct to 3NaB(OH)4, the reaction would make more sense. $\endgroup$ – Tamoghna Chowdhury Mar 19 '17 at 8:39
  • $\begingroup$ Ya I understand, I even had this as my doubt when I used to study this, but later on in various texts I noticed the above kind of ionisation i.e. Na3BO3 one . $\endgroup$ – Yb609 Mar 19 '17 at 9:15
  • $\begingroup$ Can you mention the texts in which you found the above? Anyway, even with the standard ionization the overall reaction doesn't change, just the stoichiometry of the last step and the boron byproduct. $\endgroup$ – Tamoghna Chowdhury Mar 19 '17 at 9:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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