Hydroboration of Alkenes

by James

Description: Hydroboration-oxidation transforms alkenes into alcohols. It performs the net addition of water across an alkene.

Notes: Note that the oxygen is always attached at the less substituted carbon (anti-Markovnikoff). Furthermore the stereochemistry is always syn (H and OH add to same side of the alkene).

The boron byproduct will depend on the # of equivalents of BH3 used reative to the alkene. Here their molar ratio is 1:1. One equivalent of BH3 can hydroborate up to 3 equivalents of alkene.

BH3-THF is the same as BH3. Tetrahydrofuran (THF) is merely a solvent. Sometimes B2H6 is written, which is another form of BH3. It behaves in exactly the same way as BH3.

You might also see 9-BBN or (Sia)2BH. These are hydroboration reagents in which two of the H atoms in BH3 have been replaced by carbon atoms. They will do the exact same reaction as BH3.



Notes: Note in example 2 that the hydrogen and oxygen add to the same side of the alkene (syn addition)


The reaction begins with the concerted syn addition of B and H across the double bond, with the boron adding to the less substituted carbon (Step 1, arrows A and B). In the second step, hydrogen peroxide and a base such as NaOH are added. the NaOH deprotonates the hydrogen peroxide (Step 2, arrows C and D) which makes the conjugate base of hydrogen peroxide (a better nucleophile than H2O2 itself). The resulting NaOOH then attacks the boron (Step 3, arrow E). This sets up the key migration step, where the carbon-boron bond migrates to the oxygen bound to boron, breaking the weak oxygen-oxygen bond (Step 4, arrows F and G). The OH expelled then comes back to form a bond on the boron (Step 5, arrows H and I) resulting in the deprotonated alcohol (alkoxide). The alkoxide is then protonated by water or some other comparably acidic species (Step 6, arrows J and K).

Notes: In step 1 the addition is syn and the reaction is concerted.

If excess alkene is present the two remaining B-H bonds can do subsequent hydroborations.

Note that only one enantiomer is shown here (but the product will be racemic)

The migration step (Step 4, arrows F and G) occurs with retention of stereochemistry at the carbon.

For the last step it’s reasonable to use water, HOOH or any other comparably acidic species as the acid.


{ 16 comments… read them below or add one }

DOUGLAS TEINE September 9, 2012 at 6:50 pm

very informative and helpful for students like me. Thankyou!!


Leah Harrell July 4, 2013 at 4:17 pm

Thank you SO much for this. My organic book and the other four organic texts I got from the library explain this in the most cryptic way ever. I’ve spent HOURS trying to find a coherent explanation for this mechanism. You are a LIFE saver!
Please, never take this site away! At least not until I get into med school anyways! :)


james July 8, 2013 at 4:57 pm



Jason January 10, 2014 at 11:56 pm

So I see how this can result in racemic mixtures for an alkene, and I’m wondering – Is there a way to then filter out one specific enantiomer, so I can get only the R or S enantiomer?



Jason Mathias March 23, 2014 at 10:13 pm

Hi James,
Is carbocation rearrangement possible? In adding H-Cl, there is a carbocation, but it appears that there isn’t one here.


TT May 8, 2014 at 1:25 pm

This complete mechanism is shown in few textbooks, yet it is one of the important anti-Markovnikov syn-addition. Most authors don’t bother to explain the oxidation part that yields the final OH group.


Janus May 25, 2014 at 2:33 pm

this post is like a medicine for students who are sick with the difficult bookish language. thanx james


Diana June 27, 2014 at 11:24 pm

Is there a reason borane specifically adds to the least substituted alkene side..
I was trying to reason it out. It is a electrophilic addition because Boron is electron deficient(?). But that doesn’t explain why the end-product is more stable.

I could ask my prof this but already he thinks I have too many q’s outside the scope of this class.


vibhu July 15, 2014 at 1:28 pm

Usually the less substituted carbon gets the negative charge on electromerisation. So the the empty orbital of boron comes and gets attached to the “less substituted carbon”. The moment this happens, the H-B bond breaks and the H- ion gets attached to the neighbouring C which had positive charge due to electromerisation. And all this happens in a single concerted step. The product you get is the third compound mentioned in the step 1 illustration.


Hannah Rodrigues July 31, 2014 at 3:17 pm

Just kind of confused as to the greyed out Na+, is it just kind of floating around getting transferred between the molecules, or does it only really come in at step 5 and 6?

Thank you :)


peter November 1, 2014 at 10:58 am

Hi, I’m wondering, how come O and BRH2 get swapped between step 4 and 5?
There doesn’t really seem to be an explanation that I can see for that…


James November 1, 2014 at 8:23 pm

Hi Peter, in this key rearrangement step, we’re breaking the C-B bond and forming a C-O bond, which happens also to break the weak O-O bond on the peroxide.


James Curry December 13, 2014 at 9:10 pm

In step 5, what happens to the BH2 group after it is attacked by OH-?


James December 14, 2014 at 12:19 pm

In this sequence, it becomes H2B-OH . However I should note that all the B-H bonds are available for hydroboration of other alkenes; in the end, what will happen is that we form B(OH)3 (boric acid).


Isaac Traynis March 30, 2015 at 2:10 pm

I was under the impression that for Step 5 there are a series of oxidizing reactions on organoborane until it finally leaves via E1 to form a more stable alkoxide in a basic solution. Am I correct on this or does this not happen in this reaction? Thanks.


James April 1, 2015 at 10:05 pm

There are definitely a series of oxidizing reactions on the organoborane, until it becomes a boronic ester (i.e. all B-O bonds, no B-R bonds). I’m not sure I follow re: “the E1 step”. If you mean formation of a tetrahedral, charged borate intermediate followed by loss of the alkoxide as a leaving group (somewhat similar to the first step in the SN1 or E1 reactions) then yes!


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