Hydroboration Oxidation of Alkenes Mechanism
Last updated: November 18th, 2022 |
Hydroboration – Oxidation Of Alkenes: The Mechanism
In the last post we saw that the results of hydroboration of alkenes are not in accord with any of the two families of mechanisms we’ve previously seen (carbocation pathway, 3-membered ring pathway). With hydroboration, we observe that the regiochemistry is “anti-Markovnikov” (H ends up bound to the most substituted carbon, B ends up attached to the least substituted carbon) and the stereoselectivity of the reaction is syn (both new bonds are formed on the same face of the alkene).
How to make sense of all of this? Here is our best hypothesis for the mechanism of hydroboration-oxidation of alkenes.
Table of Contents
- Hydroboration Of Alkenes: A Proposed Mechanism
- Why Does Hydroboration Exhibit So-Called “Anti-Markovnikov” Regioselectivity?
- Hydroboration-Oxidation: How Does The Oxidation Work?
- Oxidation Steps 1 And 2: Deprotonation Of Hydrogen Peroxide, And Attack Of The Peroxide Ion On Boron
- Oxidation Step 3: The Key Rearrangement Step
- Oxidation Step 4: Cleavage Of The O–B Bond
- (Advanced) References and Further Reading
1. Hydroboration Of Alkenes: A Proposed Mechanism
Instead of proceeding in multiple steps, as do reactions in the carbocation and 3 membered ring pathway, the hydroboration reaction occurs all at once. That is, the mechanism is “concerted” (Those dashed lines below represent “partial bonds”). The new bonds to carbon are forming at the same time as the C-C π bond is breaking. The result is that the stereochemistry is “syn”.
2. Why Does Hydroboration Exhibit So-Called “Anti-Markovnikov” Regioselectivity?
Why does the hydrogen end up bound to the most substituted carbon, rather than the least? Think of the transition state for this reaction. As the C–C π bond breaks, any partial positive charge from the alkene carbons would be most favorably situated on the carbon best able to accept it – the most substituted carbon, in our cases. In the transition state, this partially positive carbon would be best stabilized by interacting with the atom of BH3 that bears a partial negative charge.
Here’s the twist: in BH3, the atom that bears a partial negative charge is hydrogen, because hydrogen is more electronegative than boron. Let’s have a closer look. First, a reminder on why H-Cl and H-Br are “Markovnikov” in the first place: hydrogen bears a partial positive charge.
In BH3, this situation flips around. Hydrogen is more electronegative than boron:
The irony of all of this is that even though hydroboration is often thought of as that “exception” of a reaction which is anti-Markovnikov, it actually follows the same principle as the reactions we’ve encountered before: the more electronegative atom ends up bound to the carbon best able to stabilize positive charge. So there’s actually no real “exception” at all here!
3. Hydroboration-Oxidation: How Does The Oxidation Work?
Which brings us to a subject which is not specifically related to alkenes, but as a crucial part of the utility of hydroboration, worth discussing here. How does that oxidation step work?
After hydroboration, treatment of the organoborane with basic hydrogen peroxide leads to replacement of C-B with C-OH. Note that there is no change in stereochemistry; it has occurred with retention!
4. Oxidation Steps 1 And 2: Deprotonation Of Hydrogen Peroxide, And Attack Of The Peroxide Ion On Boron
The first step here is deprotonation of hydrogen peroxide to give NaO-OH. Since the conjugate base is a better nucleophile, this speeds up the rate of the subsequent step.
The next step is a simple Lewis acid-base reaction. The deprotonated peroxide anion then adds to the empty orbital of boron, forming a negatively charged boron species:
5. Oxidation Step 3: The Key Rearrangement Step
The next step often gives students difficulty. Here, the pair of electrons in the C–B bond migrates to oxygen, leading to breakage of C–B and formation of C–O, along with rupture of the O–O bond. It’s very similar to 1,2-hydride and alkyl shifts we’ve seen previously, except that instead of migrating to the empty p orbital on a carbocation, the electron pair is essentially performing a “backside attack” on the σ* orbital of the weak O–O bond (See post: The SN2 Mechanism)
Note how the charge on boron goes from negative to neutral.
6. Oxidation Step 4: Cleavage Of The O–B Bond
The next step can be written several different ways. Hydroxide ion attacks the empty p orbital of boron, and the O–B bond breaks. Although drawn here as a “concerted” step, where bond formation accompanies bond breakage, it need not be so, since addition of hydroxide to boron does not violate the octet rule.
Finally the negatively charged oxygen is then protonated by water (the solvent).
That sums up the key points of the hydroboration reaction.
In the next post, we’ll go through some other reactions of alkenes that might not share the exact same mechanism as hydroboration, but share a similar pattern of stereochemistry that is also a result of “concerted” reactions.
NEXT POST: Alkene Addition Pathway #3 – The “Concerted” Pathway
(Advanced) References and Further Reading
- Mechanistic studies
XXIV. Directive Effects in the Hydroboration of Some Substituted Styrenes
Herbert C. Brown and Richard L. Sharp
Journal of the American Chemical Society 1966 88 (24), 5851-5854
A very nice Physical Organic study of the hydroboration of styrenes, involving a Hammett plot (a classic tool in physical organic chemistry) to determine a relationship between the stereochemistry of the reaction and the electron density of the alkene.
- References from Organic Syntheses
A SIMPLE AND CONVENIENT METHOD FOR THE OXIDATION OF ORGANOBORANES USING SODIUM PERBORATE: (+)-ISOPINOCAMPHEOL
George W. Kabalka, John T. Maddox, Timothy Shoup, and Karla R. Bowers
Organic Syntheses, Coll. Vol. 9, p.522 (1998); Vol. 73, p.116 (1996)
A Hydroboration-oxidation procedure where the oxidizing agent is sodium perborate, a cheap, safe, easily handled oxidant commonly used in laundry detergents.
- Modern Example
Hydroboration with Pyridine Borane at Room Temperature
Julia M. Clay and and Edwin Vedejs
Journal of the American Chemical Society 2005 127 (16), 5766-5767
A modern method for doing hydroboration at room temperature using pyridine-borane, which usually requires heating to 75-100 °C to liberate the borane.
35 thoughts on “Hydroboration Oxidation of Alkenes Mechanism”
Worth noting that the B-to-O migration step is analogous to the migration step in the Baeyer-Villiger reaction (http://www.organic-chemistry.org/namedreactions/baeyer-villiger-oxidation.shtm). That one’s C-to-O, but it’s a similar idea.
Boron loves those migrations…the way it was described to me, it’s conflicted–it doesn’t know whether it wants to have an octet or neutral formal charge. A source is escaping me at the moment, but Zweifel olefination is similar (B-to-C migration).
Yes! the key step in the BV, as well as in the Beckmann, Curtius, Hoffman, Wolff, and other rearrangements. Notoriously difficult for students, even though it’s only a single arrow to draw.
I tell my students that boron is the petulant teenager of the periodic table. It mopes around the house all day whining about how it doesn’t have enough electrons. So someone comes along and gives it two more – then it can’t stop complaining about this stupid negative charge…
Excellent!!! This is the post I’ve been looking for the past two years!!
This post saved my life! Thank you so much!
OK, thank you Nayara!
You are great!! Thank you Dr.James
What does the mechanism look like for hydroboration 1. BH3, THF 2. HOOH, OH-, ROH? If we have an alcohol instead of water in the last step.
You will get a borate ester, B(OR)3 . Which is fine for our purposes.
Just wanted to say I love what you did in this series of posts, the order at which you presented the addition reactions and the reasons for why we need to look for new mechanisms to explain them. This would really help remembering it.
Couple of questions though:
1) Is there a driving force behind the backside attack of the electron pair in step 3?
2) Why is it that the hydroxide anion attacks the boron in step 4 but not in step 1? Is it a question of concentration ratios?
Anti Markonokov regioselectivity of the final product is all right, But is’t the addition of BH3 following the Markonikov rule?
Yes, it essentially is. The carbon best able to support positive charge is interacting with the most electronegative group of the electrophile and vice versa.
The link in the fourth paragraph that leads to the review on why H-Cl and H-Br are Markovnikov reactions no longer works.
Fixed. Thank you
what if both locations are equally substituted? Such as in trans-4,4-dimethylpent-2-ene Where will the OH be placed?
You’ll obtain a mixture of products. There won’t be much, if any, regioselectivity. A 55:45 mixture of products is not a great result.
you helped me a lot to clear some of my confusions. thank you.
Glad to hear it Visva.
OH- is attacking in 4th step and substituting the stronger base RO-. But leaving tendency of stronger base is low as we have seen that -OH group can not be substituted by halogen from alcohol and for that we use haloacids.
Substituting RO- for HO- is very possible with an excess of reagent. They are of comparable basicity.
What would have been the product of the oxidation after hydroboration if H2O2 is replaced by D2O2?
Good question! The answer is… R-OD. But it you use a protic solvent like H2O in the workup, the deuterium will exchange for hydrogen, and you’ll end up with R-OH !
On the other hand if you use BD3 instead of BH3, you will form a C-D bond in the product.
Hey, why do you say that the partial charge on boron of BH3 is positive?
It should be negative!!
Compare the electronegativities of B and H and tell me where the dipole should be in the B-H bond?
As we know HB oxidation used for the production of ketone and aldehyde,… why not u give as an ex for those?
Hydroboration of alkenes would not produce ketones or aldehydes directly; you are referring to hydroboration of alkynes, which is covered here. https://www.masterorganicchemistry.com/2013/05/14/hydroboration-and-oxymercuration-of-alkynes/
James…..You are just awesome…!!
I am an IIT JEE aspirant and I can’t explain it to you how much it is helping me….GREAT JOB…
Glad it helps, and good luck on your exam!
Do you have a sorted progression of these blog posts? I love them! It’s just I wish there was a way I could look at topics and then follow the sequence instead of just looking up specific subjects and then following the next blog post.
Yes – check out the new homepage at http://www.masterorganicchemistry.com !
In the transition state, the boron should have a partial negative charge (not positive). The B in borane is electrophilic with only 6 valence electrons, and so when the alkene forms a bond to B, it gains a partial negative charge. Hence, at the end of the other double bond, the carbon bears a partial positive charge. Don’t use the electronegativity analogy here, but rather the formal charge.
The partial positive charge reflects the (slight) polarization of the B-H bond, and in turn helps to rationalize the anti-Markovnikov regioselectivity.
I thought that electronegativity was a constant for an atom. You indicate H=2.2, Cl=3.2, Br=3.0.
OK, this is the value given in the periodic table.
But further down, you indicate that in BH3, boron is less electronegative than H: if electronegativity is a constant, I don’t understand anymore? Why is that?
Br is bromine (electronegativity 3.0). The symbol for boron is B (electronegativity 2.0)