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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: queeniestrudels on October 16, 2021, 10:25:36 PM

Title: Hydroboration regioselectivity
Post by: queeniestrudels on October 16, 2021, 10:25:36 PM

So we are currently covering a variety of reactions in my ochem class, one of which is hydroboration-oxidation. I am very confused by the regioselectivity of hydroboration.

So far, we were given the two possible transition states of the first step: (

The two main reasons for the regioselectivity they told us about were
1) Charge stabilization (the most substituted carbon is able to stabilize the partial + charge better through hyperconjugation)
2) Lower steric interactions between the big BH2 and alkyl groups.

Now, here's what doesn't make sense to me: why does the carbon on the left have to have a + charge in order to stabilize it through hyperconjugation?? Why can't it have a partial negative charge? Other than the sterics of the top transition state, wouldn't it be able to stabilize the negative charge just as well as the positive charge through hyperconjugation? Why is this a reason for the regioselectivity?

Also, another question I have is about the charge dipoles on the transition states. Shouldn't the boron technically have a partial negative charge because it is acting as the electrophile, and the hydrogen should have a partial positive charge because it is acting as the nucleophile? I understand that the hydrogen is slightly more electronegative in this scenario (and that's why it acts as the nucleophile), but I remember drawing transition states for electrophilic additions with HBr and the Br had a partial negative charge (since it acts as the electrophile) and the H had a partial positive charge.

I have seen something similar to what I am describing in our textbook (funnily enough, it is different from our lecture), and it looks like this: (

Can someone explain the charge dipoles in this? Are they technically the same as the ones above or no? How do they explain the induced dipole on the carbon? Thanks a ton!!!

Title: Re: Hydroboration regioselectivity
Post by: wildfyr on October 17, 2021, 09:54:08 AM
Hyperconjugation more typically stabilizes positive charges because electrons are donated in. A negative charge is destabilized by having electrons donated into it. There are some exceptions, it sort of depends what the resonance forms are. If some heteroatoms or funky aromaticity isn't going on and this is a pure C-H system then typically hyperconjugation only stabilizes a plus charge.

I think you mistyped, and meant to say the hydrogen has a partial negative charge as well. The dipole is induced because pi electrons from the C=C bond are donating into the boron, this leaves behind carbons with partially positive charges. Since hydrogen is more electronegative than boron, it then can inductively take some electron density from boron. I think it might be a bit nicer to show the electrons coming from the double bond in that photo, rather than the carbon, but I guess they are trying to also illustrate the sterics with respect to the hydrogen.

Also, for an undergraduate student you are asking some pretty astute questions showing you are really thinking about the chemistry, not just trying to check a box to pass the class. Good job :).
Title: Re: Hydroboration regioselectivity
Post by: queeniestrudels on October 17, 2021, 03:21:07 PM
Why can't hyperconjugation stabilize carbanions? All I've learned about it so far is that it is the donation of an electron from the sigma to the sigma* orbital - couldn't the carbanion technically transfer its extra lone pair to its substituents sigma* orbitals and delocalize its charge? Why is hyperconjugation specifically for positive charges?
Title: Re: Hydroboration regioselectivity
Post by: wildfyr on October 17, 2021, 08:36:53 PM
We know that hyperconjugation stabilization of a carbocation occurs because the electrons in a C-H bond donate into an empty p orbital of the adjacent carbocation right because there is orbital overlap right? And the p orbital is "happier" being partially filled and not having a full + charge.

If you stick two electrons into that same orbital, the sterics of the orbital overlap become much less favorable due to coulombic repulsion. So they won't want to line up in order to donate electron from the carbanion into the C-H bond and "relieve" some of the full negative charge. Also a C-H bond doesn't really want to make a double bond, so it isn't very accepting of electrons anyways.

Its not an ironclad rule, but I have trouble easily thinking of instances where its otherwise...