Comments on: Hydroboration Oxidation of Alkenes

By: Emilie

Emilie — Mon, 08 Dec 2025 20:45:40 +0000

There is a typo; when discussing the summary up top, instead of saying the LEAST substituted carbon for anti-markovnikov, it says the MOST. Great article, this has been so helpful!

By: James Ashenhurst

James Ashenhurst — Thu, 31 Oct 2024 15:49:39 +0000

In reply to S.

In solution, the molecule can be approached by a reactant from every direction (360 degrees). If the molecule were perfectly flat, then half (180 degrees) of the molecule would be approachable from the top and the other half (180 degrees) from the bottom.

In the tent shaped molecule the top is approachable from more like 270 degrees and the bottom something like 90 degrees. (This is an oversimplification, but it will do for our purposes).

Statistically, due to the shape of the molecule, it is more likely for reactants to approach from the top face.

By: S

Thu, 31 Oct 2024 09:59:29 +0000

In tent shaped alkene molecule why bottom face approach more hindered .can you explain more

By: Ushindi Muna

Ushindi Muna — Sun, 03 Mar 2024 04:00:21 +0000

Why are most substituted carbons more positive in charge. Seems counter intuitive since carbons are electron donating, so the carbon with more carbons attached to it should have a partial negative charge since there is a higher electron concentration right? Why is it the other way around.

By: James Ashenhurst

James Ashenhurst — Tue, 21 Mar 2023 14:33:19 +0000

In reply to Mark Springsteel. Nice! As the parent of a toddler, it resonates.

By: Mark Springsteel

Mark Springsteel — Mon, 20 Mar 2023 04:45:50 +0000

I tell my students Boron is just a whiner, never happy, always complaining; “I want an Octet, everyone else has an octet, except H but it doesn’t even want an octet. Wait now I have a Negative Charge, I don’t want a Negative Charge…”. Then I say Al is a more aggressive whiner, its just angry! “I WANT AN OCTET NOW!”
If interested see slides 10 and 11 of my animation about reductions with NaBH4 and LiAlH4. Fun stuff…

https://docs.google.com/presentation/d/1u1YSKJN2D9nRz0BkG18prvmlSnURum1fOsMoafurJ5U/edit#slide=id.p4

By: James Ashenhurst

James Ashenhurst — Wed, 15 Feb 2023 02:35:59 +0000

In reply to mikel Do. Br is bromine (electronegativity 3.0). The symbol for boron is B (electronegativity 2.0)

By: mikel Do

mikel Do — Mon, 13 Feb 2023 10:18:39 +0000

Hello!
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?

By: James Ashenhurst

James Ashenhurst — Tue, 07 Dec 2021 18:44:48 +0000

In reply to Freeman John. The partial positive charge reflects the (slight) polarization of the B-H bond, and in turn helps to rationalize the anti-Markovnikov regioselectivity.

By: Freeman John

Freeman John — Mon, 22 Nov 2021 01:56:55 +0000

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.