Comments on: Hydroboration Oxidation of Alkenes

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.

By: Hydroboration-Oxidation: Definition, Examples, and Mechanism

Hydroboration-Oxidation: Definition, Examples, and Mechanism — Sat, 01 Feb 2020 07:35:20 +0000

[…] Mechanism – Masterorganicchemistry.com […]

By: James Ashenhurst

James Ashenhurst — Mon, 08 Jul 2019 03:53:00 +0000

In reply to Nayara. OK, thank you Nayara!

By: James Ashenhurst

James Ashenhurst — Mon, 08 Jul 2019 03:52:51 +0000

In reply to Wei. Compare the electronegativities of B and H and tell me where the dipole should be in the B-H bond?