Question

1. Show the reaction, along with the transition state (‡), between 2-methyl-2-pentene and BH3 in THF. Then, show the subsequent reaction of the organoborane intermediate with H2O2 and OH-1. Explain why hydroboration is non-Markovnikov and forms a 2o alcohol. Refer to hydroboration in the chapter notes, the handout for example 8.2, and figure 8.4 in text. (2 pts)

Answer 1

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 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”.

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.

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

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!

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:

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.

Note how the charge on boron goes from negative to neutral.

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.