Why does an alkoxide steal H from water?

[CuriousBanker]

TL;DR Summary

Probably overthinking this and there's an easy solution

So I'm watching this video on preparation of alcohol using LiAlH4:

https://www.khanacademy.org/science.../preparation-of-alcohols-using-lialh4?modal=1In the final step, there is a primary alkoxide. H2O moves past it, and the Oxygen on the alkoxide steals a Hydrogen from water, neutralizing it and the water becomes OH-

But why? How, in total, is an alcohol and an oxide ion more stable than an alkoxide plus a water molecule. Carbon is more electronegative than Hydrogen, which means That the O connected to the C in alkoxide should have less electron density than the O connected to the H in water, since the C withdraws electrons a little more than H does. Which means the O in water should have more of a negative charge than the O in alkoxide, which means it should hold on to hydronium (proton) more strongly.

This is puzzling me greatly, and I'm sure I'm missing something obvious. Please help!

 

[willem2]

Can you send a better link? THe one you provided is about
Preparation of alcohols using LiAlH4, and I can't find anything else on the Khanacademy site about this.

 

[CuriousBanker]

That's the correct video. I apologize, it's not an alkoxide. It's a carboxyl group that broke its pi bond to bond to a hydrogen ...idk what it's called...Its at the 6 minute mark. CHHRO, the O has a negative 1 formal charge, 3 pairs of lone electrons and a covalent bond to carbon. In the video, it nabs an H off water to leave it with two lone pairs of electrons, a covalent bond to H, and a covalent bond to C...but then the turns the water molecule into the oxide anion...how is that thermodynamically favorable? i would think the Carbon molecule can accomdate a negative charge better than an oxide anion.

 

[TeethWhitener]

No, it’s an alkoxide. Alkoxides are quite basic, and if they’re in a fairly low concentration in water, they’ll end up being mostly protonated to form the alcohol as the conjugate acid product.

That said, your intuition is right. Normally, a reduction like this would involve an acid workup: meaning that acid would be added as a last step to shift the alcohol/alkoxide equilibrium as far as possible to the alcohol.

 

[CuriousBanker]

Thank you for your reply. Yeah, I know Alkoxides are quite basic, but the alkoxide turns the water molecule into an oxide anion...isn't oxide more basic than alkoxide? So I'm guessing its a reversible reaction that doesn't shift that way very much?

 

[TeethWhitener]

Thank you for your reply. Yeah, I know Alkoxides are quite basic, but the alkoxide turns the water molecule into an oxide anion...isn't oxide more basic than alkoxide? So I'm guessing its a reversible reaction that doesn't shift that way very much?

I’m not seeing what you’re talking about. The alkoxide in the video only abstracts one proton from the water, leaving a hydroxide ion.
Edit: I’m talking about the events that happen at ~5:45 in the video.

 

[CuriousBanker]

Yeah, sorry, I typed that incorrectly, it leaves a hydroxide ion...why doesn't that hydroxide anion steal the hydrogen right back from the newly formed alcohol?

 

[TeethWhitener]

Because alkoxides have somewhat higher pKa than water and usually you’re adding water to the reaction in a much much higher concentration than the alcohol. The equilibrium will favor the formation of alcohol over alkoxide.

 

[CuriousBanker]

I gotcha, and I figured that, but my question is why. In my head, water should have a stronger attraction to that hydrogen, because the hydride anion is an OH bond, and so O should hog the H electron more so than the electron between the O and the C in the alcohol, since C is more electronegative than H. So the O in the alkoxide should be less electrophilic than the O in the hydride, since it is easier for the alkoxide to dispere the negative charge. Clearly I'm wrong though, and am curious what factors lead to the actual real life outcome?

 

[TeethWhitener]

hydride

I'm assuming you mean hydroxide here.

The equilibrium concentrations will be determined by 1) the initial concentrations, and 2) the equilibrium constant. You seem to be asking why the equilibrium constant should favor hydroxide and alcohol over alkoxide and water. It doesn't always, but they're usually within a few pKa units of one another (see http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf for examples). As I mentioned before, if you were doing this reaction in the lab, you'd do an acid workup to ensure you got the protonated product.

Your electronegativity argument doesn't really work, as C and H have quite similar electronegativities, to the point where the C-H bond is generally considered non-polar except in special circumstances. More important will be the relative stability of the alkoxide and hydroxide in the chosen solution. Some alkoxides are stabilized by electronic structure, and are therefore much more acidic than normal (phenol is a good example, with a pKa of ~10 largely from resonance stabilization), some are stabilized by electron-withdrawing groups (like fluorous alkoxides), and some are destabilized by electron-donating groups. Often this is another way of saying that the O-H bond is stronger or weaker based on what the substituents on the carbon are.

 

[CuriousBanker]

Gotcha. If it were destabilized by electron-donating groups, then it would have even more electron density around it, and therefore by even more likely to nab H+, right?

 

[TeethWhitener]

Yes, roughly. A better way to think about it might be to say that, for an alkoxide like tert-butyl, with a bunch of electron donating groups, the CO-H bond formed is stronger than the HO-H bond in water; thus the proton transfer process is energetically favorable.

Again, there are secondary effects (e.g., choice of solvent), but the bond energetics is the main effect.