Basicity Question

[ichoai]

I got a question on Qvault about ranking basicity..

The two in question are [CH2CHO]- and CH3O-. They said that the acetaldehyde is more basic than methoxide, but I can't seem to justify this in my head using the CARDIO method.

Is it because the acetaldehyde has an alkene which is more electron donating, destabilizing the negative charge?

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[ichoai]

anyone?

 

[Daneosaurus]

Do you mean [CH3CHO]- ?

 

[ichoai]

Nope. I thought the same thing at first glance.

I guess they are specifically referring to the enolate ion

 

[Daneosaurus]

I'm sorry - I'm no help. My textbook isn't even giving me a clear answer.

 

[ichoai]

What is the general rule when comparing basicity among RO- groups? I'm aware this is probably out of the scope of the exam, however I wanna figure this out for myself haha

 

[Will 8wt2]

My reasoning is to see which compound is more ready to accept a proton --> What I mean by this is looking at it from an electron density perspective (not really seen in Gen chem through Chads videos) but more so from my organic/biochemistry courses at Uni.


The Methoxide:
-Here, there is no resonance structures.
- The Oxygen has a high electronegativity, willing to keep hold of its electrons. --> making it somewhat stable in its anion form.

The Enolate ion:
- there are Resonance structures for the enolate ion.
- the electrons are shared between the double bonded carbons and the single bonded Oxygen.
- electrons are delocalized --> this does make it stable but it also gives the opportunity of a proton to attack (the double bond) or for the electrons on the oxygen when in C=O form to attack a proton and produce the OH form.
- From wiki: Enolates can exist in quantitative amounts in strictly Brønsted acid free conditions,

They are both anions, but i suppose my reasoning is based on the delocalization of the electrons and the abiity for a proton to attack or be attacked.
I'll let you guys know if if I find other justifications.

 

[Will 8wt2]

Let me know what others think! I'd like to know the for sure answer.

 

[Daneosaurus]

I know they are both extremely strong bases, with relatively similar pka's of ~20. So I think this is a super nitpicky question, that I really hope I wouldn't see on the DAT. I would have chosen the methoxide due to the localized electrons with no resonance, but I really don't know.

 

[Cool Beans]

This is a tricky example. What makes it tricky is that we tend to make the following generalization; that resonance always makes a base more stable. And while this is almost always true here we have an exception to the contrary. Resonance typically stabilizes a base because electrons are delocalized (shared across multiple atoms). In the example above you might view it as the negative charge potentially being shared by more than one atom.

In methoxide, the negative charge (and basic electrons) are localized only on the oxygen atom.
In the enolate, the negative charge (and basic electrons) are delocalized, being shared by an oxygen atom and a carbon atom.

It turns out that it is more stable to have the basic electrons localized on a single more electronegative oxygen atom than shared between an oxygen atom and a less electronegative carbon atom. And you can kind of make sense of this.

The more typical comparison might be to compare a base where the electrons are localized on a single oxygen atom (like methoxide) to a base where the electrons are delocalized and shared by two oxygen atoms (like a carboxylate). It's not difficult to predict that the more stable base has the electrons shared by 2 oxygen atoms.

Another typical example would be comparing a normal carbanion (localized on a single carbon atom) to an allylic carbanion (delocalized and shared by 2 carbon atoms) and again it's not too difficult to predict that the more stable base has the electrons shared by 2 carbon atoms.

Since these are the typical examples we generalize that resonance "always" makes a base more stable, but again, comparing methoxide to an enolate shows that there are exceptions to the rule.

Hope this helps!

 

[ichoai]

Thanks, Coolbeans...

So the enolate is only more basic because the anion is vinyl? Had there been one extra carbon separating the double bond and the oxygen, it would make enough of a difference to make methoxide into a stronger base (relatively)?

 

[Cool Beans]

It technically isn't "vinyl" but I know what you meant. The key here is that two resonance structures can be drawn for the enolate: the major with the negative charge on the oxygen and the minor with the negative charge on the alpha carbon. If the double bond were further away (it wouldn't be an enolate anymore) there wouldn't be any resonance and the negative charge would be localized on the oxygen only just like in methoxide and the two would end up being very close in basicity.