inaccensa

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If a carboanion is attached to 3 e-withdrawing groups, wouldn't that be stable? Why is carbanion** stability opp of carbocation?
 
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loveoforganic

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Carbanion*

It would be relatively more stable than a carbanion without 3 e- withdrawing substituents. You talk about stability in relative, not absolute terms.

You're actually looking for the same thing in the stability of both carbanions and carbocations - a carbon with a normal amount of electron density around it. Carbanions are just electron rich species, so to better distribute electron density requires electron withdrawing substituents. Carbocations are electron deficient species, so to better distribute electron density requires electron donating species.
 

sleepy425

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First of all, it's carbanion, not carboanion.

Anyway, they're the opposite because carbocations and carbanions are the opposite thing. A carbocation is electron deficient (for two reasons: 1. It has less than a complete octet, and 2. it is positively charged. Carbon is not stable under either of these conditions). A carbanion has excess electron density (it is negatively charged, which carbon doesn't handle very well).

Soooo, that's why each is stabilized by opposite things:

Since a carbocation is electron deficient, it is stabilized by things that donate electron density into the carbocation (like alkyl groups or heteroatoms that can stabilize by resonance).

Since a carbanion is too electron rich, it is stabilized by things that remove electron density from the carbon center, electron withdrawing groups.
 

wanderer

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Carbanion*
Thank you.

And carbanions can be somewhat stable. Haloform reactions use a free carbanion intermediate. 1,3 diketones can also be stable in their anion form, but that is due more so to resonance than electron withdrawing groups.
 
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inaccensa

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Ok, let me restate the question

In books, carbanion stability 3< 2 < l < methyl, but why?

Forget the resonance and the electron withdrawing groups,if they are out then why a primary anion more stable than a tertiary anion. i'm thinking in terms of delocalization of the negative charge, the more no of substituents, the better spread of charge.
 

loveoforganic

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Alkyl groups are electron donating. You're supplying more electron density to your negative charge. This is why potassium tert-butoxide is such a strong base.
 

loveoforganic

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sure it does. there are a ton of threads I don't even open when I see that the last post was by loveoforganic (unless of course you started the thread and had 0 replies)
Ok, I'll qualify my statement - doesn't happen often where I don't get corrected to some extent or ask a question at the end of a thread just to have you ignore it! (kidding) :p

Out of curiosity, I was trying to look up the pkb of potassium tert-butoxide and came across schlosser's base. What's the point of it?

http://en.wikipedia.org/wiki/Schlosser%27s_base
 

sleepy425

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oh cool, I'd never heard of it, but I just looked it up. Apparently, it's basically a way of making n-butylpotassium (as opposed to n-butyllithium). n-butylpotassium is much more reactive and has some other appealing features that make it desirable sometimes. So basically, you use the KOtBu just as a source of K+, and you get exchange with the lithium of the n-buLi so you have n-buK.

One of those desirable features I mentioned is that, when you do a metalation of an alkene, if you use an organolithium to do the metalation of the alkene, the allyl lithium product usually isomerizes rapidly between cis and trans (which is usually not good), whereas if you use an organopotassium, isomerization is much slower so you can retain cis/trans. Allyl sodiums do not isomerize as fast as allyl lithiums, but isomerization is still pretty fast to the point that you will generally lose the stereochemistry.

Look up organosodium on wikipedia, that's where I got most of the information.
 
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inaccensa

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Alkyl groups are electron donating. You're supplying more electron density to your negative charge. This is why potassium tert-butoxide is such a strong base.

:thumbup:
However, if a carbon where attached to three oxygen vs a primary carbon anion, the one with o2 will be more stable correct
 

loveoforganic

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Oh ok, cool stuff. I've seen other instances of the cation exchange in superbases, and out of curiosity again, what's the driving force between the cation exchange? If I'm branching into the world of physical organic chem, you need not bother with a response because I probably won't understand it! Thanks
 

sleepy425

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Oh ok, cool stuff. I've seen other instances of the cation exchange in superbases, and out of curiosity again, what's the driving force between the cation exchange? If I'm branching into the world of physical organic chem, you need not bother with a response because I probably won't understand it! Thanks
I'm going to speculate on this one. I'm actually pretty sure this is correct, but take it with a grain of salt, just in case.

Have you ever heard of hardness/softness of nucleophiles and electrophiles? You might be more familiar with the nucleophilicity/basicity discrepancy between fluoride and iodide. In general, iodide is a better nucleophile than fluoride, but fluoride is a much stronger base than iodide. In many cases, basicity and nucleophilicity follow each other, but in this case, they don't. The reason is because iodide is bigger and more polarizable.

Hard nucleophiles are more likely to act as a bronsted base, while soft nucleophiles are more likely to act as nucleophiles. So fluoride is a hard nucleophile, iodide is a soft nucleophile. Softer nucleophiles are usually bigger and often don't have formal negative charges (so t-butanol is a softer nucleophile than t-butoxide). Harder nucleophiles are usually smaller (less polarizable) and have formal negative charges.

Similarly, there are also hard and soft electrophiles. Again, charge density and polarizability of the electrophile are important factors. Hard electrophiles usually have formal positive charges and are very small (not polarizable). Soft electrophiles have lower charge density (so alkyl halides are soft electrophiles) and are usually much bigger (polarizable).

The point of all of this is that hard electrophiles pair better with hard nucleophiles, and soft electrophiles pair better with soft nucleophiles. I believe this has to do with the fact that hard atoms can pair up better because they both interact primarily using electrostatic attraction, while soft atoms pair up better because both interact primarily because of polarizability. If you tried to pair up a soft electrophile and a hard nucleophile, the hard nucleophile, which has a strong negative charge, can't interact well with the weakly charged soft electrophile. Conversely, the soft electrophile, which is easily polarizable, doesn't interact well with the hard nucleophile that isn't very polarizable.

Anyway, so in Schlosser's base, we have potassium tertbutoxide, and n-butyl lithium. Potassium is a softer electrophile than lithium because potassium is bigger and more polarizable. Tertbutoxide is an oxide nucleophile, while the n-butyl nucleophile is a carbon nucleophile. Both have the same formal charge, but carbon is bigger. So carbon nucleophiles are softer than oxygen nucleophiles. So our hardest electrophile is lithium. Our hardest nucleophile is the oxygen nucleophile from tertbutoxide. So the lithium and and tertbutoxide prefer to pair up. Our softest electrophile is potassium, and our softest nucleophile is the carbon nucleophile of the n-butyl anion. So the potassium and the n-butyl anion pair up better. Thus, we have lithium complexed with tertbutoxide and potassium complexed with the n-butyl anion, giving us primarily lithium tertbutoxide and n-butyl potassium.
 

loveoforganic

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I'm familiar with HSAB and that's a great explanation :) Thanks for taking the time