[spirochete]

We all know that hydroxide will never be a leaving group in an SN2 reaction, no matter how strong the nucleophile.  Yet an intermediate containing R-O- is created when a nucleophile adds to a carbonyl compound.  It looks sort of like a leaving group, but it's still attatched to the molecule.

What is the apparently huge difference between these two reactions that allows this to happen?  The only difference I can see is that SN2 involves a pair of SP3 electrons and the addition to a carbonyl involves a pair of P orbital electrons.  By VSEPR the intermediate has its new electron pair in an SP3 orbital, but I know lone pair orbitals can be unpredictable.

[spirochete]

Does my question make sense?  I posted a file with a couple reactions showing what I mean.

[macman104]

The difference is that in order for the nucleophile to attack the alcohol, the OH has to be kicked out as a leaving group.  -OH is a very strong base and does not act as a leaving group.

However, for the nucleophile to attack the carbonyl, the bonds just get displaced to the oxygen atom which is already electronegative, it's not leaving at all.  If you think about it, a "resonance" form of the carbonyl is with a charge separation (positive charge on carbon and negative on oxygen).

[spirochete]

The resonance argument is convincing, that is a significant difference. 

[Hazel]

But in Claisen Condensation(for example),

Assessed from:http://en.wikipedia.org/wiki/Claisen_condensation

Why the R₄O- can remove in this way?I seem it quite similar to leaving group.

[macman104]

Indeed, but that is an alkoxide ion, not an OH.  They are completely different.

[Hazel]

Yea,that is alkoxide..but alkoxide is also a weak leaving group and almost unreactive leaving group,isn't?

[macman104]

Yea,that is alkoxide..but alkoxide is also a weak leaving group and almost unreactive leaving group,isn't?

If your alkoxide is CH₃O^- for example, you would probably run the reaction using NaOCH₃/CH₃OH.  Since the alpha proton is much more acidic, the solution is basic enough to deprotonate the alpha-carbon, and then you are simply regenerating the alkoxide that you used up to deprotonate the carbon.  Since you are running this in a weak base, it is acceptable to have an alkoxide be a leaving group.

[spirochete]

I haven't learned the claissen condensation yet so take this with a grain of salt.  In the step before R40- is kicked out by the shift of pi electrons, the molecule already has a negative charge on the other oxygen.  So perhaps there is not as high an energetic barrier to forming the R4O- leaving group for that reason.  It's just trading a one strong base for another strong base.

For SN2, wouldn't an alkoxide ion be an equivalently bad leaving group as hydroxide?

Also is it just me or does that mechanism use "R4" in 2 different ways: once as a base in the beginning of the mechanism, and again as an alkyl group in the second ester molecule?  That confused me at first.

[macman104]

I haven't learned the claissen condensation yet so take this with a grain of salt.  In the step before R40- is kicked out by the shift of pi electrons, the molecule already has a negative charge on the other oxygen.  So perhaps there is not as high an energetic barrier to forming the R4O- leaving group for that reason.  It's just trading a one strong base for another strong base.

I would not call an alkoxide a strong base.  When I think of strong bases, I think of KOH, lithium diisopropyl amide, and the like.

For SN2, wouldn't an alkoxide ion be an equivalently bad leaving group as hydroxide?

No.  However, it is not necessarily a great leaving group either.  Only under circumstances do you have an alkoxide acting as a leaving group (good thing too, can't have ethers falling apart willy-nilly ).  You'll notice that in this case, with the claisen condensation, the alkoxide that is being kicked out is already actually in solution.  The reaction is run in a mixture of the alcohol and the corresponding sodium salt.

Also is it just me or does that mechanism use "R4" in 2 different ways: once as a base in the beginning of the mechanism, and again as an alkyl group in the second ester molecule?  That confused me at first.

It does.  This is because, they want you to notice that whatever group is going to be acting as the leaving group they also want to use as a base.  This is because there is the possibility that the base can instead act like a nucleophile and attack the carbonyl and exchange the ester alkoxide out.  However, if we use the same alkyl chain then we don't really care if that happens because the structure is unchanged.

[Hazel]

Only under circumstances do you have an alkoxide acting as a leaving group (good thing too, can't have ethers falling apart willy-nilly Wink).  You'll notice that in this case, with the claisen condensation, the alkoxide that is being kicked out is already actually in solution.  The reaction is run in a mixture of the alcohol and the corresponding sodium salt.

So,the certain circumstances u mean is like when we regenerating the alkoxide back to the solution?Have any other special conditions regarding to this?



I found an explanation of the determination of good and poor leaving group goes like this : A leaving group is whether good or not based on its stability after leaving.The more stable after it left the molecule,the better leaving group it is.

Then,I am curious,why a leaving group can left the molecule in the 1st step of Sn1(Ex.Why there is a spontaneous dissociation occurred in (CH3)3CBr to generate the carbocation intermediate and bromide ion)?
Is it because the leaving group(Br-) will be more stable after leaving?Or in other words,it is a very reactive and unstable compound(alkyl bromide in this example)?

[kryptoniitti]

I found an explanation of the determination of good and poor leaving group goes like this : A leaving group is whether good or not based on its stability after leaving.The more stable after it left the molecule,the better leaving group it is.

Then,I am curious,why a leaving group can left the molecule in the 1st step of Sn1(Ex.Why there is a spontaneous dissociation occurred in (CH3)3CBr to generate the carbocation intermediate and bromide ion)?
Is it because the leaving group(Br-) will be more stable after leaving?Or in other words,it is a very reactive and unstable compound(alkyl bromide in this example)?

It also depends heavily on the solvent of choise. For S_N1 reactions, polar is solvent is good choise because it accommodates the resulting carbocationic and anionic species.

[Hazel]

It also depends heavily on the solvent of choise. For SN1 reactions, polar is solvent is good choise because it accommodates the resulting carbocationic and anionic species.

I thought the solvent mainly affects the nucleophiles.As in polar solvent mainly solvating anion and the reactivity of the nucleophile will like I->Br->Cl->F-
So,u mean the polar solvent helps stabilizing the leaving group in SN1?And how about in Sn2?

[kryptoniitti]

So,u mean the polar solvent helps stabilizing the leaving group in SN1?And how about in Sn2?

It helps stabilizing both the carbocation and the leaving group. S_N2 is a little different story. It depends whether the nucleophile is neutral or anionic. If you have anionic nucleophile, you probably shouldn't use protic solvents because the possibility of hydrogen bonding. It really depends what kind of reactants used etc. Difficult question to give an broad answer (for me at least ). It goes pretty deep into kinetics I think.

[Yarr]

Hydroxide ion can act as a leaving group: just consider the mechanism of an aldol condensation with base.