[MAFII]

Above is Hagemann's Ester.  Can somebody tell me why it alkylates preferentially at the 2 position and not the 4 position and does this necessarily mean that the 2 proton is more acidic than the 4 proton?


I appreciate your discussion.

[Dan]

Can I check two things?

1. Is position 2 the unsaturated position beta to the ketone, and is postition 4 the ketone carbonyl?

2. Under what conditions are you proposing to alkylate? The regioselectivity can be different with different alkylating methods. For example, LDA/MeI, Me₂CuLi and MeLi will all give different products.

[MAFII]

Can I check two things?

1. Is position 2 the unsaturated position beta to the ketone, and is postition 4 the ketone carbonyl?

2. Under what conditions are you proposing to alkylate? The regioselectivity can be different with different alkylating methods. For example, LDA/MeI, Me₂CuLi and MeLi will all give different products.

Thanks for responding.

The first answer is position two is the position that is alpha to the carbonyl and position 4 is gamma.  (there are 2 gamma positions but the one in question is allylic and alpha to the ester carbonyl.

The second is that the alkylations are done using sodium ethoxide followed by methyl Iodide.

Thanks for your help.

[Dan]

The first answer is position two is the position that is alpha to the carbonyl and position 4 is gamma.  (there are 2 gamma positions but the one in question is allylic and alpha to the ester carbonyl.

By "alpha to the carbonyl" do you mean alpha to the ester or to the ketone? I'm still not clear on which positions you mean. Can you tell me which positions, A, B, C or D you are referring to as positions 2 and 4?

[nj_bartel]

A and C I believe.

[orgopete]

I don't think it is a question of proton acidity in this case. It might be due to the double bond being more stable, tetrasubstituted and conjugated with the ester that makes the alpha position react faster. I believe steroids react in the same way to alkylate in at C4 if you start with an alpha-beta unsaturated ketone at C3.

A device I suggest for why enolates may alkylate on carbon rather than oxygen is that the electrons extend further from carbon than from oxygen in the enolate. That can lead to a faster reaction on carbon. That kind of argument seems difficult to apply here as I am unable to predict which bond length might be greater, a C-H of the starting material or the C-H of the isomer.

[Ulfsaar]

Well,i think the A position can't occur the alkylation because it is a C=C bond, the base can't get a alpha-H to make the carbon anion,so the 4 position if generated a anion ,it shoule be conjugated with the C=C and the carbonly group,it is more  willing to be alkylated

[orgopete]

You should first make the anion and then draw all of the resonance structures. Hagemann's ester as a cyclic vinylog of ethyl acetoacetate.

Note this common reaction of steroids.

[MAFII]

You should first make the anion and then draw all of the resonance structures. Hagemann's ester as a cyclic vinylog of ethyl acetoacetate.

Note this common reaction of steroids.

Thanks,

I have been trying to understand this http://pubs.acs.org/doi/pdf/10.1021/jo00930a001 paper because it gave the 83% yield and that did not seem like the most acidic hydrogen... and it wasn't. 

[orgopete]

…and it wasn't

Think again. This is like ethyl acetoacetate with an extra -CH=CH- inserted. While not the same as a direct connection, I'm sure it will be the most acidic.

[MAFII]

Just for clarity. I have attached one more picture.  In the scheme one of the paper I hyperlinked, it shows the following data.  I'm just trying to determine between the red and the green protons which would be the more acidic.  Intuitively, I would say the green.  I understand that the red would be sp2 making it more acidic than sp3 hydrogens.  However, I never thought that hybridization would give a greater effect than conjugation.  The results suggest that the red may be the more acidic, but I just guessed that the product came about from something other than a direct deprotonation of the red hydrogen.

[Dan]

The results suggest that the red may be the more acidic, but I just guessed that the product came about from something other than a direct deprotonation of the red hydrogen.

You are correct that the green H is much more acidic, and the product you are attributing to direct deprotonation of the red hydrogen will actually have come from intital deprotonation of the green hydrogen.

I think this is a methylation-tautomerisation sequence

[MAFII]

The results suggest that the red may be the more acidic, but I just guessed that the product came about from something other than a direct deprotonation of the red hydrogen.

You are correct that the green H is much more acidic, and the product you are attributing to direct deprotonation of the red hydrogen will actually have come from intital deprotonation of the green hydrogen.

I think this is a methylation-tautomerisation sequence

Thank you Dan,

This mechanism is what I was trying to understand.