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Diazo transfer to 1,3-dicarbonyl with tosyl azide
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Kal
Diazo transfer to 1,3-dicarbonyl with tosyl azide
This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.
Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.
The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:
In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!
Anyway, the product above is simply a resonance form of the diazo product you want.
Reference
Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc.1990,112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.
This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.
Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.
The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:
In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!
Anyway, the product above is simply a resonance form of the diazo product you want.
Reference
Evans, D. A.; Britton, T. C.; Ellman, J. A.; Dorow, R. L. The asymmetric synthesis of α-amino acids. Electrophilic azidation of chiral imide enolates, a practical approach to the synthesis of (R)- and (S)-α-azido carboxylic acids. J. Am. Chem. Soc.1990,112 (10), 4011–4030. DOI: 10.1021/ja00166a045. Non-paywall version available from the Evans group website.
This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.
Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.
The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:
In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!
Anyway, the product above is simply a resonance form of the diazo product you want.
Reference
This starting material is not actually an enone; it is a 1,3-dicarbonyl, confusingly drawn as the enol tautomer. 1,3-Dicarbonyls are mildly acidic ($\mathrm pK_\mathrm a \sim 10$) and can be deprotonated by weak bases, such as triethylamine ($\mathrm pK_\mathrm a \sim 9$), which was mentioned in the question. So, now, we have an enolate ion which is nucleophilic.
Tosyl azide, $\ce{TsN3}$, is an odd reagent. It's easy to think that it is acting as a nucleophile of some sort, because the azide ion is nucleophilic, but actually it is an electrophile.
The next part is the removal of the formyl group, which is a bit tricky. Here I am not 100% sure, so take it with a pinch of salt, but I think Mith was on the right track with the retro-(3+2) cycloaddition. This is my guess:
In fact, the cyclisation of the terminal nitrogen onto a carbonyl group has been proposed before, in mechanistic studies of electrophilic azidation by D. A. Evans.1 That makes me feel a bit more confident in my guess!
Anyway, the product above is simply a resonance form of the diazo product you want.
Reference
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