Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: 08roselx on December 04, 2019, 05:00:57 PM
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Hello!
I am trying to write out a mechanism but struggling to see which hydrogen the base will remove to kick out the leaving group (SO2Ar(Pr)3)
Usually it would be the H on the neighbouring C atom from the C=N however in this case there are only H atoms attached to a neighbouring sp3 C and I wasn't sure if it would remove one of these? Otherwise it would have to be from the benzene ring but I'm not sure how that would work with the C carrying a negative charge within the benzene ring?
Any help would be greatly appreciated!
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The standard mechanism for this (Bamford-Stevens) is:
deprotonation to form the anion you have drawn
loss of the Arylsulfinyl anion (ArSO2-) to generate the diazo compound
Loss of the diazo N2 to form the alkene (styrene) - this goes through the carbene in aprotic solvents, through the carbocation in protic solvents.
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The standard mechanism for this (Bamford-Stevens) is:
deprotonation to form the anion you have drawn
loss of the Arylsulfinyl anion (ArSO2-) to generate the diazo compound
Loss of the diazo N2 to form the alkene (styrene) - this goes through the carbene in aprotic solvents, through the carbocation in protic solvents.
Hi
As there are 2 equivalents of base used I'm not sure which hydrogen is attacked to kick out the ArSO2- group. This is how the reaction usually proceeds but I am unsure in the case of the benzene ring.
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I don't think that is quite right.... but you aren't losing an aryl proton inany case so there is only one other choice.
see this:
https://www.organic-chemistry.org/namedreactions/bamford-stevens-reaction.shtm
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I don't think that is quite right.... but you aren't losing an aryl proton inany case so there is only one other choice.
see this:
https://www.organic-chemistry.org/namedreactions/bamford-stevens-reaction.shtm
I believe it is the Shapiro Reaction rather than the bamford stevens reaction
https://www.organic-chemistry.org/namedreactions/shapiro-reaction.shtm
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the only place to take a proton is the methyl. The aromatic ring has no H at the ipso.
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the only place to take a proton is the methyl. The aromatic ring has no H at the ipso.
can I take a proton from the methyl group? I wasn't sure if that was a thing that was allowed? I'm so confused!
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You can take any proton that is more acidic than butane. And here the methyl gives an anion that is conjugated with the hydrazine double bond so it will be more like an allylic anion.
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You can take any proton that is more acidic than butane.
Thermodynamically yes, kinetically no - otherwise BuLi in hexane would be mostly HexLi!
I think there's some confusion here- can the OP clarify which drawing of the substrate is correct? In the initial post it was an acetophenone-derived tosylhydrazone, but in the later post it was cyclohexane-derived. But I think you're right, with BuLi it is a Shapiro rather than Bamford-Stevens.
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otherwise BuLi in hexane would be mostly HexLi!otherwise BuLi in hexane would be mostly HexLi!
How do you know it isn't? We assume it stays BuLi, but HexLi would do the same reactions.
Hmmm - but it might build up more excess pressure in the bottles since Bu-H is gas at 25 C..
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You can take any proton that is more acidic than butane.
Thermodynamically yes, kinetically no - otherwise BuLi in hexane would be mostly HexLi!
I think there's some confusion here- can the OP clarify which drawing of the substrate is correct? In the initial post it was an acetophenone-derived tosylhydrazone, but in the later post it was cyclohexane-derived. But I think you're right, with BuLi it is a Shapiro rather than Bamford-Stevens.
This is the scheme that I am trying to deduce the mechanisms and resultant products for. The original photo on the post is the mechanism I was doing, what I commented later was an example as to why I was confused at which proton I am removing.