Boiledcabbages
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Stevens rearrangement
Hey guys, I don't post often but I just have a quick question about the Stevens rearrangement. Does it have to be a fully substituted quaternary
ammonium ion? I.E. if you have R-CH2-NH2 and you react it with methyl bromide, can you perform a Stevens rearrangement on the quaternary ammonium salt
formed to obtain R-CH(CH3)-NH2?
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Sigmatropic
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What are the exact conditions for the Steven's rrr? According to a cursory look (Wikipedia only) the 2nd step involves treatment with strong base to
form a stabilized ylid. With non quatenary ammonium salts that would just lead to free basing of the amine.
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clearly_not_atara
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Yes, it must be fully substituted. Also, generally only benzyl or allyl groups migrate for some unclear reason. With benzyl groups, care is necessary
to ensure selectivity over the competing Sommelet-Hauser rearrangement:
https://pubs.acs.org/doi/pdf/10.1021/jo00044a050
Quaternary amines bearing an allyl group and an alpha-ketoalkyl group are particularly good substrates for the Stevens rearrangement. However, the
synthesis of these quats is not easy.
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Boiledcabbages
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Ahh damm, ok thanks, I guess it makes sense that it has to be fully substituted as that probably allows for stabilization
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