Boiledcabbages
Harmless
Posts: 4
Registered: 15-6-2017
Member Is Offline
Mood: No Mood
|
|
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?
|
|
|
Sigmatropic
Hazard to Others
Posts: 307
Registered: 29-1-2017
Member Is Offline
Mood: No Mood
|
|
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.
|
|
|
clearly_not_atara
International Hazard
Posts: 2694
Registered: 3-11-2013
Member Is Online
Mood: Big
|
|
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.
[Edited on 04-20-1969 by clearly_not_atara]
|
|
|
Boiledcabbages
Harmless
Posts: 4
Registered: 15-6-2017
Member Is Offline
Mood: No Mood
|
|
Ahh damm, ok thanks, I guess it makes sense that it has to be fully substituted as that probably allows for stabilization
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
