lastlokean - 28-2-2013 at 15:14
I am pretty well versed in basic chemistry and organic chemistry but I just can't make sense of this topic.
For example:
Propanamide + Acetaldehyde <-> ?
CH3CH2C=O(NH2) + CH3 CHO <-> ?
My initial assumption is that it would result in something like a-Acetal Propanamine? (I'm sure I named tat wrong.) I'm trying to imply it would
result in the cleavage of H2 from the N and tack on the CH3 CHO?
I have found some literature on the subject of Amide and Aldehyde condensing on ethanol in what seems similar to N-amine substitution. This just
feels 'wrong' to me though. Also all the literature I can find is with very complex amides and aldehydes.
Edit: Also if it does work for one amide and aldehyde any reason to believe it wouldn't work for any amide + aldehyde condensation? Say, Propanamide
and benzaldehyde?
[Edited on 28-2-2013 by lastlokean]
[Edited on 28-2-2013 by lastlokean]
kavu - 1-3-2013 at 00:01
This condensation can be done and there are a fair bit of examples in literature. In general
R-CHO + H2N-C(O)-R' → R-CH=NH-C(O)-R'
Nicodem - 1-3-2013 at 13:50
See DOI: 10.1021/ja01335a085
lastlokean - 3-3-2013 at 22:38
Thanks very much for the information!
I find it interesting that in some cases an aldehyde and amide will react 1:1 and other times 1:2. Is there any insight as to how to determine which
will occur before experimentation with the reaction?
Also, is it possible that the opposite will occur? In which case you would have a doubled up aldehyde onto an amide? If so, does anyone have any
examples?
Sorry if all this seems redundant I'm just trying to develop a general understanding of aldehyde chemistry.
Edit: Actually I can find no referances of amide aldehyde condensation other than the amide doubling onto the aldehyde. Does anyone have any for
the 1:1 case also?
[Edited on 4-3-2013 by lastlokean]