I am on chapter 6 of my orgo book and unfortunately the AP chem teacher at my high school recently quite his job. I already asked a uni student who is
taking some chem courses and I have yet to receive an expiation for this reaction. I also googled the problem and came up with no answers or
explanations. Help me Science Madness you are my only hope.
LearnedAmateur - 9-12-2017 at 14:15
I found this on the Wikipedia article for 'nucleophilic conjugate addition'. I know it's not the same mechanism as what is occurring here, but it
would explain why only the 3-chloro isomer is formed. Hopefully more insight and/or improvements will come this way, it's peaked my interest now.
[Edited on 9-12-2017 by LearnedAmateur]ninhydric1 - 9-12-2017 at 14:43
I would assume LearnedAmateur's mechanism is correct, as only HBr can undergo free-radical addition, ending in a product opposing Markovnikov's rule.
Another one of the many exceptions in organic chemistry .DJF90 - 9-12-2017 at 14:44
I know it's not the same mechanism as what is occurring here, but it would explain why only the 3-chloro isomer is formed.
[Edited on 9-12-2017 by LearnedAmateur]
What makes you say that?VSEPR_VOID - 9-12-2017 at 15:07
Thank you. I worked out mechanism on paper and it seems to work. LearnedAmateur - 9-12-2017 at 16:58
I know it's not the same mechanism as what is occurring here, but it would explain why only the 3-chloro isomer is formed.
[Edited on 9-12-2017 by LearnedAmateur]
What makes you say that?
Propenoic acid contains an a,b unsaturated carbonyl, which is what is required for this particular reaction. I know that the hydroxyl group generates
a resonance structure which could mean a completely different mechanism, but this is the closest thing I can find.DraconicAcid - 9-12-2017 at 17:53
Since chloride isn't that great of a nucleophile, I wouldn't be surprised if the first step was the protonation of the carboxylic acid group, which
would give a cation that would resonate and put some positive charge on the beta carbon.VSEPR_VOID - 9-12-2017 at 20:32
Could anyone help me with this problem as well? Thanks.