chemrox
International Hazard
Posts: 2961
Registered: 18-1-2007
Location: UTM
Member Is Offline
Mood: LaGrangian
|
|
reaction mechanism-dumbshit question maybe
A compound of the type R-S-CH2-CH2-X is hydrolized in water very rapidly and a cyclic sulfonium intermediate has been proposed as the rate determining
step. The writeup (this is from a textbook) indicated formation of one product: RSCH2CH2OH (+HCl). Unfortunatley I can't draw this but its a simple
diagram with the two methyls and sulfur atom making a triangle; the sulfur carrying the positive charge. If the only product is the straight
RS-ethanol, it seems that one of the methyls was attacked by water preferentially. Yet the intermediate is symmetrical so why not 50% iso-product? I
feel like I might be missing something really obvious but since I can't see it, I'm sticking my neck out (and maybe my foot in.) I'm studying up on
reactions and mechanisms. It's been a long time ...
Thanks,
CRX
PS added:
The Text is: Miller, Bernard, 1998, Advanced Organic Chemistry, Reactions and Mechanisms, Prentice Hall. The reaction is on page 177.
[Edited on 23-7-2007 by chemrox]
|
|
|
not_important
International Hazard
Posts: 3873
Registered: 21-7-2006
Member Is Offline
Mood: No Mood
|
|
Do you mean like this :
http://www.cem.msu.edu/~reusch/VirtualText/addene2.htm
the nucleophile is breaking the C-S bond as it attaches to the C, so that carbon must become 'normal' or beta to the sulfur, right?
|
|
|
chemrox
International Hazard
Posts: 2961
Registered: 18-1-2007
Location: UTM
Member Is Offline
Mood: LaGrangian
|
|
Yes and no. In the example of a sulfur mustard the "R" group is attached to the sulfur atom and the two methyls forming the bse of the cyclic
sulfonuim ion are completely equivalent.
I tried attaching the page; we'll see how it worked.
[Edited on 23-7-2007 by chemrox]
[Edited on 23-7-2007 by chemrox]
[Edited on 23-7-2007 by chemrox]
|
|
|
guy
National Hazard
Posts: 982
Registered: 14-4-2004
Location: California, USA
Member Is Offline
Mood: Catalytic!
|
|
Because the methyl group that is attacked is the one that breaks away from the sulfur. You can't have an iso-product because it would be something
like R-CH2(OH)-CH2 which is impossible.
|
|
|
not_important
International Hazard
Posts: 3873
Registered: 21-7-2006
Member Is Offline
Mood: No Mood
|
|
Same thing - the carbon being attacked gives up its bound with the sulfur, which means the sulfur will be beta to that carbon. The only way to get
iso, if by iso you mean something like S-C(OH)-CH3 is for the carbon being attacked to keep the bound to sulfur and lose a hydrogen.
|
|
|
chemrox
International Hazard
Posts: 2961
Registered: 18-1-2007
Location: UTM
Member Is Offline
Mood: LaGrangian
|
|
I knew there was something there I wasn't seeing. Now it seems obvious. Thanks to you both!
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
