tekkado
Harmless
Posts: 37
Registered: 8-12-2013
Member Is Offline
Mood: No Mood
|
|
Double bond formation location?
Given the first compound, when dehydrating the OH on the ethane chain would the double bond form from the carbon on the ring or further away from it?
Im thinking away due to stability but not entirely sure if anyone could help? The doc has images of the structure in it =]
Thanks
Attachment: Untitled document.pdf (252kB)
This file has been downloaded 363 times
|
|
|
Hexavalent
International Hazard
Posts: 1564
Registered: 29-12-2011
Location: Wales, UK
Member Is Offline
Mood: Pericyclic
|
|
The product likely depends on the reaction conditions.
Under thermodynamic control - e.g. with a mild base or in acid with heating - the reaction will have a late transition state. According to the Hammond
postulate this will resemble the alkene product so you will favour the more substituted (more stable) alkene ("on the ring"). This is called the
Zaitsev product.
Under kinetic control - e.g. with a strong base - the TS will be early and resemble a carbanion. Accordingly you will get (initially) the terminal
alkene (called the Hoffmann product). It may be the case that prolonged reaction will cause isomerisation to the internal alkene.
In either case steric effects will be important too and so bulky bases will support formation of the terminal alkene (abstract the most
sterically-accessible proton).
"Success is going from failure to failure without loss of enthusiasm." Winston Churchill
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
