71
Vivek @ Born this Way
·2012-01-19 05:26:11
It depends highly on where the C14 atom is ?
1
rishabh
·2012-01-19 07:52:36
i gues debosmit is right.
first eto- will attack c14 and c-o bond will break rendering a negative charge on o.
now internal nucleophilic rxn. => o- will attack the carbon and cl- will leave.
EDIT: ^ c14 is just for marking the carbon.doesn't affect the reaction.
71
Vivek @ Born this Way
·2012-01-20 04:02:04
Your mechanism is right.
But even if OEt attacks dirrectly at cl- attached carbon in an SN2 fasion, the product will be same ?
So why it isn't so?
1
rishabh
·2012-01-20 07:26:56
beacause the epoxide ring is more reactive.
262
Aditya Bhutra
·2012-01-20 08:42:18
because OEt will attack from least hindered site . (ie the C14 atom)
hence in the product given by debosmit, the rightmost carbon is C14
71
Vivek @ Born this Way
·2012-01-20 22:30:54
Thanks..
My doubt was cleared regarding steric hindrance with 3D viewer..