-3
$\begingroup$

What is the drawn product drawn here the major product? I thought the radical (intermediate) is more stabilized at a primary carbon than at a tertiary one. enter image description here

$\endgroup$

closed as off-topic by Mithoron, airhuff, Jan, bon, Jon Custer Feb 7 '18 at 14:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

Carbon-centered radicals are electron deficient species in general--you have 7 electrons instead of the desired 8. This means that to first order, you can gauge the stability of a series of radicals the same way as you would for a series of carbocations. In this case, the tertiary cation and the tertiary radical would be the most stable in their respective series. So, no, definitely don't make a primary radical when you can make a tertiary radical.

$\endgroup$
  • $\begingroup$ Ah thank you, that makes sense intuitively. I think I got confused because I was reading this website ( masterorganicchemistry.com/2013/04/12/… ) on free radicals and in one of the examples, it says addition of a bromine radical has anti-markovnikov selectivity. Do you know why this might be? $\endgroup$ – Sarah Smith Feb 6 '18 at 21:42
  • $\begingroup$ The link you provide has to do with the addition of HBr in the presence of peroxides (Kharasch reaction), not the broomination of an alkane. $\endgroup$ – user55119 Jun 13 at 18:04

Not the answer you're looking for? Browse other questions tagged or ask your own question.