00:01
We're asked to find which of these species is more stable.
00:03
And when considering stability, we need to think about the possible resident structures that each of these species can have, which can lend to electron delocalization, therefore lower energy.
00:15
So we're going to start by doing some arrow pushing to figure out which can have residence species and which resident species are more stable and greater contributors.
00:25
So with this one, we can do some arrow pushing here.
00:31
We've got a single arrow.
00:33
So we can move the double bond and the radical.
00:37
And we can form this resonance structure here, where we now have the radical on the secondary carbon, and we've moved the double bond.
00:54
And we can do the same thing with this, where we have single arrows, here but in this case we form a molecule with a primary radical instead of a secondary radical and remember that secondary carbons are more stable because of the hyper conjugation of the surrounding carbon hydrogen sigma bonds so actually this species here is more state this is a more resonant structure.
01:34
I'll just kind of indicate that these are resonant structures.
01:37
This is a more stable resonance structure here because we have the radical on the second carbon instead of the primary carbon here.
01:45
We can see that there's two carbons attached to it, whereas here we only have one carbon attached to that.
01:53
So that's a primary carbon.
01:55
Okay.
01:56
So moving on, what about b? so b, we can form actually a we can form an enolate, excuse me.
02:09
In both of these we can actually form the enolate.
02:12
And that, oops, oops, it's a little laggy here, sorry about that.
02:22
In this case, we have the negative charge on the oxygen, and we form a carbocan.
02:42
Whoops, so now.
02:45
So now, now, now we have the negative charge on, i forgot to draw that arrow there.
02:52
Okay, i think we're good.
02:53
I think this is the enolates.
02:56
And we have a positive charge on this position here...