00:01
Okay, so now you have to take the substitution effect information that you just got from reading the last few sections and apply that system chemistry.
00:10
So the first question here is talking about the nitration of bromobenzene and it appears that i forgot the bromine on the ring.
00:18
Okay, so identify the group.
00:20
In this case, we have a halogen, right? so halogens are slightly deactivating, but they are orthoparad directors.
00:27
So just based off the logic here, you're going to get a mixture of the ortho and the para product to nitration.
00:37
Now, remember that the group you're adding doesn't have any effect on the electronics.
00:44
It's the groups that are already on the ring.
00:48
Sorry, that's n .o2, n03.
00:50
I'm thinking about sulfination here.
00:55
So in this case, you know, bromine is pretty big.
00:59
So you might actually favor the power addition here just ever so slightly between the two, just because bromide is a big atom and hysterically speaking.
01:09
It might play a role while the power position is completely unhindered.
01:13
But you'll probably still get some degree of a mixture, but it won't be 50 -50, for example.
01:18
Okay, so now it's bromination of nitro benzene.
01:22
So nitro is one of the more deactivating.
01:25
It's one of the most deactivating groups that you will learn about.
01:29
So it's also meta -directing, right? so it's pretty classical in that sense...