00:01
Okay, so we want to give the products for the following reactions.
00:02
And the first one, we have trans -2butene, reacting with hbr and peroxide.
00:12
So we can draw like this or just like this.
00:27
So hbr is a strong acid, and it's going to act as a proton donor.
00:32
And if you react to the peroxide, it's going to add in anti -mercropony competition, where the berymine is going to be on the less stable carbon.
00:40
But in this case, they're both equally stable.
00:48
So we can really add it to either one, but it's going to be the same product regardless.
00:51
Count from right to left, it's on carbon 2.
00:57
And if you add it to this one, counting from left to right, it's on carbon 2 as well.
01:08
Next one is with 3 methyl to pentine, and it's the z confirmation.
01:28
So z meaning the same side that it'll bond, if we draw a liner here, the highest priority groups in the same side that it will bond.
01:38
We have a methyl on carbon 3, and serrathing with hbr.
01:48
So again, hbr is strong acid.
01:51
It's going to act as proton donor, and it's going to add to the more stable carbon, which is the one with the more r groups.
02:01
So we're going to put the hydrogen on the less stable carbon, so we can have a more stable carboketion.
02:12
We're going to get this intermediate, and birmingham's going to attack it.
02:53
Next one is z3 methyl to pentene, this time with hbr and peroxide.
03:00
So the only difference here is that when we're reacting with peroxide, it's going to add to the less.
03:05
Less stable carbon and as a result it's going to add to this one instead.
03:10
It is less r groups.
03:28
That's product for c...