00:01
Solving problem 2 of chapter 15.
00:08
Give upac names for the following compounds.
00:15
So let's start with the first structure.
00:33
So we have a benzene ring with the chlorine substituent and the bromine substitute.
00:40
Now we can start numbering the carbon of the benzene ring, starting from here.
00:47
1, 2, 3, 4, 5 and 6.
00:56
So we start numbering from here, and then we go in this direction in order to give to the second substituent the lower number possible.
01:06
So instead of having 1, 2, 3, 4, 5, we go 1, 2, 3.
01:13
Then we start naming the 2 substituent and finally the principal group that is the benzene.
01:21
So in position 1 we have bromo.
01:27
So it's going to be 1 bromon.
01:32
In position 3 we have a chlorine.
01:35
So it's going to be 3 chloro.
01:41
And then finally we have the principal group that is the benzene group.
01:46
So 3 chloro benzene.
01:53
The second one was this one.
02:06
Sorry.
02:12
Again and we have ch 2 ch 2 ch 2 ch 3 c h 3 ok so in this case we basically only have one substituent it is a long chain so composed by more than one group by more than one carbon atom but it is only one single substituent because the same long chains is connected to the same to only one carbon.
03:14
So we only have to name this chain and then finally the benzene.
03:20
Let's count how many carbon atom we have in this alkyne chain.
03:25
One, two, three, four, five in total.
03:29
So the name when we have five carbon is pentane.
03:33
Now we have a linear chain except that for this methyl group connected in here.
03:42
So in this case, the chain is going to be named isopentene.
03:48
Since it is a substitute on the principal group that is the benzene ring, so this chain is going to be called isopentil.
03:58
So we have isopentil, benzene.
04:08
Benzene.
04:13
In this case, we don't need to put a number in here because again, we only have one substituent, so we don't need to.
04:21
Okay.
04:22
The second molecule, the third, sorry, molecule is this one.
04:31
We have an nh2 and a bromine in here.
04:40
So here we have two different substituents...