00:01
In this problem, we need to get the iac names for all of these molecules.
00:05
So starting with a, the first thing you need to do is identify the longest carbon chain.
00:09
So our longest carbon chain here is one, two, three, four carbons long.
00:18
So the root is going to be butane.
00:21
At the one position, we have a fluorine atom.
00:23
So we have one fluoros.
00:31
And at the three position, we have two methyl groups.
00:35
So we have a 3 -3 -dymethyl group and our root is butane.
00:53
And b, once again, find the longest carbon chain.
00:58
So starting here, it's one, two, three, four, five, six carbons long.
01:06
So our root this time is hexane.
01:08
At the one position, we have an iodine.
01:12
At the two position, we have a methyl group.
01:15
And at the three position, we have an ethyl group.
01:16
So of these substituents, you put them in alphabetical order.
01:21
So our first one is three ethyl, and then one iota, and then two methyl.
01:46
See, find the longest carbon chain.
01:50
This time it's three carbons long.
01:53
So our root is propane.
01:55
At the one position, we have a bromine substituent, and at the two, two position, we have two methyl groups.
02:02
So once again, you put the substituents in alphabetical order.
02:05
So we start with one bromo and then 2 -2 dimethyl propane.
02:31
On to d, sorry...