00:01
Okay, so for this problem, you want to give a systematic name to all of these alkenes.
00:08
So let's start with problem a.
00:10
So the first thing you want to do is you want to identify the longest chain that includes your principal functional group, which in this case, your principal functional group is going to be our alkenes.
00:25
So if we start counting, we find that we have one, two, three, four, five.
00:32
If we start numbering with this carbon, we still get five.
00:35
We get one, two, three, four, five.
00:39
So we have a five carbon chain.
00:42
So this tells us the root of our word, we need to use pent.
00:47
The second thing we want to do is give this double bond the lowest possible number.
00:55
So if we start numbering from the left, we would get one, two, three, four, five.
01:03
If we started numbering from the right, we would get one, two, two, three, four, five.
01:10
So looking at this, if we start numbering from the right to the left, we give a lower number to our double bond.
01:18
So our double bond is going to be on carbon two.
01:24
Lastly, we want to look for any substituent.
01:28
In this case, we have a methyl group, and this is going to be on carbon four.
01:33
So we have four methyl.
01:36
Okay, so if we put all of this together, we start with our substituent.
01:45
So you would have four, methyl, two, pentene.
01:55
Alternatively, you could write this, methyl to in.
02:06
So this is a little bit of a newer nomenclature.
02:08
I think your book uses two pentine, but both of these are correct in terms of this answer.
02:20
Let's look at the second one.
02:23
So for this one, if we see our double bond here, we count our longest possible chain.
02:29
We have one, two, three, four, five, six.
02:33
So we have six carbons.
02:38
So we need to use hex.
02:43
Now for this one, we're actually kind of symmetric.
02:47
Right? so no matter if we number from the left or the right, our double bond is going to be on carbon three.
02:55
So now what we have to pay attention to is our substituents because that's going to tell us how to number this.
03:02
So we want our substituents to have the lowest possible number.
03:06
So if we start numbering from the left, we would have one, two, three, four, five, six, versus if we start at numbering, sorry, if we number from right to left in red or left to right in green.
03:24
Right? so we would have a methyl on three, a methyl on four, and a coral on five.
03:30
Versus in red, if we start numbering from the right, we have a chloro on two, a methyl on three, and a methyl on four.
03:37
So we want to use this nomenclature, so we have two chloro, and then we have a three, four, dye, methyl.
03:52
Okay, and so if we put everything together, we have our different substituents.
03:57
We want to make sure they're alphabetized, so chloro comes before methyl.
04:00
So we have two, chloro, three, four, dimethyl.
04:10
And i'll keep the nomenclature that your book uses.
04:14
Three, hexene.
04:25
So part c is a cyclic structure.
04:30
By convention, you want this double bond to be between carbon one and carbon two.
04:38
And because you have this bromine substituent on this carbon here, this is going to be carbon 1 and this is going to be carbon 2.
04:52
Because you also want your substituent to have the lowest possible number...