00:01
Okay, so let's start with this.
00:03
Let's break it down so it's a little bit more easier to kind of visualize.
00:08
Pentine, five carbons per pent.
00:12
E and e would mean that there is a double bond somewhere in the compound.
00:18
Cyclo, which means that these five carbons are all connected to each other.
00:25
Now here, 3 .3 dimethyl.
00:29
So that means in the third position of our cyclopens, so we're two methyl substituents.
00:38
So if we draw this out and we start with pentine, cyclopentine, we have something like this.
00:47
And now we know there is a double bond somewhere, so let's put it in there.
00:51
There you go.
00:52
And double bonds takes precedence over metal substituents.
00:57
So when numbering our carbons, this will be number one, this will be number two, and the third position would be here.
01:05
Here.
01:07
So that's where our two metals will be.
01:11
So the compound, try a little better, something like this.
01:21
Okay, so let's do the same thing would be and break it down.
01:26
Hexene, that is six, a double bond somewhere.
01:31
And from this two right here, we know that this double bond is going to be in the second position.
01:38
Die metal.
01:40
In the second and in the third position there is a metal substituent.
01:45
And in the sixth position there is a bromo.
01:48
So there's a bromine substituent at the very at the sixth position of this compound.
01:54
So let's start with the fact that it's a hexene and let's draw six carbon chain.
01:59
So one, two, three, four, five, six.
02:03
And we know in the second position there is a double bond.
02:07
So right here.
02:08
And and there is a 2 -3 die metal.
02:11
So we know that this is the second position because our double bond is found in the second position and it takes precedence over the metal groups.
02:21
So 2, 3.
02:24
And that would mean that the 6th position, which is right here, will be our bromo...