00:02
So for problem 4 .3, give the structure of each of the following two methyl propene.
00:10
We'll just start with that one.
00:12
For a, i have it drawn out all the answers here already.
00:18
So if you need a quick answer for the structures, there you go.
00:25
I'll draw it out again just to give you an example.
00:44
So 2 methyl propine, you can see that the ending in e &e is what gives it the double bond.
00:54
You have your 3 right here, 1, 1, 2, 3 with the e -en, e, and e, right there, and a methyl on the second carbon.
01:28
For b, we have four methyl -13 hexidine.
01:35
So you're going to take the hexa to be six.
01:45
One, two, three, four, five, six.
01:49
So a six -carbon backbone with adine, which is two double bonds on the one -three, so one, three, and a methyl on the fourth.
02:22
C is isopropyl penel cyclopentine.
02:30
So you can break this down into a cyclo -pent.
02:41
That would be your cyclopent.
02:47
And you have an en on the end.
02:50
So that would become a double.
02:53
Bond isoprop means you have a little y type of carbon backbone uh it'll be iso so separated and three one two and that'll be on the number one carbon carbon so i so isopropenial cyclopensine is all the substituents and double bonds are coming off of the first carbon.
03:56
So you can enable this one as number one, two, three, four, five.
04:07
And then you have your isopropyl, which you could consider this one to be one, two, three...