00:01
We are answering, trying to find constitutional isomers of these three different compounds.
00:07
So starting with a, we have c3h8.
00:10
This is nice because it means we have no loss of degrees of freedom, so we don't have any double bonds, nor do we have any ring structures.
00:18
So we can draw out our nice three carbons.
00:22
We have carbon, one, two, three.
00:26
And if you count up your hydrogens here, you have three hydrogens here.
00:30
Here, three hydrogens here, and two hydrogens here.
00:34
So this makes up the chemical structure we need.
00:39
And luckily for us, there are actually no constitutional isomers of this because this is the only way i can rearrange that those atoms.
00:48
So a is done.
00:52
So now we can move over to b, where we have c3h6, meaning that we have at least one degree of freedom we've lost, which means we either have one ring or one double bond.
01:03
So the first thing is i can draw, since i have three carbons, i can draw my little three carbon ring here.
01:11
And if you want to count your carbons, you would have, or sorry, count your hydrogens, you would have two hydrogens here, two hydrogens here, and two hydrogens here for a grand total of six hydrogens.
01:22
So that's one structure.
01:24
Notice i mentioned that you could also have a double bond.
01:27
So if i draw my carbon chain like this, you can imagine one of these carbons as an alkine, or an alkan, alkyen rather...