00:02
Okay, we have the formula c2h6o2 and we need to draw all constitutional isomers with this molecular formula that contain an oxygen -oxygen double bond.
00:23
Now the one thing that i'll point out is that from our formula of c2h6 we can derive that there are no double bonds and no rings.
00:34
How do i know this? because of the double bond equivalency formula in which you take the number of carbons, which is 2, you subtract the number of hydrogens, divided by 2, in this case 6 divided by 2, and then you add 1.
00:52
And the number you get out, in this case 0, tells you the number of double bonds or rings.
01:01
Since our double bond equivalence is 0, or it's also called degree of unsaturation, is 0, we know that we have a purely linear molecule, no double bonds, no rings.
01:14
Okay, so now if we've got 2 carbons and 6 hydrogens to deal with, we know that at some point we're going to have to have a terminal carbon.
01:30
For example, if we start with 2 oxygens next to each other, because our options here are to put a carbon on this side or a carbon on this side.
01:42
So if i put a carbon on the right hand side, it has to have a total of 4 bonds, which means i can either then put a hydrogen here, at which point i would have to continue outward this way with the other carbon, or i can put a bond to carbon here, at which point i'm out of carbons, and so i'm going to fill in the 4 bonds to this carbon with hydrogens.
02:13
And at this point i have 1 hydrogen left, which is good because all of our oxygens should have 2 bonds in the end, so this is one possibility.
02:22
Let's do that again, but now this time we'll do our 2 oxygens that are bonded together.
02:27
We'll put a carbon here like we did in the first case, but instead of putting a second carbon, we'll put a hydrogen to the right, which then forces us to put another carbon over here, and cap it off with hydrogens.
02:46
Ok, so are there any more isomers? well, let's start with our 2 oxygens bound together.
02:53
If i put a carbon on the left hand side to start with, well, in fact that's equivalent to or no different than putting a carbon on the right hand side.
03:09
These are equivalent structures.
03:11
I simply flop one over 180 degrees and it would look exactly the same.
03:16
So that is not a viable option, meaning at this point my only option would be to put a carbon on the right hand side, and then what we'll notice is that we've already exhausted all the options for a carbon on the right hand side.
03:29
I can either do a carbon with 2 hydrogens and a capping methyl group, ch3, or put a capping methyl group on either side.
03:36
So at this point we are done with isomers that contain a peroxy or a pair of oxygens bound together.
03:45
So we can move on to the second half of the question, in which we take the same molecular formula but consider structures when the oxygens are not bound together.
03:55
Now if the oxygens are not bound together, recall that because hydrogens are always terminal, then our only option is to place a carbon between the oxygens...