00:01
All right.
00:02
We've got the molecular formula c4h802.
00:06
First thing i ever do with these spectroscopy problems is i calculate double bond equivalencies.
00:11
The formula for double bond equivalencies is the number of carbons, which is four, minus the number of hydrogens divided by two, minus the number of halogens divided by two, plus the number of nitrogens divided by two, plus one.
00:28
In this case, we get 4 minus 4, minus 0, plus 0, plus 1, or 1, d, be.
00:36
One double bond equivalency tells us that we either have a cyclic structure of some kind, which is unlikely with only four carbons and two oxygens, or that we have a double bond, which, if you ask me, is rather likely when you've got oxygens around, and sort of suggests we maybe have a carbonyl.
00:53
We don't know yet, but that's a possibility.
00:55
The next information that is very useful is that our ir spectrum has a strong absorption peak near 1720 wave numbers.
01:05
Well, 1720 is right about where carbonyls show up.
01:08
So, in fact, i'm almost certain we have a carbonyl.
01:12
More than certain, we do have a carbonyl.
01:15
1720 is kind of vague.
01:17
We don't know if it's carboxylic acid, aldehyde, or ketone.
01:21
1720 is kind of right in there.
01:23
It could be an ester as well.
01:24
That's another possibility.
01:26
Okay, the next thing that i'm going to look at is our fact that when dissolved in d2o, the compound gives the same spectrum, with the exception that the signal at 3 .75 has disappeared.
01:40
Now, this is what happens when you have exchangeable protons.
01:44
If you mix a compound with exchangeable protons in d2o, since deuterium isn't nmr active, if you have o8, is and they are in equilibrium with d2o, they can easily become deuterated and not give a signal.
02:06
And if your concentration of your product is low enough and you've got enough d2o, this will tend to dominate and you won't see a signal for your protons because all of it will have changed into deuterium or such a large amount will have that the signal for the protonated form will be too weak to see at all.
02:25
So we know that we have an oh somewhere.
02:29
We have got a double bond oxygen somewhere.
02:32
We've got an oh somewhere.
02:36
And now, once i've established that, i think it's appropriate to start looking at the information from the nmr.
02:44
So off to the side here, i will note there's some missing information from the nmr.
02:50
It doesn't tell us exactly how many protons belong to each signal.
02:55
For all of them, just some of them.
02:57
But either way.
02:58
So at sigma equals 1 .35, we've got a doublet.
03:05
At sigma equals this chemical shift at 2 .15.
03:10
We've got a singlet.
03:11
At sigma equals 3 .75.
03:16
We've got a singlet.
03:18
Remember, this is the one that disappeared.
03:20
So this is our oh.
03:21
It's also a broad singlet, which is another sign that it's likely in oh.
03:26
That's due to hydrogen bonding that makes the signal more broad.
03:31
And we've got a quartet at sigma equals 4 .25 is quartet.
03:43
And we do know that this is one hydrogen.
03:46
And we do know that this is one hydrogen.
03:49
Okay.
03:51
So right off the bat, i'm looking at the fact that we have a quartet and a doublet.
03:59
The only way that we can get a quartet is if there's a proton which neighbors three other protons.
04:07
The only way we can get a doublet is if there's a proton that neighbors a proton.
04:12
So i think, or rather i know, we have something that looks like this in our molecule.
04:19
Let me finish writing out these other pieces that i know to be there.
04:31
We can glue these together later.
04:39
We've got a hydrogen, neighboring three hydrogens.
04:45
We also know that there's something that's not.
04:48
This is saying it's like it can be attached to anything, but it's not a hydrogen...