Let's discuss the first piece of information that we can derive from a proton NMR and that's
the total number of signals.
On a typical proton NMR, there's going to be as many signals on the spectrum as there
are unique non-equivalent types of protons.
For that, we need to understand what's an equivalent or non-equivalent proton.
Well, an equivalent proton is going to be a proton that has the same perspective on
the molecule as another proton.
If two protons are in pretty much the same place on a molecule, like for example, let's
say the three protons that are attached to a methyl group.
A methyl group usually has three H's on it, so all three of those would be said to have
the same position on the molecule, so all three of them would be what we call equivalent.
For right now, how are we going to be able to determine if something's equivalent or
non-equivalent?
We're just going to go with a really easy rule which is that let's go ahead and assume
that hydrogen is bound to the same atom or equivalent.
So like I just said, the three hydrogens on a methyl group or on a carbon would be equivalent,
so that would apply to other atoms as well, not just carbon.
In general, a rule that we can go by is that any type of symmetry is going to reduce the
total number of signals.
This is because if you have any planes of symmetry then you're by definition going to
have some protons that are the same as other protons on the other side of the molecule.
Symmetry is something you have to watch for when we're using this type of information.
What we're going to do is I'm going to go ahead and do practice problem (a) as a worked
example and then I'll save the other three for you guys to do on your own.
Let's just go ahead and read this question.
It says how many different types of protons or signals are there on each molecule.
Let's look at (a).
What we notice is that (a) has obviously a bunch of hydrogens on it that aren't drawn,
but it has four different atoms.
What I'm wondering is how many different signals do you think this is going to have.
Now notice three of them are carbon, one of them is oxygen.
Is there any plane of symmetry, etcetera.
Those are the things we need to be thinking about.
Well, I'm just going to tell you right now.
The answer is that there are going to be four different types of hydrogens here.
Let's see why.
The reason is because we could just start counting from the oxygen.
Let's say that the oxygen is going to be letter A so the oxygen has a hydrogen that's attached
to it.
And there's no other hydrogen like that, so for sure that's one type of hydrogen.
No other hydrogens on that molecule look like that one.
Now we have all the rest of these hydrogens.
You might have through that we could group them all together since they're all on carbon,
so maybe you're thinking you have one type of hydrogen with the O and the second type
is on the carbons.
But it turns out that no, there are actually more separated than that because, for example,
the hydrogens that are attached to this carbon are closer to the oxygen than the hydrogens
attached to this carbon.
That means that theoretically, the red hydrogens, the two hydrogens that are on this red carbon,
are going to be a little bit more deshielded than the blue ones because they're going to
be closer to something electronegative.
So I would actually expect that the red ones would be a little bit more downfield because
remember those words downfield and up-field.
Anyway, because of the fact those two red hydrogens are attached to the same atom, we're
going to say that's the second type.
The two hydrogens attached to this atom are the third type and then finally the three
hydrogens on this last carbon are the fourth.
So in total, we get one, two, three, four different types of protons.
So not so bad.
Another thing to note is you might have been thinking maybe there was symmetry here, but
really this molecule isn't symmetrical.
The way that it's drawn, the oxygen is on one side and then you've got this asymmetry
that goes through the whole molecule so that's why every single atom needed its own peak
or its own signal.
Now go ahead with that knowledge, try to do question (b) and then I'll go ahead and solve
it.