00:01
Okay, we have got an nmr spectrum for synamic acid.
00:05
What does synamic acid look like? and what we can do is assign each of these signals to a proton in synamic acid.
00:25
So what may be good is to draw in protons here.
00:40
And if possible, let us group them, right? because we have some symmetry here, which means that these two proteins, will share a signal, and these two protons will share a signal, right? because they are chemically equivalent, which is also represented by the fact that we could rotate freely about that bond, meaning that these hydrogens aren't spatially unique.
01:07
They could easily take each other's places by rotating about that bond.
01:12
Okay.
01:13
So let's just go ahead and give these letters.
01:16
Let's call this ha.
01:17
It's called that's b, c, and then both of these will be d.
01:21
This will be e.
01:24
This will also be e and this will be f.
01:26
Okay.
01:27
So we see right away that we have six protons, which is good because if we look at our spectrum, we see that the total integration value of each of the signals, the total of the integrations, is six, which is what we would expect for synamic acid.
01:46
Having, sorry, the total integration value is not six.
01:54
It's, sorry, right, we have six signals, which corresponds to having six protons, and then the total integration value should be one, two, three, four, five, six, seven, eight.
02:08
And we do find that we have signals going from one plus one is two, plus two is four, plus three is seven, plus one is eight.
02:17
Right, the total integration values, when we look at the values, for example, if we look at the right most peak, we see they have this one point zero, zero.
02:32
That means that this corresponds to a single proton.
02:36
And so the signal that is a little messier that's in between 7 .4 and 7 .5, they say 3 .00.
02:45
That means that there's a total of three protons located in this signal, right? and the sum of all of these totals up to 9.
02:54
Okay.
02:56
So let's continue onward.
02:59
Okay, the first thing that we can do is we can assign this proton.
03:02
Right to the leftmost peak at 12 .19.
03:10
And that's on the basis of chemical shifts for hydrogens and carboxylic acids are in the range of 12 and higher.
03:18
Okay, so that's how we get this one assigned.
03:24
The next one that we could do would be something like hb.
03:31
How can we do that? well, hb will notice is not aromatic.
03:34
It's not on our benzene ring.
03:38
And while it is next to a carbonyl and participating in a double bond, it's chemical shift.
03:43
We should expect to be the least deshielded compared to all of the phenyl or benzal locations on the ring.
03:54
So based on chemical shift, we should expect this to be the peak at 6 .5.
03:59
We'll also notice that the peak at 6 .5 is a doublet, right? which we would expect given that hb has a single neighbor, right? hc.
04:10
Hb neighbors hc, so hb and hc split each other, right? we also may notice that this splitting is the j value is 6 .49 minus 6 .47, right? so that's 0 .02.
04:37
And that's in ppm.
04:38
If they told us what this is a 600 megahertz machine.
04:45
So we could, in fact, say that the value in hertz is simply going to be the splitting constant in ppm multiplied by the 600 megahertz that the machine is operating at, which is meaning that the j value in hertz is 12.
05:03
And a j value of 12 is commensurate with two protons, which are trans across the double bond.
05:10
These splitting values are usually in the range of 12 hertz.
05:14
Okay.
05:15
So that's another evidence that the peak at 6 .5 is proton b...