00:01
Solving problem 47 of chapter 15.
00:06
So we have some empirical formula.
00:08
We need to construct a possible structure for this molecule, having the protein and mr and one of the fingerprint in the ir.
00:21
So let's start from the first one.
00:23
We have c8, h9, br.
00:33
We know from the proton and mr.
00:36
We basically have a second.
00:42
So we have a triplet, a quadruplet and two signals at about seven.
00:51
Three plet, quadruplet.
00:54
So let's say between zero and three, chemical shift between zero and three.
01:06
We have a triplet and a quadruplet and then between seven and eight we have two different signals let's do one in here one in here so at about seven we know that the benzene ring proto's shift is there so let's start drawing a benzene since we have two different signals in this region it means that we have two different sets of protons experiencing different environments.
01:47
So we have six carbon in here since we have eight in total we need to add two more.
01:52
And from this region we know that we have two different signals.
01:56
We have a triplet and a quadruplet.
01:58
That means it corresponds to triplet bch2 and quadruplet ch3.
02:05
In fact, we always need to take the number of protons and add one in order to generate the nmr signal that we expect.
02:16
So c2, two protons plus one will give a triplet.
02:22
C3, three protons plus plus one will give a quadruplet.
02:27
So let's try to add this chain in here that would account for this part of the nmr.
02:33
And then let's put the bromine on this side.
02:36
So now we have eight carbons, nine hydrogen, one bromine.
02:41
And we have two different sets of hydrogen of the benzene ring, one and two.
02:47
That would account for the two different signals in this region.
02:53
Then the second molecule that we have is with formula c9h12.
03:08
H12 and we do know that we have in the region between 0 and 4 we have a triplet and singlet and a quadruplet.
03:21
So let's do region between 0 and 4 we have a triplet, a singlet and a quadruplet.
03:35
In the region between 7 and 8 instead this time we only have one signal that is a multiplat.
03:48
So again in the region between 7 and 8 is the characteristic shift of the the benzene protons.
03:57
So let's start drawing a benzene ring.
04:00
Now we have six carbon.
04:02
We need to add three more carbons and we know that in this region so where we should have the the shift of the protons of whatever alkaliic chain connecting to here.
04:16
We have three different signals...