00:01
Hello, students.
00:02
So in this question, you are given with the molecular formula.
00:06
This is the molecular formula that is given c7, h7no, and you have to determine the structure of compound.
00:13
Okay.
00:13
So now we are given with the data of the spectro.
00:17
This is the ir data that is given.
00:19
So before that you should find out, before that you should find out the degree of unsaturation.
00:28
Okay.
00:29
So what is the formula for degree of unsaturation? writing it over here degree of and saturation so what we do is we add it into two carbon minus two high minus hydrogen plus nitrogen so we have to add the number of these and divided by two all right so how many numbers of carbons are present over here there are seven carbons so we are multiplying 2 multiply with 7 minus 2 sorry this was plus okay so this was plus and here also plus 2 and then we have the hydrogens we will minus the sign number of hydrogen that are 7 so we will add the number of nitrogen how many nitrogen is there 1 so it would be 1 only and divided by 2 so this comes out to be 5 means we have 5 either one double either 1 ring structure or we have four double bonds.
01:39
Okay, means we have a kind of five, have five degree of this degree of unsaturation in our compound.
01:50
So what kind of that is, we will get it solved but through our data that is given in the graph.
01:57
So if you see this ir graph, so while going to the ir graph, you can see this, the peak coming out, this peak is of us.
02:07
1690 okay so if we see the i for the ii graph for i r graph i r spectrograph right we get a peak of 1690 this is a very good peak in 9 .0 centumptan inverse this comes for the c double bond o group okay so this can be this can be our ketone this can be our ketone carboxlic ketone carboglic conjurc conjugated conjugated to aromatic ring.
02:51
Aromatic ring.
02:56
So this is the conclusion that we had finder from the ir spectrum.
03:00
Now moving on to the next spectro that is c13 and h hydrogen spectra.
03:05
For the c13, i'm writing the data's over here.
03:09
For c13 nmr, what is given over here? if you see the c13, you can find out that there's a peak i'll write the peaks only peaks number only okay so you can match that from the data so we have a kind of two informations first one is decoupled if we see the decoupled that is given in the this side okay this side of the prolep the side of the graph so if you see this graph you have the peak at 28 you have it 1 2 4 you have it 1 3 2 you have it 1 3 2 you have it 1 sorry it is 1 137 and then we have it as 150 and then 154 and 194 3p3.
04:03
Alright so here with the data indicates the compound so this data indicates that the compound has seven steps seven sets of non equivalent non equic valent carbon atom okay so we have different types of carbon atom that is present in the seven steps seven sets sorry okay so we have atoms so now if we see the this this one part okay so this was our a and this is a b part so if we see the b part of the graph right so we can say at 28 it is in the upside so up upside we can see say it has the compound of ch3 group.
05:04
Next we have 1, 2, 4, it is also in the upside.
05:08
It has the compound which says, which says the group double bond c .h.
05:16
Okay.
05:17
So next we have 1, 3, 7.
05:21
Okay.
05:22
It is also in the upside.
05:23
So it has the compound, it has the group of double bond c, c, c, h.
05:32
Again, same.
05:33
Then again we have 1 5 -0 upside only then we have this same again this group and then we have 1 5 4 upside which says this is also same compound so we have 1 2 3 and 4 kind of same group and the different amount okay so so here we have next we can say next we can go to the nmr structure, proton nmr, this structure.
06:08
Okay, so in this structure we can get a major information.
06:12
See, in this structure, what we can see is we have these different peaks.
06:16
Okay, these are the different peaks which we are getting between 6 and 7, right, 6 .7 something over here.
06:22
This is between 8 .2 you can say and this is 8 .7 or 8 .8 .8 and this is 9 .1 and dot.
06:28
So we can consider the functional group from these points.
06:32
So if you see, if we are dealing with proton nmr, right? so in proton nmr, we have a kind of, we have a kind of signal over here.
06:47
So first time i'm talking about this one.
06:48
So it is occurring at 2 .5 and it is singlet in nature and this corresponds to three hydrojures.
06:57
So this corresponds to what group? this corresponds to the group, which is c double point o.
07:03
And it is attached to ch3...