Complete Question
The complete question is shown on the first , second , third and fourth
Answer:
The compound that fit the spectra data are compound B, E and H as shown in the question
Explanation:
From the question we are given that the 13C NMR data is
13C [tex]NMR: 17.2,20.4,115.1,122.4,137.3,127.6,131.1,142.1[/tex] pmm
From the spectral diagram on the question we can see that the signal at 17.2 ppm and 20.4 ppm indicates the presence of two different [tex]sp^{3}[/tex] carbon atoms.
Also the signals at 115.1 , 122.4 , 137.3, 127.6,131.1,142.1 ppm indicates the molecule with six different [tex]sp^{2}[/tex] carbon atoms.
Thus , the compound with two different [tex]sp^{3}[/tex] carbon atoms and six different [tex]sp^{2}[/tex] carbon atoms are shown on the third uploaded image
Now let us take a look at the 1H NMR as follows
For NMR spectra the integration values are in the ration 1.99:1.00:2.19: 5.93
Now we can look at this above ratio like this 2:1:2:6 by reduction
Hence the relative number of proton is in the ratio of 2H: 1H: 2H: 6H
2H at 6-7 ppm and 1H at 6-7 ppm are due to the presence of three aromatic proton which are present on the benzene ring.
2H at 3.5 ppm are due to proton in amino group [tex]NH_{2}[/tex] protons.
6H gives two signals at 2.2 ppm due to the presence of the two [tex]CH_{3}[/tex] protons
The compound A is not matched with 1 H NMR spectra, thus remove the compound A .The other compounds are B,E,F
Thus ,the compound with the proton in the ratio 2H : 1H :2H : 6H are shown on the fourth uploaded image
Hence the compounds with the given NMR spectra are B,E and H
