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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 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 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 carbon atoms.
Thus , the compound with two different carbon atoms and six different
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 protons.
6H gives two signals at 2.2 ppm due to the presence of the two 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
