Question

Given the 1H and 13C data below, which isomers of C8H11N fit the spectral data? 13C NMR: δ 17.2, 20.4, 115.1, 122.4, 137.3, 127.6, 131.1, 142.1 ppm

Answer 1

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 $NMR: 17.2,20.4,115.1,122.4,137.3,127.6,131.1,142.1$ 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 $sp^{3}$ 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 $sp^{2}$ carbon atoms.

Thus , the compound with two different $sp^{3}$  carbon atoms and six different $sp^{2}$ 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 $NH_{2}$ protons.

6H gives two signals at 2.2 ppm due to the presence of the two $CH_{3}$  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