Question

please explain and label the signals of ethylbenzene c6h14o proton nmr and IR. thank you..

Answer 1

Hi, I cannot see any spectrum uploaded from you in this question. To explain this I am using a simulated spectrum for NMR. I have generated this spectrum from ChemSketch. Here it is:

The CH3 group at the terminal gives a triplet since it is next to a CH2 group. The triplet appears at 1.18 ppm.

Since the CH2 group is next to a CH3 it gives a quartet and the quartet appears at 2.72. This signal is downfield since the CH2 group is next to the aromatic ring.

The aromatic ring protons all appear as a multiplet at 7.19-7.23. Aromatic protons usually appear at ppm values around 6.5 to 7.5. Now come to the IR.

I took the IR spectrum from the NIST website which contains the data for many chemicals.

Since the molecule does not have any functional group, the IR spectrum is not of much help. All that we can interpret is the following:

A monosubstituted ring pattern (1667-2000) which is very decisive and the aromatic C-H stretch which occurs above 3000.