Q14-66P 聽How can the signals in the 6.5... [FREE SOLUTION]

Chapter 14: Q14-66P (page 677)

How can the signals in the 6.5 to 6.8 ppm region of their ¹H NMR spectra distinguish the following compounds?

Short Answer

Expert verified

The following compounds can be distinguished based on the number of signal and peak type.

Step by step solution

Proton spectrum reading

A single peak indicates the presence of a zero neighbor H atom. If the peak splits into two it indicates only one neighbor H is present and if the peak is split into three then shows two neighbors H are present and if four peaks then three neighbors H and so on.

If a molecule has an equivalent proton i.e. symmetry then they don鈥檛 give separate signals.

Table

The table given below has some values observed in the chemical shift for various proton:

Chemical shifts

Result interpretation:

The signals in the region 6.5-8.1 ppm for all three compounds are as follows:

From the table, it can be seen that the aryl proton is within the given range 6.5-8.1 ppm.
The below molecule will give 4 signals due to the presence of non-equivalent protons: two doublets as two carbon have only one neighboring hydrogen in the ring and two doublet of doublets, as a doublet of doublets is the result of splitting by two nonequivalent adjacent protons.

Structure of molecule A

The below-given molecule will give 4 signals due to the presence of non-equivalent protons: one singlet as one carbon without any neighboring hydrogen, two doublets as two carbon only have one neighboring hydrogen, and one doublet of doublets as is the result of splitting by two nonequivalent adjacent protons.

Structure of molecule B

Below given molecule has only two signals that is doublet because here molecule has symmetry and each carbon has only one neighboring hydrogen. The two hydrogen present in vicinity of each group are equivalent so will not give separate signals.

Structure of molecule C