Question

A sample of tert-butyl bromide (2-bromo-2-methyl propane) has been contaminated with some methyl bromide. The NMR spectrum of this sample shows two signals, one at 2.2 ppm and one at 1.8 ppm, corresponding to these two compounds respectively. The integrals of the two signals equate to 6:1 respectively. What is the mole percent of each compound in the mixture, or phrasing it another way, what percentage of the mixture is tert-butyl bromide and what percentage is methyl bromide?

Answer 1

Answer:

In tert-butyl bromide there are 9 equivalent protons. ( 3 from each methyl group) peaks at 2.2 ppm

In methyl bromide, there are 3 equivalent protons.     have peaks at 1.8 ppm

The signal ratio of 6:1 is only possible if the ration of tert-butyl bromide to methyl bromide is 2:1.

So that for every 2 x 9 = 18 proton signal of teert butyl bromide at 2.2 ppm, there are 3 signals of methyl bromide at 1.8 ppm