Question

You will work as a group on this exercise. A sample of tert-butyl bromide (2-bromo-2-methylpropane) has been contaminated with some methyl bromide. The NMR spectrum of this sample shows two signals, one at 1.8 ppm and one at 2.2 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? Your objective as a group is to answer this nulection aunlaininn...

Answer 1

ANSWER :

The given information is

NMR signal in NMR spectra depends upon number of atoms of same type.

As in this case two types of ¹H present.

Each corresponds to each compound present in mixture.

• A) In tert-butyl bromide 'A' nine(9) Protons( ¹H ) present each have same environment and show only one signal.
• B) Similarly in methyl bromide 'B' Three proton present each have same environment and show one signal.

Now NMR based upon the corresponding amount of signals.
We can calculate their ratio using this formula.  

molar ratio of two compounds (M_A/B)=

(Integral_A/N_A)/ (Integral_B/N_B)

Where,

Integral_A - corrsponds to signal integral of A in spectra

N_A- Number of protons of A corresponds to signal from molecular formula.

Integral_B - corrsponds to signal integral of B in soectra

N_B - Number of protons of B .

Now here to solution.

M_A/B = (6/9) / (1/3)

=2/1

   Or 2:1

Moles of A and B is in 2:1

Now, mole percentage of each.

Let Total number of moles = 2+1 = 3

=> For A​​ Number of moles = 2

Mole percentage of A  = (2÷3) ×100 %

= 66.67%

So, Mole percentage of tert-butyl bromide = 66.67%

Now,

For B Number of moles = 1

Mole percentage = (1÷3) × 100%

= 33.33 %

So, Mole percentage of methyl bromide is 33.33%