Question

The 1H NMR spectrum of chloroethane (CH3CH2Cl) recorded on a 500 MHz NMR spectrometer consists of signals at chemical shifts of 1.48 ppm and 3.57 ppm. Calculate the frequency, downfield from TMS, of each absorption. Be sure to answer all parts.

Answer 1

Answer:The frequency of absorption for the proton having chemical shift 1.48 ppm is 740 Hz downfield from TMS.

The frequency of absorption for the proton having chemical shift 3.57 ppm is 1785 Hz downfield from TMS.

Explanation:

We are given with the following data:

Frequency of the instrument(NMR spectrometer)=500MHz=500×10⁶Hz

Chemical shift (δ) value  for 1st proton=1.48PPM=1.48×10⁻⁶

Chemical shift (δ) value  for 2nd proton=3.57PPM=3.57×10⁻⁶

We know that frequency of reference that is of TMS(Tetramethylsilane) is assumed to be 0.

We have to calculate the frequency of absorption of each protons downfield from TMS.

The formula for the chemical shift (δ) is:

δ=[Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)

So using the above formula we can calculate the frequency of absorption for the two protons whose δ value is given.

1. For the proton having δ value 1.48ppm:

1.48= [Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)

1.48×10⁻⁶=[Frequency of sample(νₐ)-0]÷[500×10⁶Hz]

1.48×10⁻⁶×500×10⁶Hz=[Frequency of sample(νₐ)]

740Hz=[Frequency of sample(νₐ)]

2. For the proton having δ value 3.57ppm:

3.57= [Frequency of sample(νₐ)-Frequency of TMS(νₓ)]÷Frequency of instrument MHz(Mega Hertz)

3.57×10⁻⁶=[Frequency of sample(νₐ)-0]÷[500×10⁶Hz]

3.57×10⁻⁶×500×10⁶Hz=[Frequency of sample(νₐ)]

1785Hz=[Frequency of sample(νₐ)]

So the frequency of absorption for the proton having  δ value 1.48ppm is 740 Hz and for the proton having  δ value 3.57 ppm is 1785Hz.