Q2P Question: The \(^{\bf{1}}{\bf{H}... [FREE SOLUTION]

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Chapter 14: Q2P (page 527)

Question: The \(^{\bf{1}}{\bf{H}}\) NMR spectrum 1,2-dimethoxyethane \(\left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OC}}{{\bf{H}}_{\bf{2}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OC}}{{\bf{H}}_{\bf{3}}}} \right)\) recorded on a 300 MHz NMR spectrometer consists of signals at 1017 Hz and 1065 Hz downfield from TMS. (a) Calculate the chemical shift of each absorption. (b) At what frequency would each absorption occur if the spectrum were recorded on a 500 MHz NMR spectrometer?

Short Answer

Expert verified

a) The chemical shifts for signals at 1017 Hz and 1065 Hz are 3.39 ppm and 3.55 ppm, respectively.

(b) The absorbed frequencies for 3.39 ppm and 3.55 ppm are 1695 Hz and 1775 Hz, respectively

Step by step solution

Chemical shift formula

The formula for the chemical shift is given as follows:

\({\bf{\delta = }}\frac{{{\bf{Absorbed}}\;{\bf{frequency}}\left( {{\bf{Hz}}} \right)}}{{{\bf{Frequency}}\;{\bf{of}}\;{\bf{the}}\;{\bf{operator}}\left( {{\bf{MHz}}} \right)}}\)

Calculation of chemical shift in a

(a)The frequency of the NMR spectrometer is 300 MHz.

The signals are 1017 Hz and 1065 Hz. Substitute the values in the above formula.

\(\begin{array}{c}{\rm{Chemical}}\;{\rm{shift}} = \frac{{{\rm{1017}}\;{\rm{Hz}}}}{{{\rm{300}}\;{\rm{MHz}}}}\\{\rm{\delta }} = 3.39\;{\rm{ppm}}\\\\{\rm{Chemical}}\;{\rm{shift}} = \frac{{{\rm{1065}}\;{\rm{Hz}}}}{{{\rm{300}}\;{\rm{MHz}}}}\\{\rm{\delta }} = 3.55\;{\rm{ppm}}\end{array}\)

Thus, the chemical shifts for signals at 1017 Hz and 1065 Hz are 3.39 ppm and 3.55 ppm, respectively.

Calculation for b

(b) The chemical shifts forsignals at 1017 Hz and 1065 Hzare 3.39 ppm and 3.55 ppm, respectively. Thefrequency of the NMR spectrometer is 500 MHz.

The absorbed frequency for 3.39 ppm is calculated as follows:

\(\begin{array}{c}{\rm{\delta }} = \frac{{{\rm{Absorption}}\;{\rm{frequency}}}}{{{\rm{500}}\;{\rm{MHz}}}}\\3.39\;{\rm{ppm}} = \frac{{{\rm{Absorption}}\;{\rm{frequency}}}}{{{\rm{500}}\;{\rm{MHz}}}}\\{\rm{Absorption}}\;{\rm{frequency}} = 3.39\;{\rm{ppm}}\left( {{\rm{500}}\;{\rm{MHz}}} \right)\\ = 1695\;{\rm{Hz}}\end{array}\)

The absorbed frequency for 3.55 ppm is calculated as follows:

\(\begin{array}{c}{\rm{\delta }} = \frac{{{\rm{Absorption}}\;{\rm{frequency}}}}{{{\rm{500}}\;{\rm{MHz}}}}\\3.55\;{\rm{ppm}} = \frac{{{\rm{Absorption}}\;{\rm{frequency}}}}{{{\rm{500}}\;{\rm{MHz}}}}\\{\rm{Absorption}}\;{\rm{frequency}} = 3.55\;{\rm{ppm}}\left( {{\rm{500}}\;{\rm{MHz}}} \right)\\ = 1775\;{\rm{Hz}}\end{array}\)

Thus, the absorbed frequencies for 3.39 ppm and 3.55 ppm are 1695 Hz and 1775 Hz, respectively.