Question

The following compounds all show a single line in their ${}^{\mathbf{1}}\mathbf{H}$NMR spectra. List them in order of expected increasing chemical shift:

${\mathbf{CH}}_{\mathbf{4}\mathbf{,}}{\mathbf{CH}}_{\mathbf{2}}{\mathbf{Cl}}_{\mathbf{2}}$,cyclohexane ,${\mathbf{CH}}_{\mathbf{3}}{\mathbf{COCH}}_{\mathbf{3}}\mathbf{,}{\mathbf{H}}_{\mathbf{2}}\mathbf{C}\mathbf{=}{\mathbf{H}}_{\mathbf{2}}\mathbf{C}$,benzene

Short answer

Increasing order is as follows:

${\mathrm{CH}}_4<\mathrm{cyclohexane}<{\mathrm{CH}}_3{\mathrm{COCH}}_3<{\mathrm{CH}}_2{\mathrm{Cl}}_2{\mathrm{H}}_2\mathrm{C}={\mathrm{H}}_2\mathrm{C}<\mathrm{benzene}$

Step-by-step solution

Step 1: Chemical shift

It is defined as the resonant frequency of the nucleus in relation to the standard magnetic field which is defined to be at 0 ppm (parts per million).

Causes for a chemical shift in NMR

1. Due to the electronegative atom the electron density is reduced which leads to the deshielding of electrons and it occurs when the net magnetic field is increased.

2.Anisotropy due to the pi-bonds that coupling the double bond and “n” hydrogens splits the signal by the n+1 rule.

Highest chemical shift

The molecule shows a deshielding effect because the direction of the induced magnetic field is the same as the direction of the applied magnetic field in that the net magnetic field is increased and here, the benzene shows a high chemical shift.

Lowest chemical shift

The molecule shows a shielding effect because the direction of the induced magnetic field is different from the direction of the applied magnetic field that the net magnetic field is decreased and here, the methane shows a low chemical shift.

Increasing order of chemical shift

Hence, as per the effect of shielding and deshielding the more reactive molecule due to de-shielding and the less reactive molecule due to shielding the order is as follow:

${\mathrm{CH}}_4<\mathrm{cyclohexane}<{\mathrm{CH}}_3{\mathrm{COCH}}_3<{\mathrm{CH}}_2{\mathrm{Cl}}_2{\mathrm{H}}_2\mathrm{C}={\mathrm{H}}_2\mathrm{C}<\mathrm{benzene}$