Question: How many peaks are present in the NMR signal of each labelled proton?

a.

b.

c.

d.

NMR involves the interaction of radiowaves with magnetic nuclei. Only those nuclei which have non-zero values of spin exhibit magnetic property. The spin quantum number of nuclei is represented by I, which has \({\rm{2I + 1}}\) degenerate magnetic levels.

If \({\rm{n}}\)non-equivalent protons are present in the neighborhood of a given set of protons, then \(\left( {{\rm{n + 1}}} \right)\)peaks will arise in the NMR spectrum.

If two sets of non-equivalent protons \(\left( {{\rm{a and b}}} \right)\)are present, then the multiplicity is given by \(\left( {{{\rm{n}}_{\rm{a}}}{\rm{ + 1}}} \right)\left( {{{\rm{n}}_{\rm{b}}}{\rm{ + 1}}} \right)\). But if the coupling constant of a and b protons are approximately the same, the rule reduces to \(\left( {{{\rm{n}}_{\rm{a}}}{\rm{ + }}{{\rm{n}}_{\rm{b}}}{\rm{ + 1}}} \right)\).

The condensed formula of a

6 equivalent hydrogen atoms split the labeled proton. So, the number of peaks is given by \(\left( {{\rm{6 + 1}}} \right) = 7\)peaks.

The condensed formula of b

The labeled proton \({{\rm{H}}_{\rm{a}}}\) is split by 2 hydrogen atoms. So, the number of peaks due to \({{\rm{H}}_{\rm{a}}}\) is given by \(\left( {{\rm{2 + 1}}} \right) = 3\)peaks. The proton \({{\rm{H}}_{\rm{c}}}\) is split by 4 equivalent hydrogen atoms. So, the number of peaks due to \({{\rm{H}}_{\rm{c}}}\) is given by \(\left( {{\rm{4 + 1}}} \right) = 5\)peaks.

The proton \({{\rm{H}}_{\rm{b}}}\) is split by 2 sets of hydrogen atoms. So, the maximum number of peaks will be \(\left( {{\rm{3 + 1}}} \right)\left( {{\rm{2 + 1}}} \right) = 12\)peaks. But the signal due to \({{\rm{H}}_{\rm{b}}}\)will have a peak overlap as this is a flexible alkyl chain. And the number of peaks that will be likely visible is \(3 + 2 + 1 = 6\)peaks.

The condensed formula of c

The labeled proton \({{\rm{H}}_{\rm{a}}}\) is split by 1 hydrogen atom. So, the number of peaks due to \({{\rm{H}}_{\rm{a}}}\) is given by \(\left( {{\rm{1 + 1}}} \right) = 2\)peaks. The proton \({{\rm{H}}_{\rm{b}}}\) is split by 2 sets of hydrogen atoms. So, the number of peaks due to \({{\rm{H}}_{\rm{b}}}\)will be \(\left( {{\rm{1 + 1}}} \right)\left( {{\rm{2 + 1}}} \right) = 6\)peaks.

The condensed formula of d

The proton \({{\rm{H}}_{\rm{a}}}\) is split by 2 different hydrogen atoms. So, the number of peaks due to \({{\rm{H}}_{\rm{a}}}\) will be \(\left( {{\rm{1 + 1}}} \right)\left( {{\rm{1 + 1}}} \right) = 4\) peaks.

The proton \({{\rm{H}}_{\rm{b}}}\) is split by 2 different hydrogen atoms. So, the number of peaks due to \({{\rm{H}}_{\rm{b}}}\) will be \(\left( {{\rm{1 + 1}}} \right)\left( {{\rm{1 + 1}}} \right) = 4\) peaks.

The proton \({{\rm{H}}_{\rm{c}}}\) is split by 2 different hydrogen atoms. So, the number of peaks due to \({{\rm{H}}_{\rm{c}}}\) will be \(\left( {{\rm{1 + 1}}} \right)\left( {{\rm{1 + 1}}} \right) = 4\) peaks.