91Ó°ÊÓ

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.