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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 2I +1 degenerate magnetic levels.

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

If two sets of non-equivalent protons (a and b) are present, then the multiplicity is given by (n_a + 1)(n_b + 1) . But if the coupling constant of a and b protons are approximately the same, the rule reduces to (n_a + n_b + 1) .

The condensed formula of a

6 equivalent hydrogen atoms split the labeled proton. So, the number of peaks is given by (6 + 1) = 7 peaks.

The condensed formula of b

The labeled proton H_a is split by 2 hydrogen atoms. So, the number of peaks due to H_a is given by (2 + 1) = 3 peaks. The proton H_c is split by 4 equivalent hydrogen atoms. So, the number of peaks due to H_c is given by (4 + 1) = 5 peaks.

The proton H_b is split by 2 sets of hydrogen atoms. So, the maximum number of peaks will be (3 + 1)(2 + 1) = 12 peaks. But the signal due to H_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 + 3+ 2+ 1 = 6 peaks.

The condensed formula of c

The labeled proton H_a is split by 1 hydrogen atom. So, the number of peaks due to H_a is given by (1 + 1) = 2peaks. The proton H_b is split by 2 sets of hydrogen atoms. So, the number of peaks due to H_b will be (1 +1)(2 +1) = 6 peaks.

The condensed formula of d

The proton H_a is split by 2 different hydrogen atoms. So, the number of peaks due to H_a will be (1 + 1)(1 + 1) = 4 peaks.

The proton H_b is split by 2 different hydrogen atoms. So, the number of peaks due to H_b will be (1 + 1)(1 + 1) = 4 peaks.

The proton H_c is split by 2 different hydrogen atoms. So, the number of peaks due to H_c will be (1 + 1)(1 + 1) = 4 peaks.