Propose a structure for a compound ${\mathbf{C}}_{\mathbf{15}}{\mathbf{H}}_{\mathbf{24}}\mathbf{O}$that has the following${}^{\mathbf{1}}\mathbf{H}$NMR spectrum. The peak marked by an asterisk disappears when${\mathbf{D}}_{\mathbf{2}}\mathbf{O}$is added to the sample.

The proton nuclear magnetic resonance spectroscopy is used to analyze compounds based on the frequency of vibration of the atomic nucleus. The protons in compounds undergo three bonded coupling to give the spectrum containing singlet, doublet, and triplet peaks.

From the given molecular formula, the degree of unsaturation in the given organic compound is:

$\begin{array}{rcl}\mathrm{DU}& =& \left({\mathrm{C}}_{\mathrm{n}}+1\right)-\frac{{\mathrm{H}}_{\mathrm{n}}}{2}-\frac{{\mathrm{X}}_{\mathrm{n}}}{2}+\frac{{\mathrm{N}}_{\mathrm{n}}}{2}\\ & =& \left(15+1\right)-\frac{24}{2}\\ & =& 16-12\\ & =& 4\end{array}$

Thus, the degree of unsaturation 4 indicates the presence of the benzene ring.

The molecular formula of the compound ${\text{C}}_{\text{15}}{\text{H}}_{\text{24}}\text{O}$shows that the compound with the resulting spectrum is given below.

Butylated hydroxyl toluene(BHT)

Where, a = 1.4$\delta$, b = 2.24$\delta$, c = 5.00$\delta$and d = 6.97$\delta$

Since this is a symmetric compound, it only has four peaks in the spectrum, and the same protons cause the intensity of the peaks.