For C_{7}H_{16}O:

(1) Calculate for the IHD:

$\overline{){\mathbf{IHD}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{-}\mathbf{H}}{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}\mathbf{IHD}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{2}\left(7\right)\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{-}\mathbf{16}}{\mathbf{2}}$

IHD = 0

where H or X = 1; N = -1 and O = 0

Structure is not cyclic and has 0 double bonds

(2) Inspecting the signals:

a. 0.9 ppm → triplet → next to -CH_{2}-

b. 1.38 ppm → singlet peak is observed during D_{2}O shaking → -OH present

c. 1.42 ppm → quartet → next to -CH_{3}

Give the structure that corresponds to the following molecular formula and ^{1}H NMR spectrum:

The magenta numbers in the spectrum are the relative integrals in arbitrary units.

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