Propose structures for ketones or aldehydes that have the following ¹H

NMR spectra:

(a) ${\mathbf{C}}_4{\mathbf{H}}_7\mathbf{ClO}$

IR: 1715 ${\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$

(b)${\mathbf{C}}_7{\mathbf{H}}_{14}\mathbf{O}$

IR: 1710 ${\mathbf{cm}}^{\mathbf{-}\mathbf{1}}$

NMR stands for Nuclear Magnetic Resonance Spectroscopy is the technique used for determining the structure of the organic compounds. It is of two types ¹H NMR and¹³C NMR.

Proton nuclear magnetic resonance (¹H NMR) is the application of nuclear magnetic resonance in NMR spectroscopy with respect to hydrogen-1 nuclei within in the molecule of substance in order to determine the structure of its molecules.

The number of signals present in NMR reflects the number of magnetically different protons. Thus the number of signal reflects the number of chemically different protons or sets of protons. For example methanol has two different set of protons (methyl and hydroxyl).

Degree of unsaturation is also known as double bond equivalent (DBE). If molecular formula is given put them into these formula to obtain degree of unsaturation. The formula for degree of unsaturation is given below:

$\mathrm{DBE}=\frac{2\mathrm{C}+2+\mathrm{N}-\mathrm{X}-\mathrm{H}}{2}$

Where, C is the number of Carbons

N is the number of nitrogen’s

H is the number of hydrogens

X is the number of halogens

Firstly, we have to calculate degree of unsaturation of the given compound

For ${\mathrm{C}}_4{\mathrm{H}}_7\mathrm{ClO}$

$\begin{array}{l}\mathrm{DBE}=\frac{2\left(4\right)+2+0-1-7}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=\frac{2}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=1\end{array}$

The actual formula,${\mathrm{C}}_4{\mathrm{H}}_7\mathrm{ClO}$corresponds to one degree of unsaturation.

The available IR data (IR: 1715${\mathrm{cm}}^{-1}$)assign that the given compound has carbonyl group (C=O) as a principal functional group.

Now, the structure of compound (a)following ¹HNMR spectra is shown below:

(a)

Structure of (a)

a = 1.6$\mathrm{\delta }$

b = 2.33$\delta$

c = 4.32 $\delta$

Firstly, we have to calculate degree of unsaturation of the given compound

For ${\mathrm{C}}_7{\mathrm{H}}_{14}\mathrm{O}$,

$\begin{array}{l}\mathrm{DBE}=\frac{2\left(7\right)+2+0-0-14}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=\frac{2}{2}\\ \underset{}{}\underset{}{}\underset{}{}\underset{}{}=1\end{array}$

The actual formula, ${\mathrm{C}}_7{\mathrm{H}}_{14}\mathrm{O}$corresponds to one degree of unsaturation.

The available IR data (IR: 1710${\mathrm{cm}}^{-1}$)assign that the given compound has carbonyl group (C=O) as a principal functional group.

Now, the structure of compound (a) following ¹HNMR spectra is shown below:

(b)

Structure of (b)

a = 1.02$\mathrm{\delta }$

b = 2.12$\mathrm{\delta }$

c = 2.33$\delta$