Chapter 16, Problem 16.84P

Interpretation Introduction

Interpretation:

The structure of the compound of the formula ${\text{C}}_4{\text{H}}_{\text{10}}{\text{O}}_2$ is to be determined using its IR, ${}^{\text{1}}\text{H}$ NMR, and ${}^{\text{13}}\text{C}$ NMR spectra.

Concept introduction:

The structure of a compound can be determined from the IR and NMR spectra.

The IR spectrum of a molecule consists of several absorption peaks/bands. These bands, particularly those above about 1200 ${\text{cm}}^{-1}$, are characteristic of different functional groups and the nature of $\text{C}-\text{H}$ bonds (hybridization of carbon).

The ${}^{\text{1}}\text{H}$ NMR spectrum provides information about the types of structurally distinct hydrogens and the number and types of coupled hydrogens. This helps identify fragments of the molecule.

The ${}^{\text{13}}\text{C}$ NMR spectrum provides information about the structurally distinct types of carbon atoms in the compound.

The formula of the compound, if known, can provide information about the unsaturation, number of double or triple bonds, and rings in the structure. This is known as the Index of Hydrogen Deficiency (IHD), and for a molecule containing only C, H, and O, it is calculated as ${\text{IHD = (C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{ - C}}_{\text{n}}{\text{H}}_{\text{x}}\text{)/2}$.

Using this information, the possible ways in which the fragments and functional groups are connected can be determined, leading to the probable structure of the molecule.

Answer to Problem 16.84P

The structure of the compound is

Explanation of Solution

The formula of the compound is ${\text{C}}_4{\text{H}}_{\text{10}}{\text{O}}_2$. The IHD for this compound is

$\begin{array}{rll}\text{IHD }& =& {\text{ (C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{ - C}}_{\text{n}}{\text{H}}_{\text{x}}\text{)/2}\\ & =& {\text{ (C}}_4{\text{H}}_{\text{10}}{\text{ - C}}_4{\text{H}}_{\text{10}}\text{)/2}\\ & =& \text{ 0}\end{array}$

This possibly means the oxygen atoms are part of alcohol or ether groups.

The ${}^{\text{1}}\text{H}$ NMR spectrum shows four peaks, i.e., four types of hydrogens. The integration trace shows heights in the ratio $\text{3:1:4:2}$, starting from the lowest $\text{δ}$ value. Assuming an equal number of protons, the total matches that in the formula. It is likely that the first two fragments are the same. The broad peak at 1.9 ppm could be the result of splitting by similar protons. The third peak is a four hydrogen doublet at a relatively high 3.6 ppm. This is likely the result of two ${\text{OCH}}_{\text{2}}\text{CH}$ groups. The 4.1 ppm triplet would then be two hydroxyl groups.

The data from the spectrum can then be summarized as follows:

δ ppm	No. of H’s	Splitting	Coupled H’s	Fragment
0.85	3	Doublet	1
1.9	1	Broad	-
3.6	4	Doublet	1
4.1	2	Triplet	2	OH

The fragments can then be put together as

The presence of the alcoholic OH group is supported by the broad IR absorption band at about 3500 ${\text{cm}}^{\text{-1}}$. This structure also shows three distinct carbons, as seen in the ${}^{\text{13}}\text{C}$ NMR spectrum.

Conclusion

The structure of a compound was determined from the ${}^{\text{1}}\text{H}$ NMR and IR spectra along with the chemical formula.