Chapter 16, Problem 16.82P

Interpretation Introduction

Interpretation:

The structure of the compound of the formula ${\text{C}}_{\text{9}}{\text{H}}_{\text{10}}{\text{O}}_{\text{3}}$ 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 16.82P

The structure of the compound is

Explanation of Solution

The formula of the compound is ${\text{C}}_{\text{9}}{\text{H}}_{\text{10}}{\text{O}}_{\text{3}}$. The IHD for this compound is

$\begin{array}{rll}\text{IHD }& =& {\text{ (C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{ - C}}_{\text{n}}{\text{H}}_{\text{x}}\text{)/2}\\ & =& {\text{ (C}}_{\text{9}}{\text{H}}_{\text{20}}{\text{ - C}}_{\text{9}}{\text{H}}_{\text{10}}\text{)/2}\\ & =& \text{ 10/2}\\ & =& \text{ 5}\end{array}$

Considering the high value of IHD, it is likely that the compound contains a benzene ring. This will account for four, leaving one double bond in the substituent.

The presence of the two peaks at 6.9 and 7.0 ppm in the ${}^{\text{1}}\text{H}$ NMR spectrum supports the presence of the benzene ring.

The ${}^{\text{1}}\text{H}$ NMR spectrum shows five peaks, i.e., five types of hydrogens. The integration trace shows heights in the ratio $\text{3:2:2:2:1}$, starting from the lowest $\text{δ}$ value. Assuming the number of hydrogens at the highest $\text{δ}$ value of 12.2 to be one, the total of the hydrogens will be $\text{1+2+2+2+3 = 10}$, which matches the number in the formula. Since none of the peaks shows a complex splitting pattern, it is possible to directly determine the number of coupled protons using the $\text{n+1}$ rule. Then, based on the number of hydrogen atoms and the number of coupled hydrogen atoms, it is possible to determine the fragment of the molecule responsible for each signal. For example, there are three hydrogens at 1.4 ppm, with a signal that is split into a triplet. This means the group responsible for this signal must be a methyl ($-{\text{CH}}_{\text{3}}$) group that is bonded to a carbon with two hydrogens, $-{\text{CH}}_2-$. The second of these groups will then produce a signal that is split into a quartet, the 4.1 ppm signal. The relatively higher chemical shift of the $-{\text{CH}}_2-$ groups suggests the presence of the electronegative oxygen bonded to this carbon. The two doublets at 6.9 and 7.9 ppm are characteristic of para-substituted benzene. Finally, the singlet at 12.2 ppm is an isolated proton, most likely from a carboxylic acid group.

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

δ ppm	No. of H’s	Splitting	Coupled H’s	Fragment
1.4	3	Triplet	2
4.1	2	Quartet	3
6.9	2	Doublet	1
7.9	2	Doublet	1
12.2	1	Singlet	0

The three fragments – a di-substituted benzene ring ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}$, a ${\text{CO}}_{\text{2}}\text{H}$, and an ${\text{OCH}}_{\text{2}}{\text{CH}}_{\text{3}}$– account for all the atoms in the formula.

The presence of the carboxylic acid group is supported by the broad peak from about 2500 ${\text{cm}}^{\text{-1}}$ to 3200 ${\text{cm}}^{\text{-1}}$ along with the $\text{C}=\text{O}$ stretch at about 1700 ${\text{cm}}^{\text{-1}}$. The $\text{C}=\text{C}$ stretch at about 1450-1600 ${\text{cm}}^{\text{-1}}$ is indicative of an aromatic ring.

Thus, the molecular structure of the compound can be constructed by putting together the fragments so as to get para-disubstituted benzene. This structure shows seven structurally distinct carbons, matching the seven signals seen in the ${}^{\text{13}}\text{C}$ NMR spectrum.

Conclusion

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