Chapter 23, Problem 35P

Interpretation Introduction

Interpretation:

Two compounds are to be identified on the basis of their chemical formulas, and IR and ${}^{\text{13}}\text{C}$ NMR spectra.

Concept introduction:

Peaks in an IR spectrum help in determining functional groups in a compound.

In ¹³C NMR spectroscopy, we get information about the different types of carbon present in the given molecular formula.

¹³C NMR spectroscopy can even differentiate between primary, secondary, tertiary, and quaternary carbons.

If the compound only contains C, H, and O, the index of hydrogen deficiency is calculated as

$\text{Index of hydrogen deficiency = }\frac{\text{1}}{\text{2}}({\text{C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{ - C}}_{\text{n}}{\text{H}}_{\text{x}})$

If a compound contains C, H, O, and a halogen, the halogen can be treated as if it were a hydrogen for the purpose of calculating index of hydrogen deficiency.

Answer to Problem 35P

Solution:

a) The structure of the compound ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}\text{O}$ is

b) The structure of the compound ${\text{C}}_{\text{9}}{\text{H}}_{\text{11}}\text{BrO}$ is

Explanation of Solution

The chemical formula of the compound is ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}\text{O}$. As it only contains C, H, and O, the index of hydrogen deficiency can be calculated as follows:

$\begin{array}{c}\text{Index of hydrogen deficiency = }\frac{\text{1}}{\text{2}}{\text{(C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{ - C}}_{\text{n}}{\text{H}}_{\text{x}}\text{)}\\ \text{= }\frac{\text{1}}{\text{2}}\left({\text{C}}_{\text{9}}{\text{H}}_{\text{20}}{\text{ - C}}_{\text{9}}{\text{H}}_{\text{12}}\right)\\ \text{= }\frac{\text{1}}{\text{2}}\text{ × 8}\\ \text{= 4}\end{array}$

The count of four indicates the possibility of a benzene ring. If a benzene ring is present, the oxygen atom would be singly bonded, either as a hydroxyl group or as an ether functional group.

The IR and ¹³C NMR spectra of the compound are as follows:

The presence of a broad peak around 3200 ${\text{cm}}^{-1}$ in the IR spectrum supports the presence of a hydroxyl group. The peak at about 800 ${\text{cm}}^{-1}$ suggests the presence of a para disubstituted benzene ring. This is also supported by the ¹³C NMR signals in the range of 110 to 150 ppm. The remaining NMR signals are between 15 and 40 ppm, characteristic of alkyl (${\text{sp}}^3$) carbons. If the hydroxyl group was attached to the alkyl chain, the signal for that carbon would appear above 50 ppm. Absence of such a signal indicates that the hydroxyl group is attached directly to the benzene ring, not to the alkyl group. This also supports a disubstituted benzene ring. Subtracting the $\text{OH}$ and ${\text{C}}_{\text{6}}{\text{H}}_{\text{4}}$ for the disubstituted benzene ring from the chemical formula leaves ${\text{C}}_3{\text{H}}_7$. Taking into account the two ${\text{CH}}_2$ peaks and one ${\text{CH}}_3$ peak, this matches a propyl substituent. Therefore, the structure of the compound must be

Chemical formula of the compound is ${\text{C}}_{\text{9}}{\text{H}}_{\text{11}}\text{BrO}$. Treating the bromine atom as a hydrogen, the equivalent formula would be ${\text{C}}_{\text{9}}{\text{H}}_{\text{12}}\text{O}$. The index of hydrogen deficiency is then as follows:

$\begin{array}{c}\text{Index of hydrogen deficiency = }\frac{\text{1}}{\text{2}}{\text{(C}}_{\text{n}}{\text{H}}_{\text{2n+2}}{\text{ - C}}_{\text{n}}{\text{H}}_{\text{x}}\text{)}\\ \text{= }\frac{\text{1}}{\text{2}}\left({\text{C}}_{\text{9}}{\text{H}}_{\text{20}}{\text{ - C}}_{\text{9}}{\text{H}}_{\text{12}}\right)\\ \text{= }\frac{\text{1}}{\text{2}}\text{ × 8}\\ \text{= 4}\end{array}$

Therefore this compound also is likely to have a benzene ring.

The IR and ¹³C NMR spectra of the compound are as follows:

Two peaks in the IR spectrum, at about 700 and 750 ${\text{cm}}^{-1}$, suggest the presence of a monosubstituted benzene ring. Presence of a benzene ring is also supported by the ¹³C NMR signals of two $\text{CH}$ and one $\text{C}$ in the range of 115 to 160 ppm. The absence of a broad peak around 3200 ${\text{cm}}^{-1}$ in the IR spectrum rules out a hydroxyl group. The IR peak at 1200 ${\text{cm}}^{-1}$ points to the presence of an ether linkage. After subtracting the monosubstituted benzene ring from the chemical formula, ${\text{C}}_{\text{3}}{\text{H}}_{\text{6}}\text{BrO}$ is left. If the ether linkage (oxygen atom) was between two alkyl carbons, there would be two ¹³C NMR signals in the range of about 50 to 80 ppm. The NMR spectrum shows only one such signal. Also, there are three alkyl ${\text{CH}}_2$ signals. This suggests the oxygen is directly bonded to the benzene ring on one side, with a three carbon alkyl chain on the other side. Two of the alkyl ${\text{CH}}_2$ signals are near 30 ppm, one of them slightly higher at 33 about ppm. This indicates the bromine is bonded to the terminal carbon in the alkyl chain.

Therefore, the structure of the compound is