Question

How can you distinguish the following compounds using:

a. their infrared spectra?

b. their 1 H NMR spectra?

Short answer

a. Compounds A, B, and C can be differentiated from the rest by the absence of a peak corresponding to the carbonyl group. The absence of the phenyl group can be used to separate A from B and C. Compound C can be distinguished by the absence of the hydroxyl group.

Compounds D, E, F, and G can be separated based on the peaks corresponding to functional groups in each compound.

b. The absence of phenyl protons can separate compound A. Compounds B and C produce two peaks in addition to phenyl protons, and they can be differentiated by integrating the peaks.

Compounds D, E, F, and G produce one peak in addition to the phenyl protons, and chemical shift values can be used to differentiate them.

Step-by-step solution

Step-by-Step Solution Step 1: Infrared spectroscopy

IR spectroscopy studies the interaction of organic compounds with Infrared radiations.

The covalent bonds in a molecule behave like spring and vibrate at a specific natural frequency. These vibrations are quantized, and therefore, covalent bonds have vibrational energy levels.

Suppose the energy difference between the vibrational levels falls within the IR region. They can absorb the radiation and move to higher energy where it vibrates with higher amplitude and constant frequency.

The frequency of IR absorbed will be equal to the natural vibrational frequency of the molecule and therefore, can be used for identifications.

Proton NMR spetroscoppy

The proton NMR spectroscopy studies the interaction of the nucleus with radiowaves.

The nuclei of atoms rotate on their axis and behave like small magnets. The magnetic moments of nuclei are placed randomly. In the presence of an external magnetic field, these magnetic moments either align with or against the direction of the external field.

This causes the nucleus to lose its degeneracy, with the nuclei aligned with the magnetic field having lower energy. The transition of nuclei between the energy levels occurs by absorption of radio waves. The frequency of radio waves absorbed depends on the chemical environment of the proton.

Differentiating the compounds using NMR and

(a) Using IR spectroscopy

The given compounds can first be divided based on the presence of a carbonyl group.

Consider compounds A, B, and C. These compounds will not show an absorption band at 1700${\mathrm{cm}}^{-1}$due to the absence of the carbonyl group.

Compounds A, B, and C

These compounds can be separated as follows:

Compound B is benzyl alcohol and, therefore, will have bands at 3300${\mathrm{cm}}^{-1}$ corresponding to the hydroxyl group and 1600${\mathrm{cm}}^{-1}$ corresponding to the phenyl group.

Compound A only has a hydroxyl group and not a phenyl group. Therefore, it will have a band at 3300${\mathrm{cm}}^{-1}$ , and the band at 1600${\mathrm{cm}}^{-1}$ will be absent.

Compound C has only the benzyl group and not the hydroxyl group. Therefore, the peak at 3300${\mathrm{cm}}^{-1}$ will be absent. It will have a peak at 1600${\mathrm{cm}}^{-1}$ .

Consider compounds D, E, F, and G. These compounds will have a strong band at 1700${\mathrm{cm}}^{-1}$ due to the presence of the carbonyl group.

Compounds D, E, F, and G

The above compounds can be differentiated as follows:

Compound D will have a strong band at approximately 2500-3500${\mathrm{cm}}^{-1}$ due to the presence of the carboxyl group. The bands corresponding to the ester group and aldehyde group will be absent in this compound.

Compound E will have a strong band at 2700${\mathrm{cm}}^{-1}$ due to the presence of the aldehyde group. The bands corresponding to the ester group and carboxyl group will be absent in this compound.

Compound F will have a strong band at 1250${\mathrm{cm}}^{-1}$ due to the presence of the ester group. The bands corresponding to the carboxyl and aldehyde groups will be absent in this compound.

Compound G will not have bands corresponding to the carboxyl, ester, or aldehyde group. It will only have the band corresponding to the carbonyl group at 1700${\mathrm{cm}}^{-1}$ .

(b)

Of the given compounds, only compound A will not have a characteristic peak at 7-8 ppm due to the absence of benzene protons. Therefore, compound A can be distinguished using that peak.

Compound A

Compounds B and C will have two other signals than the benzene protons. These compounds can be distinguished by integrating the peaks. The ratio of protons of the peaks in compound B will be 2:3, and for compound C, it will be 2:1 based on the number of protons of each type.

Also, the peaks can be distinguished by their width. Compound B is ether and, therefore, will have 2 sharp signals, while the hydroxyl proton in compound C will show a broad peak.

Compounds B and C

The other four compounds will have one signal in addition to the benzene ring signal.

Compounds D, E, F, and G

Compound D has a hydroxyl proton and will show a peak at approximately 10-12 ppm. The aldehyde proton in compound E shows a peak at approximately 9-10 ppm.

Compound F contains methoxy protons which show a peak at approximately 4 ppm. The methyl protons in compound G show a signal at approximately 2 ppm.