MCAT Organic Chemistry Review

Spectroscopy

Practice Questions

1.    IR spectroscopy is most useful for distinguishing:

1.    double and triple bonds.

2.    C–H bonds.

3.    chirality of molecules.

4.    composition of racemic mixtures.

2.    Oxygen (O2) does not exhibit an IR spectrum because:

1.    it has no molecular motions.

2.    it is not possible to record IR spectra of a gaseous molecule.

3.    molecular vibrations do not result in a change in the dipole moment of the O2 molecule.

4.    molecular oxygen contains four lone pairs overall.

3.    If IR spectroscopy were employed to monitor the oxidation of benzyl alcohol to benzaldehyde, which of the following would provide the best evidence that the reaction was proceeding as planned?

1.    Comparing the fingerprint region of the spectra of starting material and product

2.    Noting the change in intensity of the peaks corresponding to the benzene ring

3.    Noting the appearance of a broad absorption peak in the region of 3100–3500 cm–1

4.    Noting the appearance of a strong absorption in the region of 1750 cm–1

4.    Which of the following chemical shifts could correspond to an aldehydic proton signal in a 1H–NMR spectrum?

1.    9.5 ppm

2.    7.0 ppm

3.    11.0 ppm

4.    1.0 ppm

5.    The isotope 12C is not useful for NMR because:

1.    it is not abundant in nature.

2.    its resonances are not sensitive to the presence of neighboring atoms.

3.    it has no magnetic moment.

4.    the signal-to-noise ratio in the spectrum is too low.

6.    In 1H–NMR, splitting of spectral lines is due to:

1.    coupling between a carbon atom and protons attached to that carbon atom.

2.    coupling between a carbon atom and protons attached to adjacent carbon atoms.

3.    coupling between adjacent carbon atoms.

4.    coupling between protons on adjacent carbon atoms.

7.    Compared to IR and NMR spectroscopy, UV spectroscopy is preferred for detecting:

1.    aldehydes and ketones.

2.    unconjugated alkenes.

3.    conjugated alkenes.

4.    aliphatic acids and amines.

8.    Considering only the 0 to 4.5 ppm region of a 1H–NMR spectrum, how could ethanol and isopropanol be distinguished?

1.    They cannot be distinguished from 1H–NMR alone.

2.    A triplet and quartet are observed for ethanol, whereas a doublet and septet are observed for isopropanol.

3.    A triplet and quartet are observed for isopropanol, whereas a doublet and septet are observed for ethanol.

4.    The alcohol hydrogen in ethanol will appear within that region, whereas the alcohol hydrogen in isopropanol will appear downfield of that region.

9.    Before absorbing an ultraviolet photon, electrons can be found in:

1.    the HOMO only.

2.    the LUMO only.

3.    both the HOMO and the LUMO.

4.    neither the HOMO nor the LUMO.

10.In an IR spectrum, how does extended conjugation of double bonds affect the absorbance band of carbonyl (C=O) stretches compared with normal absorption?

1.    The absorbance band will occur at a lower wavenumber.

2.    The absorbance band will occur at a higher wavenumber.

3.    The absorbance band will occur at the same wavenumber.

4.    The absorbance band will disappear.

11.Wavenumber is directly proportional to:

1.    wavelength.

2.    frequency.

3.    percent transmittance.

4.    absorbance.

12.Two enantiomers will:

1.    have identical IR spectra because they have the same functional groups.

2.    have identical IR spectra because they have the same specific rotation.

3.    have different IR spectra because they are structurally different.

4.    have different IR spectra because they have different specific rotations.

13.In a molecule containing a carboxylic acid group, what would be expected in a 1H–NMR spectrum?

1.    A deshielded hydrogen peak for the hydroxyl hydrogen, shifted left

2.    A deshielded hydrogen peak for the hydroxyl hydrogen, shifted right

3.    A shielded hydrogen peak for the hydroxyl hydrogen, shifted left

4.    A shielded hydrogen peak for the hydroxyl hydrogen, shifted right

14.The coupling coefficient, J, is:

1.    the value of n + 1 when determining splitting in NMR spectra.

2.    measured in parts per million (ppm).

3.    corrected for by calibration with tetramethylsilane.

4.    a measure of the degree of splitting caused by other atoms in the molecule.

15.The IR spectrum of a fully protonated amino acid would likely contain which of the following peaks?

1.    A sharp peak at 1750 cm–1

2.    A sharp peak at 3300 cm–1

3.    A broad peak at 3300 cm–1

1.    I only

2.    I and II only

3.    II and III only

4.    I, II, and III

PRACTICE QUESTIONS