Answers and Explanations - Spectroscopy - MCAT Organic Chemistry Review

MCAT Organic Chemistry Review


Answers and Explanations

1. AInfrared spectroscopy is most useful for distinguishing between different functional groups. Almost all organic compounds have C–H bonds, choice (B), so except for fingerprinting a compound, these absorptions are not useful. Little information about the optical properties of a compound, such as choices (C) and (D), can be obtained by IR spectroscopy.

2. CBecause molecular oxygen is homonuclear (composed of only one element) and diatomic, there is no net change in its dipole moment during vibration or rotation; in other words, the compound does not absorb in a measurable way in the infrared region. IR spectroscopy is based on the principle that, when the molecule vibrates or rotates, there is a change in dipole moment. Choice (A) is incorrect because oxygen does have molecular motions; they are just not detectable in IR spectroscopy. Choice (B) is incorrect because it is possible to record the IR of a gaseous molecule as long as it shows a change in its dipole moment when it vibrates. Choice (D) is incorrect because lone pairs do not have an effect on the ability to generate an IR spectrum of a compound.

3. DIn this reaction, the functional group is changing from a hydroxyl to an aldehyde. This means that a sharp peak will appear around 1750 cm–1, which corresponds to the carbonyl functionality. Choice (C) is the opposite of what occurs; the reaction will be characterized by the disappearance of the O–H peak at 3100 to 3500 cm–1, not its appearance. Comparing the fingerprint regions, as in choice (A), will provide evidence that a reaction is occurring, but is not as useful for knowing that the reaction that occurred was indeed the one that was desired.

4. AThe peak at 9.5 ppm corresponds to an aldehydic proton. This signal lies downfield because the carbonyl oxygen is electron-withdrawing and deshields the proton. Choice (C) corresponds to a carboxyl proton and is even further downfield because the acidic proton is deshielded to a greater degree than the aldehyde proton. Choice (B) corresponds to aromatic protons. Choice (D) is characteristic of an alkyl proton on an sp3-hybridized carbon.

5. CThis isotope has no magnetic moment and will therefore not exhibit resonance with an applied magnetic field. Nuclei with odd mass numbers (1H, 11B, 13C, 15N, 19F, and so on) or those with an even mass number but an odd atomic number (2H, 10B) will have a nonzero magnetic moment.

6. DSpin–spin coupling (splitting) is due to influence on the magnetic environment of one proton by protons on the adjacent atom. These protons are three bonds away from each other. Splitting in other NMR spectra can include coupling with carbon atoms, but not in 1H–NMR.

7. CMost conjugated alkenes have an intense ultraviolet absorption. Aldehydes, ketones, acids, and amines, mentioned in choices (A) and (D), all absorb in the ultraviolet range. However, other forms of spectroscopy (mainly IR and NMR) are more useful for precise identification. Isolated alkenes, choice (B), can rarely be identified by UV.

8. BThe region in question often gives information about the types of alkyl groups present. Specifically, ethanol will give a characteristic triplet for the methyl group (which is coupled to –CH2–) and a quartet for –CH2– (which is coupled to the methyl group). Isopropanol will have a septet for the –CH– group (which is coupled to both methyl groups combined) and a doublet for the two methyl groups (which are coupled to –CH–). In both cases, the proton in the alcohol does not participate in coupling. The alcohol hydrogen likely lies downfield for both compounds because it is bound to such an electronegative element.

9. AThe HOMO is the highest occupied molecular orbital. Only after absorbing ultraviolet light is an electron is excited from the HOMO to the LUMO, the lowest unoccupied molecular orbital.

10.ACarbonyl groups (C=O) in conjugation with double bonds tend to absorb at lower wavenumbers because the delocalization of π electrons causes the C=O bond to lose double-bond character, shifting the stretching frequency closer to C–O stretches. Remember that higher-order bonds tend to have higher absorption frequencies, so loss of double-bond character should decrease the absorption frequency of the group.

11.BWavenumber is directly proportional to frequency It is inversely proportional to wavelength, choice (A), and has no proportionality to percent transmittance or absorbance, choices (C) and (D).

12.AEnantiomers will have identical IR spectra because they have the same functional groups and will therefore have the exact same absorption frequencies. Enantiomers have opposite specific rotations, but specific rotation actually has no effect on the IR spectrum.

13.AThe oxygen of the hydroxyl group will deshield the hydroxyl hydrogen, shifting it downfield, or left. Hydrogens in carboxylic acids can have some of the most downfield absorbances, around 10.5 to 12 ppm.

14.DThe coupling constant is a measured of the degree of splitting introduced by other atoms in a molecule, and is the frequency of the distance between subpeaks. It is measured in hertz, eliminating choice (B). The coupling constant is independent of the value of n + 1, and is not changed by calibration with tetramethylsilane, eliminating choices (A) and (B).

15.BAmino acids in their fully protonated form contain all three of the peaks that should be memorized for Test Day: C–O, N–H, and O–H. While statements I and II correctly give the peaks for the C=O bond (sharp peak at 1750 cm–1) and the N–H bond (sharp peak at 3300 cm–1), the peak for the O–H bond is in the wrong place. In a carboxylic acid, the C=O bond withdraws electron density from the O–H bond, shifting the absorption frequency down to about 3000 cm–1. Statement III is therefore incorrect.