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 . B
Wavenumber 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 . A
Enantiomers 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 . A
The 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 . D
The coupling constant is a measure 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 . B
Amino 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.