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Test: Infrared and Ultraviolet/Visible Spectroscopy - MCAT MCQ


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10 Questions MCQ Test - Test: Infrared and Ultraviolet/Visible Spectroscopy

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Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 1
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Which of the following statements accurately describes the molecular vibrations characteristic of IR spectroscopy?

Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 1
  • Stretching vibrations occur at higher frequencies than bending vibrations. Generally, bonds more easily bend than stretch or compress.
  • The fingerprint region is in the 1450 to 600 cm-1 region, and most of the bending frequencies appear in that region. Absorption in the diagnostic 4000 to 1450 cm-1 region are usually due to stretching vibrations.
  • Bond order and the types of atoms joined by the bond are the two most important factors in determining where a chemical bond will absorb.
  • The higher the bond order, the higher the frequency needed to stretch. Triple bonds have higher stretching frequencies than corresponding double bonds, especially bonds to hydrogen have higher stretching frequencies than those to heavier atoms.
  • Therefore, the correct answer is that bonds to hydrogen have higher stretching frequencies than those to heavier atoms.
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 2
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Which of the following absorption peak or peaks would be most useful for identifying the compound diethylamine?

Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 2
  • All carbonyl compounds absorb strongly around 1700-1750 cm−1 due to the stretching vibration of the C=O bond. Most carbonyl compounds (saturated esters, aldehydes, and ketones and unsaturated esters) absorb above 1700 cm−1 so diethylamine would not have a sharp peak at 1731 cm−1
  • A very broad stretch ranging from 3500-2500 cm−1 is usually indicative of a carboxylic acid group. The hydrogen bonding of the hydroxyl groups to each other lowers and broadens the stretching frequencies of the participating O-H bonds, and for carboxylic acids, there is dimerization. Each molecule may be H-bonded to a varying extent, and each will absorb at a slightly different frequency, which is what makes the peak broad.
  • The N-H stretches of amines are above 3000 cm-1, and the peaks are weaker and sharper than that of the O-H stretches. Two medium peaks would be indicative of a primary amine.
  • Secondary amines would only have one medium stretch from 3500 to 3300 cm-1, so that stretch would identify the sample as diethylamine.
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Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 3
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Which of the following statements about the IR spectrograph could indicate the successful completion of the reaction of pentanal with Tollen’s reagent?

Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 3
  • Tollen’s reagent is a compound used to determine whether a sample contains an aldehyde or an alpha-hydroxy ketone. A positive result produces the carboxylate ion from the aldehyde.
  • The gain of a sharp peak at 1721 cm-1 would indicate the appearance of a C=O bond.
  • Maintaining a sharp peak at 2971 cm-1 or below 3000 cm-1 is indicative of an alkane. Since it remains the same, it would be useful to indicate the completion of the reaction.
  • The appearance of small peaks at 2827 and 2725 cm-1 are usually indicative of the C-H stretch of an aldehyde (Fermi doublet).
  • Only the gain of a broad peak from 2500-3500 would unambiguously indicate the completion of the reaction from aldehyde to carboxylate ion or carboxylic acid. This peak is especially broad because carboxylic acids usually exist as hydrogen-bonded dimers.
  • The correct answer is that there is a gain of broad peak in the region of 3310-2540 cm-1
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 4
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What is the order of the following carbonyl compounds in decreasing wavenumber?
I. butanoyl chloride
II. ethyl butanoate
III. pentanal
IV. propanoic acid
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 4
  • Increasing the double bond character of the carbonyl will shift the absorption to a higher frequency, and decreasing the character will shift to a lower frequency.
  • Inductively-withdrawing groups will strengthen the double bond as the electron density of the lone pair is pulled into the carbonyl. Butanoyl chloride and ethyl butanoate has inductively-withdrawing groups and will absorb at higher frequencies.
  • Inductively-donating groups, as well as resonance, will weaken the double bond as the electron density will tend to shift the carbonyl π cloud to a lone pair localized on the carbonyl oxygen. Pentanal has only inductively-donating groups, and will absorb at a lower frequency.
  • Carboxylic acids are an interesting case since they exist in dimers, so the electron density participates in the hydrogen-bonding, which weakens the double bond.
  • The order in decreasing wavenumber is butanoyl chloride, ethyl butanoate, pentanal, and finally propanoic acid.
  • The correct order is the following: I, II, III, IV.
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 5
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Based on the absorption spectrum, what is the likely name of the compound below?
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 5
  • The spectrum has a strong peak at 1619 cm-1. While it may be tempting to attribute this peak to C=O stretch, which is the range of 1760-1670 cm-1, this is the bending frequency of the N-H bond, so this compound is not butanoic acid. 
  • The spectrum has two peaks at 3360 and 3442 cm−1. Alcohols usually have a broad peak above 3000 cm−1, so the compound is not pentanol.
  • These peaks are indicative of a primary amine, one for the asymmetrical stretch and the other for the symmetrical stretch. This correct compound is not diethylamine, but aniline.
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 6
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In the absorption spectrum provided below, assuming that the two plots represent the same compound, which of the following statements most accurately describes the graph?
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 6
  • The blue plot has the most identifiable peaks, which is the broad peak above 3000 cm−1 and the sharp peak below 3000 cm-1. 
  • The broad peak indicates the presence of an O-H bond, and the sharp peak indicates the presence of a C-H bond, so this compound is not a carboxylic acid, but an alcohol.
  • IR spectroscopy allows for samples in solid, liquid, and gas phase. An alcohol in the gaseous phase would not be able to hydrogen bond to each other, and the broad peak would disappear. That is not the case for the solid and liquid phase.
  • If we consider the two solvents, putting the alcohol in chloroform would essentially eliminate hydrogen bonding, like in the gaseous phase. In such an instance, the broad peak is replaced by a sharp peak at 3600 cm-1
  • The correct answer is that the plots represent the compound in two different solvents: water and chloroform.
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 7
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Which of the following compounds can be characterized by UV-Vis spectroscopy?
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 7
  • There are three main groups of compounds that can be characterized by UV-Vis spectroscopy: transition metal ions, organic compounds with a high degree of conjugation, and charge transfer complexes.
  • Since only transition metal ions can be identified by UV-Vis, chromium metal and hexaaquaaluminum ion cannot be identified.
  • Conjugated organic compounds can also be characterized . Penta-1,4-diene is not conjugated, but trans-1,2-dibenzoylethylene is, so it can be characterized by UV-Vis. The molecule is depicted below:
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 8
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Which of the following statements most accurately describes UV-Vis spectroscopy?
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 8
  • According to Beer’s law, the absorbance of a compound is proportional to the concentration of the sample and also proportional to the path length.
  • According to Beer’s law, the absorbance of a compound is proportional to the concentration of the sample and also proportional to the path length.
  • For molecules containing π or non-bonding electrons, the more easily excited the electrons or the smaller the HOMO-LUMO gap, the longer the wavelength it can absorb.
  • UV-Vis spectroscopy indeed can be used to measure the kinetics or the rate constant of a reaction, provided a reagent or product shows distinctive absorbance in the UV or visible spectral region.
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 9
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Based on the absorption spectrum below, which of the following plant pigments does the graph represent?
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 9
  • The graph indicates that absorption occurs strongly in the lower wavelengths of the visible spectrum, mostly blue.
  • When a compound absorbs a particular color, the complementary color is observed. The complementary color to blue is orange.
  • Chlorophyll is green, phycoerythrin is some hue of red, and phycocyanin is some hue of blue. Beta-carotene is the compound represented in the graph since it appears orange to the human eye.
Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 10
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Based on the absorption spectrum below, which of the following statements about pigment B is true?
Detailed Solution for Test: Infrared and Ultraviolet/Visible Spectroscopy - Question 10
  • Pigment B absorbs strongly in the longer wavelengths of light, but not the longest wavelengths. Orange or yellow would most likely be absorbed by this compound.
  • Chlorophyll appears green, for instance, and absorbs in the red and violet wavelengths. Pigment B appears blue based on the absorption of orange light.
  • Since pigment B appears blue, it emits photons with higher energy than pigment A, which transmits orange. Are these two (below) contradictory?)
  • Since pigment B absorbs orange and pigment A blue, pigment B has a smaller HOMO-LUMO gap than pigment A. The smaller the gap, the longer the wavelength that is absorbed.
  • We know that pigment B has a smaller HOMO-LUMO gap, which can be attributed to a greater level of conjugation or delocalization. Therefore, Pigment B has more conjugated double bonds than pigment A.
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