CHEM 219 | CHEM219 Module 9: Principles of
Organic Chemistry with Lab - Portage Learning
Updated and Latest Questions and Correct
Answers with Rationale
1. A strong, broad absorption band centered at 3300 cm-1 in an IR spectrum most likely
indicates the presence of which functional group?
A. Carbonyl group
B. Alkyne (C-C triple bond)
C. Alcohol (O-H)
D. Nitrile group
Correct Answer: C
Expert Explanation: The absorption band at 3300 cm-1 is highly characteristic of the O-H
stretching vibration found in alcohols. This broadness is primarily due to the presence of
intermolecular hydrogen bonding in the liquid phase. Carbonyl groups typically appear as
sharp peaks around 1700 cm-1, while nitriles appear near 2250 cm-1. Alkanes usually
show much narrower C-H stretches just below 3000 cm-1. Identifying this broad peak is a
fundamental step in determining the presence of hydroxyl groups during lab analysis.
2. In a 1H NMR spectrum, a quartet at 2.5 ppm and a triplet at 1.1 ppm with a 2:3 integration
ratio suggest the presence of which alkyl group?
A. Methyl group
B. Isopropyl group
C. Ethyl group
D. Tert-butyl group
Correct Answer: C
Expert Explanation: An ethyl group is characterized by a triplet and a quartet pattern due
to spin-spin coupling between the methyl and methylene groups. The triplet at 1.1 ppm
represents the three methyl protons being split by the two methylene protons. The quartet
at 2.5 ppm represents the two methylene protons being split by the three methyl protons
according to the n+1 rule. The integration ratio of 2:3 confirms the relative number of
protons in each environment. This specific pattern is one of the most recognizable features
in small molecule NMR interpretation.
3. Which peak in a mass spectrum represents the intact molecule that has lost one electron?
A. Base peak
,B. Fragment peak
C. Molecular ion peak (M+)
D. M+2 peak
Correct Answer: C
Expert Explanation: The molecular ion peak, or M+ peak, corresponds to the radical
cation formed when the molecule loses a single electron. This peak is crucial because its
m/z value provides the molecular weight of the compound being analyzed. It is often found
at the highest m/z value in the spectrum, excluding isotope peaks. However, the molecular
ion may be weak or absent if the molecule undergoes rapid fragmentation. Recognizing this
peak allows the chemist to establish the empirical and molecular formulas of an unknown.
4. A compound with the molecular formula C3H6O shows a strong IR peak at 1715 cm-1 and a
single sharp peak in the 1H NMR at 2.1 ppm. What is the compound?
A. Propanal
B. Allyl alcohol
C. Cyclopropanol
D. Acetone
Correct Answer: D
Expert Explanation: The IR peak at 1715 cm-1 indicates a carbonyl group, specifically a
ketone or aldehyde. Acetone (propanone) has six equivalent protons, which results in a
single sharp singlet in the 1H NMR spectrum. Propanal would show multiple signals,
including an aldehyde proton near 9-10 ppm and a triplet/quartet pattern for the ethyl
group. Cyclopropanol and allyl alcohol would show O-H signals and different C-H
environments. Therefore, the symmetry of acetone perfectly matches the single NMR signal
and the carbonyl IR stretch.
5. In 13C NMR, at approximately what chemical shift range do carbonyl carbons typically
appear?
A. 0-50 ppm
B. 160-220 ppm
C. 100-150 ppm
D. 50-100 ppm
Correct Answer: B
Expert Explanation: Carbonyl carbons are highly deshielded due to the electronegativity
of the oxygen atom and the double bond. They typically appear in the 160 to 220 ppm
range, which is distinct from most other carbon environments. Aldehydes and ketones
, usually appear further downfield (190-220 ppm) compared to esters and amides (160-185
ppm). Aliphatic carbons usually appear below 50 ppm, while aromatic carbons are found
between 100 and 150 ppm. Identifying the 13C carbonyl signal is vital for confirming the
oxidation state of the carbon.
6. The presence of two peaks of roughly equal intensity at M+ and M+2 in a mass spectrum
indicates the presence of which element?
A. Chlorine
B. Bromine
C. Nitrogen
D. Sulfur
Correct Answer: B
Expert Explanation: Bromine has two naturally occurring isotopes, 79Br and 81Br, which
exist in an approximate 1:1 ratio. This results in a mass spectrum showing two molecular
ion peaks separated by two mass units with nearly equal heights. Chlorine also shows an
M+2 peak, but its ratio is 3:1 due to the abundance of 35Cl and 37Cl. Nitrogen and sulfur
have different isotopic signatures that do not produce a 1:1 M+ and M+2 pattern. This
isotope pattern is a definitive diagnostic tool in mass spectrometry for halogen
identification.
7. Where would you expect to find the C-H stretching vibrations for an sp2 hybridized carbon
(alkene) in an IR spectrum?
A. Below 3000 cm-1
B. Above 3000 cm-1 (3000-3100 cm-1)
C. 1700-1750 cm-1
D. 2200-2300 cm-1
Correct Answer: B
Expert Explanation: C-H bonds involving sp2 hybridized carbons are stronger and stiffer
than those involving sp3 carbons. Consequently, these vibrations occur at higher
frequencies, typically appearing just above 3000 cm-1. In contrast, sp3 C-H stretches
appear just below 3000 cm-1, providing a useful distinction for identifying unsaturation.
Carbonyl stretches occur much lower, around 1700 cm-1, and triple bonds appear around
2200 cm-1. Lab students use this boundary at 3000 cm-1 to quickly assess whether a
compound is purely aliphatic or contains alkenes/aromatics.
8. What information is provided by the integration of signals in a 1H NMR spectrum?
A. The number of neighboring protons
B. The chemical environment of the carbon
Organic Chemistry with Lab - Portage Learning
Updated and Latest Questions and Correct
Answers with Rationale
1. A strong, broad absorption band centered at 3300 cm-1 in an IR spectrum most likely
indicates the presence of which functional group?
A. Carbonyl group
B. Alkyne (C-C triple bond)
C. Alcohol (O-H)
D. Nitrile group
Correct Answer: C
Expert Explanation: The absorption band at 3300 cm-1 is highly characteristic of the O-H
stretching vibration found in alcohols. This broadness is primarily due to the presence of
intermolecular hydrogen bonding in the liquid phase. Carbonyl groups typically appear as
sharp peaks around 1700 cm-1, while nitriles appear near 2250 cm-1. Alkanes usually
show much narrower C-H stretches just below 3000 cm-1. Identifying this broad peak is a
fundamental step in determining the presence of hydroxyl groups during lab analysis.
2. In a 1H NMR spectrum, a quartet at 2.5 ppm and a triplet at 1.1 ppm with a 2:3 integration
ratio suggest the presence of which alkyl group?
A. Methyl group
B. Isopropyl group
C. Ethyl group
D. Tert-butyl group
Correct Answer: C
Expert Explanation: An ethyl group is characterized by a triplet and a quartet pattern due
to spin-spin coupling between the methyl and methylene groups. The triplet at 1.1 ppm
represents the three methyl protons being split by the two methylene protons. The quartet
at 2.5 ppm represents the two methylene protons being split by the three methyl protons
according to the n+1 rule. The integration ratio of 2:3 confirms the relative number of
protons in each environment. This specific pattern is one of the most recognizable features
in small molecule NMR interpretation.
3. Which peak in a mass spectrum represents the intact molecule that has lost one electron?
A. Base peak
,B. Fragment peak
C. Molecular ion peak (M+)
D. M+2 peak
Correct Answer: C
Expert Explanation: The molecular ion peak, or M+ peak, corresponds to the radical
cation formed when the molecule loses a single electron. This peak is crucial because its
m/z value provides the molecular weight of the compound being analyzed. It is often found
at the highest m/z value in the spectrum, excluding isotope peaks. However, the molecular
ion may be weak or absent if the molecule undergoes rapid fragmentation. Recognizing this
peak allows the chemist to establish the empirical and molecular formulas of an unknown.
4. A compound with the molecular formula C3H6O shows a strong IR peak at 1715 cm-1 and a
single sharp peak in the 1H NMR at 2.1 ppm. What is the compound?
A. Propanal
B. Allyl alcohol
C. Cyclopropanol
D. Acetone
Correct Answer: D
Expert Explanation: The IR peak at 1715 cm-1 indicates a carbonyl group, specifically a
ketone or aldehyde. Acetone (propanone) has six equivalent protons, which results in a
single sharp singlet in the 1H NMR spectrum. Propanal would show multiple signals,
including an aldehyde proton near 9-10 ppm and a triplet/quartet pattern for the ethyl
group. Cyclopropanol and allyl alcohol would show O-H signals and different C-H
environments. Therefore, the symmetry of acetone perfectly matches the single NMR signal
and the carbonyl IR stretch.
5. In 13C NMR, at approximately what chemical shift range do carbonyl carbons typically
appear?
A. 0-50 ppm
B. 160-220 ppm
C. 100-150 ppm
D. 50-100 ppm
Correct Answer: B
Expert Explanation: Carbonyl carbons are highly deshielded due to the electronegativity
of the oxygen atom and the double bond. They typically appear in the 160 to 220 ppm
range, which is distinct from most other carbon environments. Aldehydes and ketones
, usually appear further downfield (190-220 ppm) compared to esters and amides (160-185
ppm). Aliphatic carbons usually appear below 50 ppm, while aromatic carbons are found
between 100 and 150 ppm. Identifying the 13C carbonyl signal is vital for confirming the
oxidation state of the carbon.
6. The presence of two peaks of roughly equal intensity at M+ and M+2 in a mass spectrum
indicates the presence of which element?
A. Chlorine
B. Bromine
C. Nitrogen
D. Sulfur
Correct Answer: B
Expert Explanation: Bromine has two naturally occurring isotopes, 79Br and 81Br, which
exist in an approximate 1:1 ratio. This results in a mass spectrum showing two molecular
ion peaks separated by two mass units with nearly equal heights. Chlorine also shows an
M+2 peak, but its ratio is 3:1 due to the abundance of 35Cl and 37Cl. Nitrogen and sulfur
have different isotopic signatures that do not produce a 1:1 M+ and M+2 pattern. This
isotope pattern is a definitive diagnostic tool in mass spectrometry for halogen
identification.
7. Where would you expect to find the C-H stretching vibrations for an sp2 hybridized carbon
(alkene) in an IR spectrum?
A. Below 3000 cm-1
B. Above 3000 cm-1 (3000-3100 cm-1)
C. 1700-1750 cm-1
D. 2200-2300 cm-1
Correct Answer: B
Expert Explanation: C-H bonds involving sp2 hybridized carbons are stronger and stiffer
than those involving sp3 carbons. Consequently, these vibrations occur at higher
frequencies, typically appearing just above 3000 cm-1. In contrast, sp3 C-H stretches
appear just below 3000 cm-1, providing a useful distinction for identifying unsaturation.
Carbonyl stretches occur much lower, around 1700 cm-1, and triple bonds appear around
2200 cm-1. Lab students use this boundary at 3000 cm-1 to quickly assess whether a
compound is purely aliphatic or contains alkenes/aromatics.
8. What information is provided by the integration of signals in a 1H NMR spectrum?
A. The number of neighboring protons
B. The chemical environment of the carbon
