CHM 2210 Exam 2 V1 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 2) | University of Central Florida
1. Which of the following describes a pair of enantiomers?
A. Molecules that are superimposable mirror images.
B. Molecules that are non-mirror image stereoisomers.
C. Molecules that have different molecular formulas.
D. Molecules that are non-superimposable mirror images.
Answer: D
Explanation: Enantiomers are a specific type of stereoisomer where the molecules are
related as mirror images that cannot be superimposed. This characteristic arises from the
presence of one or more chiral centers within the molecular structure. Understanding the
distinction between enantiomers and diastereomers is critical for predicting biological
activity and optical properties.
2. Assign the absolute configuration (R or S) to the chiral center of (S)-2-chlorobutane if the
chlorine atom is replaced by a hydroxyl group without changing the spatial arrangement.
A. R
B. S
C. Racemic
D. Achiral
Answer: B
Explanation: The configuration remains S because the priority sequence of the
substituents does not change when replacing chlorine with a hydroxyl group. Both chlorine
and oxygen have higher priority than the ethyl and methyl groups, maintaining the
counter-clockwise orientation of the Cahn-Ingold-Prelog priorities. This demonstrates how
molecular substitution affects nomenclature while preserving stereochemical integrity.
3. Which statement is true regarding a meso compound?
A. It is a chiral molecule with no planes of symmetry.
B. It contains chiral centers but is achiral due to an internal plane of symmetry.
C. It rotates plane-polarized light to the right.
D. It is always a liquid at room temperature.
Answer: B
,Explanation: A meso compound is characterized by having multiple stereocenters while
remaining optically inactive as a whole. This occurs because the internal plane of symmetry
allows the two halves of the molecule to cancel out each other’s optical rotation. Students
must identify these symmetry elements to correctly classify stereoisomers in exams.
4. What is the relationship between (2R,3R)-2,3-dibromobutane and (2R,3S)-2,3-
dibromobutane?
A. Enantiomers
B. Constitutional isomers
C. Identical compounds
D. Diastereomers
Answer: D
Explanation: Diastereomers are stereoisomers that are not mirror images of each other,
occurring when at least one but not all stereocenters differ in configuration. In this case,
the configuration at C3 is inverted while C2 remains the same, precluding a mirror-image
relationship. This differentiation is a core concept tested in the UCF organic chemistry
curriculum.
5. Calculate the degrees of unsaturation for a molecule with the molecular formula C6H10.
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: The formula for degrees of unsaturation is (2C + 2 - H)/2, which for C6H10
yields (12 + 2 - 10)/2 = 2. This value indicates the presence of either two double bonds, one
triple bond, two rings, or one ring and one double bond. Identifying unsaturation is the first
step in determining the structure of unknown organic compounds.
6. Which of the following is the most stable carbocation?
A. Primary carbocation
B. Tertiary carbocation
C. Secondary carbocation
D. Methyl carbocation
Answer: B
, Explanation: Tertiary carbocations are the most stable because of the inductive effect and
hyperconjugation provided by three surrounding alkyl groups. These factors help
delocalize the positive charge, lowering the overall energy of the intermediate. Stabilization
trends are vital for predicting the regiochemistry of SN1 and E1 reactions.
7. What is the rate-determining step in an SN1 reaction?
A. Attack of the nucleophile on the alkyl halide.
B. Rearrangement of the carbocation.
C. Deprotonation of the intermediate.
D. Loss of the leaving group to form a carbocation.
Answer: D
Explanation: The SN1 mechanism proceeds via a unimolecular rate-determining step
involving the heterolytic cleavage of the bond between carbon and the leaving group. This
step is slow because it requires the formation of a high-energy carbocation intermediate.
Consequently, the rate of reaction depends solely on the concentration of the substrate.
8. Which solvent is most favorable for an SN2 reaction?
A. Water
B. Dimethyl sulfoxide (DMSO)
C. Ethanol
D. Methanol
Answer: B
Explanation: Polar aprotic solvents like DMSO are ideal for SN2 reactions because they
solvate cations well but do not form strong hydrogen bonds with anions. This leaves the
nucleophile ‘naked’ and more reactive toward the electrophilic carbon. In contrast, protic
solvents hinder nucleophilicity by surrounding the nucleophile with a solvent shell.
9. Which of the following alkyl halides reacts fastest in an SN2 reaction?
A. CH3Br
B. (CH3)2CHBr
C. (CH3)3CBr
D. CH3CH2Br
Answer: A
Explanation: SN2 reactions are highly sensitive to steric hindrance because the
nucleophile must attack from the back side of the electrophilic carbon. Methyl halides
with Rationale (CHM2210 Exam 2) | University of Central Florida
1. Which of the following describes a pair of enantiomers?
A. Molecules that are superimposable mirror images.
B. Molecules that are non-mirror image stereoisomers.
C. Molecules that have different molecular formulas.
D. Molecules that are non-superimposable mirror images.
Answer: D
Explanation: Enantiomers are a specific type of stereoisomer where the molecules are
related as mirror images that cannot be superimposed. This characteristic arises from the
presence of one or more chiral centers within the molecular structure. Understanding the
distinction between enantiomers and diastereomers is critical for predicting biological
activity and optical properties.
2. Assign the absolute configuration (R or S) to the chiral center of (S)-2-chlorobutane if the
chlorine atom is replaced by a hydroxyl group without changing the spatial arrangement.
A. R
B. S
C. Racemic
D. Achiral
Answer: B
Explanation: The configuration remains S because the priority sequence of the
substituents does not change when replacing chlorine with a hydroxyl group. Both chlorine
and oxygen have higher priority than the ethyl and methyl groups, maintaining the
counter-clockwise orientation of the Cahn-Ingold-Prelog priorities. This demonstrates how
molecular substitution affects nomenclature while preserving stereochemical integrity.
3. Which statement is true regarding a meso compound?
A. It is a chiral molecule with no planes of symmetry.
B. It contains chiral centers but is achiral due to an internal plane of symmetry.
C. It rotates plane-polarized light to the right.
D. It is always a liquid at room temperature.
Answer: B
,Explanation: A meso compound is characterized by having multiple stereocenters while
remaining optically inactive as a whole. This occurs because the internal plane of symmetry
allows the two halves of the molecule to cancel out each other’s optical rotation. Students
must identify these symmetry elements to correctly classify stereoisomers in exams.
4. What is the relationship between (2R,3R)-2,3-dibromobutane and (2R,3S)-2,3-
dibromobutane?
A. Enantiomers
B. Constitutional isomers
C. Identical compounds
D. Diastereomers
Answer: D
Explanation: Diastereomers are stereoisomers that are not mirror images of each other,
occurring when at least one but not all stereocenters differ in configuration. In this case,
the configuration at C3 is inverted while C2 remains the same, precluding a mirror-image
relationship. This differentiation is a core concept tested in the UCF organic chemistry
curriculum.
5. Calculate the degrees of unsaturation for a molecule with the molecular formula C6H10.
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: The formula for degrees of unsaturation is (2C + 2 - H)/2, which for C6H10
yields (12 + 2 - 10)/2 = 2. This value indicates the presence of either two double bonds, one
triple bond, two rings, or one ring and one double bond. Identifying unsaturation is the first
step in determining the structure of unknown organic compounds.
6. Which of the following is the most stable carbocation?
A. Primary carbocation
B. Tertiary carbocation
C. Secondary carbocation
D. Methyl carbocation
Answer: B
, Explanation: Tertiary carbocations are the most stable because of the inductive effect and
hyperconjugation provided by three surrounding alkyl groups. These factors help
delocalize the positive charge, lowering the overall energy of the intermediate. Stabilization
trends are vital for predicting the regiochemistry of SN1 and E1 reactions.
7. What is the rate-determining step in an SN1 reaction?
A. Attack of the nucleophile on the alkyl halide.
B. Rearrangement of the carbocation.
C. Deprotonation of the intermediate.
D. Loss of the leaving group to form a carbocation.
Answer: D
Explanation: The SN1 mechanism proceeds via a unimolecular rate-determining step
involving the heterolytic cleavage of the bond between carbon and the leaving group. This
step is slow because it requires the formation of a high-energy carbocation intermediate.
Consequently, the rate of reaction depends solely on the concentration of the substrate.
8. Which solvent is most favorable for an SN2 reaction?
A. Water
B. Dimethyl sulfoxide (DMSO)
C. Ethanol
D. Methanol
Answer: B
Explanation: Polar aprotic solvents like DMSO are ideal for SN2 reactions because they
solvate cations well but do not form strong hydrogen bonds with anions. This leaves the
nucleophile ‘naked’ and more reactive toward the electrophilic carbon. In contrast, protic
solvents hinder nucleophilicity by surrounding the nucleophile with a solvent shell.
9. Which of the following alkyl halides reacts fastest in an SN2 reaction?
A. CH3Br
B. (CH3)2CHBr
C. (CH3)3CBr
D. CH3CH2Br
Answer: A
Explanation: SN2 reactions are highly sensitive to steric hindrance because the
nucleophile must attack from the back side of the electrophilic carbon. Methyl halides
