CHM 2210 Exam 2 V3 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 2) | University of Central Florida
1. Which of the following describes the rate-determining step in an SN1 reaction?
A. The attack of the nucleophile on the carbocation.
B. The deprotonation of the solvent molecule.
C. The dissociation of the leaving group to form a carbocation.
D. The backside attack of the nucleophile on the substrate.
Answer: C
Explanation: In an SN1 mechanism, the rate-determining step is unimolecular and
involves the departure of the leaving group. This step generates a high-energy carbocation
intermediate that is subsequently attacked by the nucleophile. Because this step has the
highest activation energy, the overall rate of the reaction depends only on the
concentration of the substrate.
2. Which factor most significantly increases the rate of an SN2 reaction?
A. Using a bulky base like potassium tert-butoxide.
B. Increasing the concentration of the nucleophile.
C. Switching from a polar aprotic solvent to a polar protic solvent.
D. Increasing the steric hindrance around the electrophilic carbon.
Answer: B
Explanation: The SN2 reaction follows second-order kinetics, meaning the rate is
proportional to both the substrate and the nucleophile concentrations. By increasing the
nucleophile concentration, the frequency of successful collisions with the substrate
increases. Conversely, steric hindrance and polar protic solvents generally decrease the
SN2 reaction rate by obstructing the backside attack or solvating the nucleophile.
3. What is the stereochemical relationship between a pair of enantiomers?
A. They are superimposable mirror images.
B. They are non-superimposable mirror images.
C. They have different physical properties like boiling point.
D. They are isomers that differ by rotation around a single bond.
Answer: B
,Explanation: Enantiomers are a specific type of stereoisomer that exist as non-
superimposable mirror images of one another. They possess identical physical properties,
such as melting point and solubility, in an achiral environment. However, they rotate plane-
polarized light in opposite directions and interact differently with other chiral molecules.
4. In the E2 elimination of 2-bromobutane with sodium ethoxide, what is the major product?
A. 1-butene
B. 2-ethoxybutane
C. cis-2-butene
D. trans-2-butene
Answer: D
Explanation: According to Zaitsev’s rule, the more substituted alkene is the major product
in an elimination reaction using a small base like ethoxide. Between the possible internal
alkenes, the trans-isomer is more stable than the cis-isomer due to reduced steric strain
between the methyl groups. Therefore, trans-2-butene is formed preferentially as the
thermodynamic product.
5. Which of the following molecules is a meso compound?
A. (2R, 3R)-dibromobutane
B. (2R, 3S)-dibromobutane
C. (2S, 3S)-dibromobutane
D. (2R, 3R)-2-bromo-3-chlorobutane
Answer: B
Explanation: A meso compound is a molecule that contains chiral centers but is achiral
overall due to an internal plane of symmetry. In (2R, 3S)-dibromobutane, the two halves of
the molecule are mirror images of each other, causing the optical activity to cancel out. This
results in a compound that does not rotate plane-polarized light despite having
stereocenters.
6. Which solvent is most suitable for an SN2 reaction between methyl iodide and sodium
cyanide?
A. Water
B. Dimethyl sulfoxide (DMSO)
C. Ethanol
D. Acetic acid
Answer: B
, Explanation: Polar aprotic solvents like DMSO, DMF, and acetone are ideal for SN2
reactions because they solvate cations well but leave anions relatively ‘naked’ and reactive.
Protic solvents like water or ethanol engage in hydrogen bonding with the nucleophile,
creating a solvent shell that hinders the attack on the substrate. Consequently, using DMSO
significantly enhances the nucleophilicity of the cyanide ion.
7. What is the expected stereochemical outcome of an SN2 reaction at a chiral center?
A. Retention of configuration
B. Complete inversion of configuration
C. Racemization
D. Formation of a meso compound
Answer: B
Explanation: The SN2 mechanism involves a backside attack by the nucleophile on the
carbon-leaving group bond. This simultaneous bond-breaking and bond-forming process
forces the other three substituents to ‘flip’ to the opposite side, much like an umbrella
blowing inside out. This result is known as Walden inversion, where the configuration of
the product is the opposite of the starting material.
8. Which of the following is the most stable carbocation?
A. CH3+
B. CH3CH2+
C. (CH3)2CH+
D. (CH3)3C+
Answer: D
Explanation: Carbocation stability increases with the degree of substitution due to
hyperconjugation and inductive effects from adjacent alkyl groups. A tertiary carbocation
like the tert-butyl cation is more stable than secondary, primary, or methyl carbocations
because it has three alkyl groups donating electron density to the vacant p-orbital. This
stability is a key factor in determining the speed and feasibility of SN1 and E1 reactions.
9. Identify the reagent that would convert 1-methylcyclohexene to 1-methylcyclohexanol via
Markovnikov addition without rearrangement.
A. 1. Hg(OAc)2, H2O; 2. NaBH4
B. H2O, H2SO4
C. 1. BH3-THF; 2. H2O2, NaOH
D. OsO4, NMO
with Rationale (CHM2210 Exam 2) | University of Central Florida
1. Which of the following describes the rate-determining step in an SN1 reaction?
A. The attack of the nucleophile on the carbocation.
B. The deprotonation of the solvent molecule.
C. The dissociation of the leaving group to form a carbocation.
D. The backside attack of the nucleophile on the substrate.
Answer: C
Explanation: In an SN1 mechanism, the rate-determining step is unimolecular and
involves the departure of the leaving group. This step generates a high-energy carbocation
intermediate that is subsequently attacked by the nucleophile. Because this step has the
highest activation energy, the overall rate of the reaction depends only on the
concentration of the substrate.
2. Which factor most significantly increases the rate of an SN2 reaction?
A. Using a bulky base like potassium tert-butoxide.
B. Increasing the concentration of the nucleophile.
C. Switching from a polar aprotic solvent to a polar protic solvent.
D. Increasing the steric hindrance around the electrophilic carbon.
Answer: B
Explanation: The SN2 reaction follows second-order kinetics, meaning the rate is
proportional to both the substrate and the nucleophile concentrations. By increasing the
nucleophile concentration, the frequency of successful collisions with the substrate
increases. Conversely, steric hindrance and polar protic solvents generally decrease the
SN2 reaction rate by obstructing the backside attack or solvating the nucleophile.
3. What is the stereochemical relationship between a pair of enantiomers?
A. They are superimposable mirror images.
B. They are non-superimposable mirror images.
C. They have different physical properties like boiling point.
D. They are isomers that differ by rotation around a single bond.
Answer: B
,Explanation: Enantiomers are a specific type of stereoisomer that exist as non-
superimposable mirror images of one another. They possess identical physical properties,
such as melting point and solubility, in an achiral environment. However, they rotate plane-
polarized light in opposite directions and interact differently with other chiral molecules.
4. In the E2 elimination of 2-bromobutane with sodium ethoxide, what is the major product?
A. 1-butene
B. 2-ethoxybutane
C. cis-2-butene
D. trans-2-butene
Answer: D
Explanation: According to Zaitsev’s rule, the more substituted alkene is the major product
in an elimination reaction using a small base like ethoxide. Between the possible internal
alkenes, the trans-isomer is more stable than the cis-isomer due to reduced steric strain
between the methyl groups. Therefore, trans-2-butene is formed preferentially as the
thermodynamic product.
5. Which of the following molecules is a meso compound?
A. (2R, 3R)-dibromobutane
B. (2R, 3S)-dibromobutane
C. (2S, 3S)-dibromobutane
D. (2R, 3R)-2-bromo-3-chlorobutane
Answer: B
Explanation: A meso compound is a molecule that contains chiral centers but is achiral
overall due to an internal plane of symmetry. In (2R, 3S)-dibromobutane, the two halves of
the molecule are mirror images of each other, causing the optical activity to cancel out. This
results in a compound that does not rotate plane-polarized light despite having
stereocenters.
6. Which solvent is most suitable for an SN2 reaction between methyl iodide and sodium
cyanide?
A. Water
B. Dimethyl sulfoxide (DMSO)
C. Ethanol
D. Acetic acid
Answer: B
, Explanation: Polar aprotic solvents like DMSO, DMF, and acetone are ideal for SN2
reactions because they solvate cations well but leave anions relatively ‘naked’ and reactive.
Protic solvents like water or ethanol engage in hydrogen bonding with the nucleophile,
creating a solvent shell that hinders the attack on the substrate. Consequently, using DMSO
significantly enhances the nucleophilicity of the cyanide ion.
7. What is the expected stereochemical outcome of an SN2 reaction at a chiral center?
A. Retention of configuration
B. Complete inversion of configuration
C. Racemization
D. Formation of a meso compound
Answer: B
Explanation: The SN2 mechanism involves a backside attack by the nucleophile on the
carbon-leaving group bond. This simultaneous bond-breaking and bond-forming process
forces the other three substituents to ‘flip’ to the opposite side, much like an umbrella
blowing inside out. This result is known as Walden inversion, where the configuration of
the product is the opposite of the starting material.
8. Which of the following is the most stable carbocation?
A. CH3+
B. CH3CH2+
C. (CH3)2CH+
D. (CH3)3C+
Answer: D
Explanation: Carbocation stability increases with the degree of substitution due to
hyperconjugation and inductive effects from adjacent alkyl groups. A tertiary carbocation
like the tert-butyl cation is more stable than secondary, primary, or methyl carbocations
because it has three alkyl groups donating electron density to the vacant p-orbital. This
stability is a key factor in determining the speed and feasibility of SN1 and E1 reactions.
9. Identify the reagent that would convert 1-methylcyclohexene to 1-methylcyclohexanol via
Markovnikov addition without rearrangement.
A. 1. Hg(OAc)2, H2O; 2. NaBH4
B. H2O, H2SO4
C. 1. BH3-THF; 2. H2O2, NaOH
D. OsO4, NMO
