CHM 2210 Exam 3 V3 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 3) | University of Central Florida
1. Which of the following describes the rate-determining step in an SN1 reaction mechanism?
A. The attack of the nucleophile on the electrophilic carbon.
B. The simultaneous formation of a bond and breaking of a bond.
C. The dissociation of the leaving group to form a carbocation intermediate.
D. The deprotonation of the solvent by a base.
Correct Answer: C
Explanation: In an SN1 mechanism, the rate-determining step is the unimolecular
ionization of the substrate to form a carbocation. This step has the highest activation
energy because it involves breaking a covalent bond without forming a new one. The rate
of the reaction is therefore independent of the concentration of the nucleophile.
2. When (S)-2-bromobutane reacts with sodium cyanide in DMSO, what is the stereochemical
outcome?
A. Racemization of the stereocenter.
B. Inversion of configuration to yield (R)-2-cyanobutane.
C. Retention of configuration.
D. Formation of a meso compound.
Correct Answer: B
Explanation: The reaction between a secondary alkyl halide and a strong nucleophile like
cyanide in a polar aprotic solvent like DMSO favors the SN2 mechanism. SN2 reactions
proceed via backside attack, which results in a complete inversion of configuration at the
chiral center. Therefore, the (S) starting material is converted specifically into the (R)
product.
3. Which of the following alkyl halides will react fastest in an SN2 reaction?
A. tert-Butyl fluoride
B. Ethyl bromide
C. Isopropyl chloride
D. Methyl iodide
Correct Answer: D
,Explanation: SN2 reactivity is governed by steric hindrance and the quality of the leaving
group. Methyl halides are the least sterically hindered, allowing for the easiest backside
attack by a nucleophile. Furthermore, iodide is a superior leaving group compared to
bromide, chloride, or fluoride due to its large size and low basicity.
4. What is the effect of using a polar aprotic solvent like DMF on an SN2 reaction?
A. It increases the reaction rate by leaving the nucleophile relatively ‘naked’ and more
reactive.
B. It slows down the reaction by solvating the nucleophile through hydrogen bonding.
C. It has no effect on the rate because the solvent does not appear in the rate law.
D. It promotes carbocation formation, favoring the SN1 pathway.
Correct Answer: A
Explanation: Polar aprotic solvents such as DMF or DMSO do not have acidic hydrogens
and therefore cannot form hydrogen bonds with anions. This lack of solvation makes the
nucleophile more energetic and ‘naked,’ significantly increasing its reactivity in SN2
processes. Conversely, polar protic solvents stabilize nucleophiles, thereby raising the
activation energy for the reaction.
5. Which rule predicts that the most substituted alkene will be the major product in an
elimination reaction?
A. Hund’s Rule
B. Markovnikov’s Rule
C. Hofmann’s Rule
D. Zaitsev’s Rule
Correct Answer: D
Explanation: Zaitsev’s rule states that in an elimination reaction, the most stable alkene is
formed as the major product. Stability in alkenes is generally correlated with the degree of
substitution, meaning more alkyl groups attached to the double bond lead to higher
stability. This is typically observed in E2 reactions using small, non-bulky bases.
6. Which of the following bases would favor the formation of the Hofmann product in an E2
reaction of 2-bromo-2-methylbutane?
A. Sodium methoxide
B. Sodium hydroxide
C. Sodium ethoxide
D. Potassium tert-butoxide
, Correct Answer: D
Explanation: Potassium tert-butoxide is a sterically bulky base that finds it difficult to
access internal, more hindered protons. As a result, it tends to abstract the most accessible
primary protons on the periphery of the molecule. This leads to the formation of the less
substituted alkene, known as the Hofmann product.
7. What is the expected major product when 1-bromo-1-methylcyclohexane is treated with a
strong, bulky base?
A. 1-methylcyclohexene
B. Methylenecyclohexane
C. 1-methylcyclohexanol
D. 1-methoxy-1-methylcyclohexane
Correct Answer: B
Explanation: When a tertiary alkyl halide reacts with a bulky base like t-butoxide, the E2
mechanism occurs. The bulky base prefers to abstract a proton from the less hindered
methyl group rather than the ring carbons. This results in the exocyclic double bond,
forming methylenecyclohexane as the major Hofmann product.
8. In the dehydration of alcohols using concentrated sulfuric acid, which type of alcohol reacts
the fastest?
A. Primary alcohols
B. Secondary alcohols
C. Tertiary alcohols
D. Methanol
Correct Answer: C
Explanation: Acid-catalyzed dehydration follows an E1 mechanism for secondary and
tertiary alcohols, involving the formation of a carbocation. Tertiary carbocations are the
most stable and therefore form the fastest due to the lower activation energy of the
transition state. Primary alcohols react much slower because they must proceed through
an E2-like mechanism to avoid unstable primary carbocations.
9. Which reagent is best suited for converting a primary alcohol to an alkyl chloride with
minimal rearrangement?
A. SOCl2 and Pyridine
B. NaCl in Acetone
C. Concentrated HCl
with Rationale (CHM2210 Exam 3) | University of Central Florida
1. Which of the following describes the rate-determining step in an SN1 reaction mechanism?
A. The attack of the nucleophile on the electrophilic carbon.
B. The simultaneous formation of a bond and breaking of a bond.
C. The dissociation of the leaving group to form a carbocation intermediate.
D. The deprotonation of the solvent by a base.
Correct Answer: C
Explanation: In an SN1 mechanism, the rate-determining step is the unimolecular
ionization of the substrate to form a carbocation. This step has the highest activation
energy because it involves breaking a covalent bond without forming a new one. The rate
of the reaction is therefore independent of the concentration of the nucleophile.
2. When (S)-2-bromobutane reacts with sodium cyanide in DMSO, what is the stereochemical
outcome?
A. Racemization of the stereocenter.
B. Inversion of configuration to yield (R)-2-cyanobutane.
C. Retention of configuration.
D. Formation of a meso compound.
Correct Answer: B
Explanation: The reaction between a secondary alkyl halide and a strong nucleophile like
cyanide in a polar aprotic solvent like DMSO favors the SN2 mechanism. SN2 reactions
proceed via backside attack, which results in a complete inversion of configuration at the
chiral center. Therefore, the (S) starting material is converted specifically into the (R)
product.
3. Which of the following alkyl halides will react fastest in an SN2 reaction?
A. tert-Butyl fluoride
B. Ethyl bromide
C. Isopropyl chloride
D. Methyl iodide
Correct Answer: D
,Explanation: SN2 reactivity is governed by steric hindrance and the quality of the leaving
group. Methyl halides are the least sterically hindered, allowing for the easiest backside
attack by a nucleophile. Furthermore, iodide is a superior leaving group compared to
bromide, chloride, or fluoride due to its large size and low basicity.
4. What is the effect of using a polar aprotic solvent like DMF on an SN2 reaction?
A. It increases the reaction rate by leaving the nucleophile relatively ‘naked’ and more
reactive.
B. It slows down the reaction by solvating the nucleophile through hydrogen bonding.
C. It has no effect on the rate because the solvent does not appear in the rate law.
D. It promotes carbocation formation, favoring the SN1 pathway.
Correct Answer: A
Explanation: Polar aprotic solvents such as DMF or DMSO do not have acidic hydrogens
and therefore cannot form hydrogen bonds with anions. This lack of solvation makes the
nucleophile more energetic and ‘naked,’ significantly increasing its reactivity in SN2
processes. Conversely, polar protic solvents stabilize nucleophiles, thereby raising the
activation energy for the reaction.
5. Which rule predicts that the most substituted alkene will be the major product in an
elimination reaction?
A. Hund’s Rule
B. Markovnikov’s Rule
C. Hofmann’s Rule
D. Zaitsev’s Rule
Correct Answer: D
Explanation: Zaitsev’s rule states that in an elimination reaction, the most stable alkene is
formed as the major product. Stability in alkenes is generally correlated with the degree of
substitution, meaning more alkyl groups attached to the double bond lead to higher
stability. This is typically observed in E2 reactions using small, non-bulky bases.
6. Which of the following bases would favor the formation of the Hofmann product in an E2
reaction of 2-bromo-2-methylbutane?
A. Sodium methoxide
B. Sodium hydroxide
C. Sodium ethoxide
D. Potassium tert-butoxide
, Correct Answer: D
Explanation: Potassium tert-butoxide is a sterically bulky base that finds it difficult to
access internal, more hindered protons. As a result, it tends to abstract the most accessible
primary protons on the periphery of the molecule. This leads to the formation of the less
substituted alkene, known as the Hofmann product.
7. What is the expected major product when 1-bromo-1-methylcyclohexane is treated with a
strong, bulky base?
A. 1-methylcyclohexene
B. Methylenecyclohexane
C. 1-methylcyclohexanol
D. 1-methoxy-1-methylcyclohexane
Correct Answer: B
Explanation: When a tertiary alkyl halide reacts with a bulky base like t-butoxide, the E2
mechanism occurs. The bulky base prefers to abstract a proton from the less hindered
methyl group rather than the ring carbons. This results in the exocyclic double bond,
forming methylenecyclohexane as the major Hofmann product.
8. In the dehydration of alcohols using concentrated sulfuric acid, which type of alcohol reacts
the fastest?
A. Primary alcohols
B. Secondary alcohols
C. Tertiary alcohols
D. Methanol
Correct Answer: C
Explanation: Acid-catalyzed dehydration follows an E1 mechanism for secondary and
tertiary alcohols, involving the formation of a carbocation. Tertiary carbocations are the
most stable and therefore form the fastest due to the lower activation energy of the
transition state. Primary alcohols react much slower because they must proceed through
an E2-like mechanism to avoid unstable primary carbocations.
9. Which reagent is best suited for converting a primary alcohol to an alkyl chloride with
minimal rearrangement?
A. SOCl2 and Pyridine
B. NaCl in Acetone
C. Concentrated HCl
