CHM 2210 Exam 3 V2 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 3) | University of Central Florida
1. For a standard SN2 reaction, what is the effect on the reaction rate if the concentration of
the nucleophile is tripled and the concentration of the alkyl halide is doubled?
A. The rate increases by a factor of 5.
B. The rate remains unchanged.
C. The rate increases by a factor of 3.
D. The rate increases by a factor of 6.
Answer: D
Explanation: The SN2 mechanism follows second-order kinetics, meaning the rate law is
defined as Rate = k[Substrate][Nucleophile]. By tripling the nucleophile concentration and
doubling the substrate concentration, the rate is multiplied by 3 times 2. Therefore, the
overall rate of the reaction increases by a factor of 6.
2. Which of the following alkyl halides would react most rapidly in an SN2 reaction with
sodium ethoxide in ethanol?
A. 2-bromo-2-methylpropane
B. 1-bromopropane
C. 2-bromopropane
D. 1-bromo-2,2-dimethylpropane
Answer: B
Explanation: SN2 reactions are highly sensitive to steric hindrance because the
nucleophile must attack from the backside of the carbon-leaving group bond. 1-
bromopropane is a primary alkyl halide with minimal steric crowding, making it the most
reactive among the choices. In contrast, tertiary substrates like 2-bromo-2-methylpropane
are essentially inert to SN2 due to extreme steric repulsion.
3. What is the primary stereochemical outcome of an SN2 reaction at a chiral center?
A. Racemization of the stereocenter
B. Retention of configuration
C. Formation of a 2:1 mixture of diastereomers
D. Inversion of configuration
Answer: D
,Explanation: The SN2 mechanism involves a concerted backside attack by the nucleophile,
which forces the other three substituents to ‘umbrella’ over to the opposite side. This
results in the complete inversion of the absolute configuration at the electrophilic carbon
center. This stereospecificity is a hallmark of the bimolecular nucleophilic substitution
process.
4. Which of the following solvents is considered polar aprotic and would best promote an SN2
reaction?
A. Methanol
B. Water
C. Ammonia
D. Dimethylformamide (DMF)
Answer: D
Explanation: Polar aprotic solvents like DMF, DMSO, and acetone are ideal for SN2
reactions because they lack hydroxyl or amino groups that can hydrogen-bond to the
nucleophile. Protic solvents solvate nucleophiles strongly through hydrogen bonding,
creating a ‘solvent shell’ that reduces their reactivity. By using aprotic solvents, the
nucleophile remains ‘naked’ and more energetic, thus lowering the activation energy for
the reaction.
5. What is the correct order of leaving group ability for the following halogens?
A. F- > Cl- > Br- > I-
B. I- > Br- > Cl- > F-
C. Cl- > Br- > I- > F-
D. Br- > I- > Cl- > F-
Answer: B
Explanation: Leaving group ability is generally inversely proportional to basicity; weaker
bases make better leaving groups. The iodide ion is the largest and most polarizable halide,
making it a very weak base and an excellent leaving group. Fluoride is a small, highly basic
ion that holds onto the carbon atom tightly, making it a very poor leaving group in
substitution reactions.
6. Which mechanism is most likely for the reaction of tert-butyl bromide with methanol at
room temperature?
A. SN2
B. Addition
C. E2
, D. SN1
Answer: D
Explanation: Tert-butyl bromide is a tertiary alkyl halide, which precludes an SN2
mechanism due to steric hindrance. Methanol is a weak nucleophile and a protic solvent,
which favors the ionization of the substrate to form a stable carbocation. Therefore, the
reaction proceeds via the SN1 pathway, leading to the formation of tert-butyl methyl ether.
7. Which of the following carbocations is the most stable?
A. Secondary carbocation
B. Primary carbocation
C. Tertiary carbocation
D. Methyl carbocation
Answer: C
Explanation: Carbocation stability is determined by inductive effects and
hyperconjugation from adjacent alkyl groups. Tertiary carbocations have three alkyl
groups that donate electron density to the positively charged carbon, stabilizing it more
effectively than secondary or primary centers. The methyl carbocation is the least stable
because it lacks any stabilizing alkyl substituents.
8. What is the major product of the E2 elimination of 2-bromo-2-methylbutane using a small
base like sodium ethoxide?
A. 2-methyl-1-butene
B. 2-methyl-2-butene
C. 3-methyl-1-butene
D. 2-methyl-2-butanol
Answer: B
Explanation: According to Zaitsev’s rule, the major product of an elimination reaction is
the most substituted alkene. Sodium ethoxide is a relatively small base that can easily
access internal protons to form the more stable, trisubstituted alkene. Consequently, 2-
methyl-2-butene is the predominant Zaitsev product in this E2 reaction.
9. Which requirement must be met for an E2 elimination to occur in a cyclohexane ring
system?
A. The leaving group and the beta-hydrogen must be anti-periplanar (trans-diaxial).
B. The leaving group must be equatorial.
C. The base must attack the alpha-carbon directly.
with Rationale (CHM2210 Exam 3) | University of Central Florida
1. For a standard SN2 reaction, what is the effect on the reaction rate if the concentration of
the nucleophile is tripled and the concentration of the alkyl halide is doubled?
A. The rate increases by a factor of 5.
B. The rate remains unchanged.
C. The rate increases by a factor of 3.
D. The rate increases by a factor of 6.
Answer: D
Explanation: The SN2 mechanism follows second-order kinetics, meaning the rate law is
defined as Rate = k[Substrate][Nucleophile]. By tripling the nucleophile concentration and
doubling the substrate concentration, the rate is multiplied by 3 times 2. Therefore, the
overall rate of the reaction increases by a factor of 6.
2. Which of the following alkyl halides would react most rapidly in an SN2 reaction with
sodium ethoxide in ethanol?
A. 2-bromo-2-methylpropane
B. 1-bromopropane
C. 2-bromopropane
D. 1-bromo-2,2-dimethylpropane
Answer: B
Explanation: SN2 reactions are highly sensitive to steric hindrance because the
nucleophile must attack from the backside of the carbon-leaving group bond. 1-
bromopropane is a primary alkyl halide with minimal steric crowding, making it the most
reactive among the choices. In contrast, tertiary substrates like 2-bromo-2-methylpropane
are essentially inert to SN2 due to extreme steric repulsion.
3. What is the primary stereochemical outcome of an SN2 reaction at a chiral center?
A. Racemization of the stereocenter
B. Retention of configuration
C. Formation of a 2:1 mixture of diastereomers
D. Inversion of configuration
Answer: D
,Explanation: The SN2 mechanism involves a concerted backside attack by the nucleophile,
which forces the other three substituents to ‘umbrella’ over to the opposite side. This
results in the complete inversion of the absolute configuration at the electrophilic carbon
center. This stereospecificity is a hallmark of the bimolecular nucleophilic substitution
process.
4. Which of the following solvents is considered polar aprotic and would best promote an SN2
reaction?
A. Methanol
B. Water
C. Ammonia
D. Dimethylformamide (DMF)
Answer: D
Explanation: Polar aprotic solvents like DMF, DMSO, and acetone are ideal for SN2
reactions because they lack hydroxyl or amino groups that can hydrogen-bond to the
nucleophile. Protic solvents solvate nucleophiles strongly through hydrogen bonding,
creating a ‘solvent shell’ that reduces their reactivity. By using aprotic solvents, the
nucleophile remains ‘naked’ and more energetic, thus lowering the activation energy for
the reaction.
5. What is the correct order of leaving group ability for the following halogens?
A. F- > Cl- > Br- > I-
B. I- > Br- > Cl- > F-
C. Cl- > Br- > I- > F-
D. Br- > I- > Cl- > F-
Answer: B
Explanation: Leaving group ability is generally inversely proportional to basicity; weaker
bases make better leaving groups. The iodide ion is the largest and most polarizable halide,
making it a very weak base and an excellent leaving group. Fluoride is a small, highly basic
ion that holds onto the carbon atom tightly, making it a very poor leaving group in
substitution reactions.
6. Which mechanism is most likely for the reaction of tert-butyl bromide with methanol at
room temperature?
A. SN2
B. Addition
C. E2
, D. SN1
Answer: D
Explanation: Tert-butyl bromide is a tertiary alkyl halide, which precludes an SN2
mechanism due to steric hindrance. Methanol is a weak nucleophile and a protic solvent,
which favors the ionization of the substrate to form a stable carbocation. Therefore, the
reaction proceeds via the SN1 pathway, leading to the formation of tert-butyl methyl ether.
7. Which of the following carbocations is the most stable?
A. Secondary carbocation
B. Primary carbocation
C. Tertiary carbocation
D. Methyl carbocation
Answer: C
Explanation: Carbocation stability is determined by inductive effects and
hyperconjugation from adjacent alkyl groups. Tertiary carbocations have three alkyl
groups that donate electron density to the positively charged carbon, stabilizing it more
effectively than secondary or primary centers. The methyl carbocation is the least stable
because it lacks any stabilizing alkyl substituents.
8. What is the major product of the E2 elimination of 2-bromo-2-methylbutane using a small
base like sodium ethoxide?
A. 2-methyl-1-butene
B. 2-methyl-2-butene
C. 3-methyl-1-butene
D. 2-methyl-2-butanol
Answer: B
Explanation: According to Zaitsev’s rule, the major product of an elimination reaction is
the most substituted alkene. Sodium ethoxide is a relatively small base that can easily
access internal protons to form the more stable, trisubstituted alkene. Consequently, 2-
methyl-2-butene is the predominant Zaitsev product in this E2 reaction.
9. Which requirement must be met for an E2 elimination to occur in a cyclohexane ring
system?
A. The leaving group and the beta-hydrogen must be anti-periplanar (trans-diaxial).
B. The leaving group must be equatorial.
C. The base must attack the alpha-carbon directly.
