CHM 2210 Exam 2 V2 | CHM 2210 Organic Chemistry I | Actual Q&A
with Rationale (CHM2210 Exam 2) | University of Central Florida
1. Which of the following describes the relationship between a pair of molecules that are
non-superimposable mirror images of each other?
A. Constitutional isomers
B. Diastereomers
C. Enantiomers
D. Meso compounds
Answer: C
Explanation: Enantiomers are defined as stereoisomers that are non-superimposable
mirror images of one another. This relationship requires at least one chiral center and the
absence of an internal plane of symmetry. Understanding this distinction is fundamental
for predicting the optical activity of organic samples.
2. Determine the maximum number of stereoisomers possible for a molecule with 3 distinct
chiral centers.
A. 8
B. 6
C. 3
D. 9
Answer: A
Explanation: The maximum number of stereoisomers for a compound is calculated using
the formula 2^n, where n is the number of stereocenters. For a molecule with 3 chiral
centers, 2^3 equals 8 total possible stereoisomers. If the molecule contains symmetry, the
actual number of stereoisomers may be fewer due to meso compounds.
3. In the Cahn-Ingold-Prelog system, which of the following groups has the highest priority?
A. -CH2OH
B. -CHO
C. -CH2SH
D. -OCH3
Answer: D
,Explanation: Priority is assigned based on the atomic number of the atom directly
attached to the chiral center. Oxygen (atomic number 8) has a higher priority than Carbon
(atomic number 6). Between -OCH3 and -CH2OH, the oxygen is directly attached in -OCH3,
whereas a carbon is directly attached in -CH2OH.
4. Which reagent is used for the anti-Markovnikov hydration of an alkene?
A. H3O+
B. Hg(OAc)2, H2O followed by NaBH4
C. BH3-THF followed by H2O2, NaOH
D. OsO4 followed by NaHSO3
Answer: C
Explanation: Hydroboration-oxidation is the standard method for achieving anti-
Markovnikov addition of water across a double bond. The boron atom adds to the less
substituted carbon, and the subsequent oxidation replaces it with a hydroxyl group. This
reaction is also stereospecific, resulting in syn-addition of the H and OH.
5. What is the degree of unsaturation for a compound with the molecular formula C6H8Br2?
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: The degree of unsaturation is calculated as (2C + 2 + N - H - X)/2. For
C6H8Br2, the calculation is (2*6 + 2 - 8 - 2)/2, which simplifies to (14 - 10)/2 = 2. A degree
of unsaturation of 2 indicates the presence of two rings, two double bonds, or one triple
bond.
6. Which of the following alkenes is the most stable?
A. Ethylene
B. cis-2-Butene
C. trans-2-Butene
D. 2,3-Dimethyl-2-butene
Answer: D
Explanation: Alkene stability increases with the degree of substitution due to
hyperconjugation and electronic effects. 2,3-Dimethyl-2-butene is a tetrasubstituted
, alkene, making it more stable than mono-, di-, or tri-substituted alkenes. Furthermore,
trans isomers are generally more stable than cis isomers due to reduced steric strain.
7. What is the major product formed when 1-methylcyclohexene reacts with HBr in the
presence of peroxides?
A. 1-bromo-2-methylcyclohexane
B. 1-bromo-1-methylcyclohexane
C. 2-bromo-1-methylcyclohexane
D. 1,2-dibromo-1-methylcyclohexane
Answer: A
Explanation: Reaction of an alkene with HBr in the presence of peroxides leads to anti-
Markovnikov addition via a radical mechanism. The bromine radical adds to the less
substituted carbon to form the more stable tertiary radical at the 1-position. Therefore, the
bromine ends up on the carbon adjacent to the methyl group.
8. Which solvent is best suited for an E2 reaction to promote the formation of the Zaitsev
product?
A. Water
B. Hexane
C. Dimethyl sulfoxide (DMSO)
D. Ethanol
Answer: D
Explanation: E2 reactions are typically favored by strong bases in protic solvents like
ethanol when using ethoxide. While polar aprotic solvents like DMSO are excellent for SN2,
E2 often employs the conjugate acid of the base as the solvent. The Zaitsev product is the
more substituted, more stable alkene formed preferentially.
9. Which of the following is a meso compound?
A. (2R, 3R)-2,3-dibromobutane
B. (2R, 3R)-2,3-pentanediol
C. (2S, 3S)-2,3-dibromobutane
D. (2R, 3S)-2,3-dibromobutane
Answer: D
Explanation: A meso compound contains chiral centers but is achiral overall due to an
internal plane of symmetry. In (2R, 3S)-2,3-dibromobutane, the two halves of the molecule
with Rationale (CHM2210 Exam 2) | University of Central Florida
1. Which of the following describes the relationship between a pair of molecules that are
non-superimposable mirror images of each other?
A. Constitutional isomers
B. Diastereomers
C. Enantiomers
D. Meso compounds
Answer: C
Explanation: Enantiomers are defined as stereoisomers that are non-superimposable
mirror images of one another. This relationship requires at least one chiral center and the
absence of an internal plane of symmetry. Understanding this distinction is fundamental
for predicting the optical activity of organic samples.
2. Determine the maximum number of stereoisomers possible for a molecule with 3 distinct
chiral centers.
A. 8
B. 6
C. 3
D. 9
Answer: A
Explanation: The maximum number of stereoisomers for a compound is calculated using
the formula 2^n, where n is the number of stereocenters. For a molecule with 3 chiral
centers, 2^3 equals 8 total possible stereoisomers. If the molecule contains symmetry, the
actual number of stereoisomers may be fewer due to meso compounds.
3. In the Cahn-Ingold-Prelog system, which of the following groups has the highest priority?
A. -CH2OH
B. -CHO
C. -CH2SH
D. -OCH3
Answer: D
,Explanation: Priority is assigned based on the atomic number of the atom directly
attached to the chiral center. Oxygen (atomic number 8) has a higher priority than Carbon
(atomic number 6). Between -OCH3 and -CH2OH, the oxygen is directly attached in -OCH3,
whereas a carbon is directly attached in -CH2OH.
4. Which reagent is used for the anti-Markovnikov hydration of an alkene?
A. H3O+
B. Hg(OAc)2, H2O followed by NaBH4
C. BH3-THF followed by H2O2, NaOH
D. OsO4 followed by NaHSO3
Answer: C
Explanation: Hydroboration-oxidation is the standard method for achieving anti-
Markovnikov addition of water across a double bond. The boron atom adds to the less
substituted carbon, and the subsequent oxidation replaces it with a hydroxyl group. This
reaction is also stereospecific, resulting in syn-addition of the H and OH.
5. What is the degree of unsaturation for a compound with the molecular formula C6H8Br2?
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: The degree of unsaturation is calculated as (2C + 2 + N - H - X)/2. For
C6H8Br2, the calculation is (2*6 + 2 - 8 - 2)/2, which simplifies to (14 - 10)/2 = 2. A degree
of unsaturation of 2 indicates the presence of two rings, two double bonds, or one triple
bond.
6. Which of the following alkenes is the most stable?
A. Ethylene
B. cis-2-Butene
C. trans-2-Butene
D. 2,3-Dimethyl-2-butene
Answer: D
Explanation: Alkene stability increases with the degree of substitution due to
hyperconjugation and electronic effects. 2,3-Dimethyl-2-butene is a tetrasubstituted
, alkene, making it more stable than mono-, di-, or tri-substituted alkenes. Furthermore,
trans isomers are generally more stable than cis isomers due to reduced steric strain.
7. What is the major product formed when 1-methylcyclohexene reacts with HBr in the
presence of peroxides?
A. 1-bromo-2-methylcyclohexane
B. 1-bromo-1-methylcyclohexane
C. 2-bromo-1-methylcyclohexane
D. 1,2-dibromo-1-methylcyclohexane
Answer: A
Explanation: Reaction of an alkene with HBr in the presence of peroxides leads to anti-
Markovnikov addition via a radical mechanism. The bromine radical adds to the less
substituted carbon to form the more stable tertiary radical at the 1-position. Therefore, the
bromine ends up on the carbon adjacent to the methyl group.
8. Which solvent is best suited for an E2 reaction to promote the formation of the Zaitsev
product?
A. Water
B. Hexane
C. Dimethyl sulfoxide (DMSO)
D. Ethanol
Answer: D
Explanation: E2 reactions are typically favored by strong bases in protic solvents like
ethanol when using ethoxide. While polar aprotic solvents like DMSO are excellent for SN2,
E2 often employs the conjugate acid of the base as the solvent. The Zaitsev product is the
more substituted, more stable alkene formed preferentially.
9. Which of the following is a meso compound?
A. (2R, 3R)-2,3-dibromobutane
B. (2R, 3R)-2,3-pentanediol
C. (2S, 3S)-2,3-dibromobutane
D. (2R, 3S)-2,3-dibromobutane
Answer: D
Explanation: A meso compound contains chiral centers but is achiral overall due to an
internal plane of symmetry. In (2R, 3S)-2,3-dibromobutane, the two halves of the molecule
