CHM 2210 Final Exam V1 | CHM 2210 Organic Chemistry I | Actual
Q&A with Rationale (CHM2210 Final Exam) | University of Central
Florida
1. Which of the following molecules contains an sp2 hybridized carbon atom?
A. Ethane
B. Ethene
C. Ethyne
D. Methane
Answer: B
Explanation: In ethene, each carbon is double-bonded to another carbon and single-
bonded to two hydrogens. This results in three sigma bonds and one pi bond for each
carbon atom. Consequently, three sp2 hybrid orbitals are formed to accommodate the
trigonal planar geometry required for the sigma framework.
2. Which factor contributes most significantly to the increased acidity of carboxylic acids
compared to alcohols?
A. Inductive effect of the alkyl group
B. Resonance stabilization of the conjugate base
C. Electronegativity of the hydroxyl oxygen
D. Hybridization of the carbon atom
Answer: B
Explanation: Carboxylic acids lose a proton to form a carboxylate anion, which is
stabilized by resonance between two equivalent oxygen atoms. In contrast, the alkoxide ion
formed from an alcohol lacks this resonance stabilization, localizing the negative charge on
a single oxygen. This difference in stability makes the carboxylic acid much more willing to
donate a proton.
3. Identify the relationship between the two molecules: (R)-2-chlorobutane and (S)-2-
chlorobutane.
A. Constitutional isomers
B. Diastereomers
C. Enantiomers
D. Identical compounds
,Answer: C
Explanation: The (R) and (S) designations indicate that these molecules are non-
superimposable mirror images of each other. Because they have the same connectivity but
opposite configurations at every stereocenter, they fit the definition of enantiomers.
Diastereomers would require at least two stereocenters where at least one, but not all,
configurations differ.
4. In a Newman projection, which conformation of butane is the most stable?
A. Gauche
B. Eclipsed
C. Totally eclipsed
D. Anti
Answer: D
Explanation: The anti conformation places the two bulky methyl groups at a dihedral
angle of 180 degrees, minimizing steric repulsion. The gauche conformation involves a 60-
degree angle, which introduces some steric strain known as gauche interaction. Eclipsed
conformations are higher in energy due to both torsional strain and steric hindrance.
5. What is the major product of the reaction between 2-methyl-2-butene and HBr?
A. 2-bromo-2-methylbutane
B. 1-bromo-2-methylbutane
C. 2-bromo-3-methylbutane
D. 2,3-dibromo-2-methylbutane
Answer: A
Explanation: This reaction follows Markovnikov’s rule, where the electrophilic hydrogen
adds to the less substituted carbon to form the most stable carbocation. The 2-methyl-2-
butene substrate forms a tertiary carbocation at the C2 position. The bromide nucleophile
then attacks this tertiary carbocation, resulting in 2-bromo-2-methylbutane.
6. Which solvent is most suitable for an SN2 reaction?
A. Water
B. DMSO (Dimethyl sulfoxide)
C. Methanol
D. Ethanol
Answer: B
, Explanation: SN2 reactions are favored by polar aprotic solvents like DMSO because they
solvate cations well but leave anions relatively ‘naked’ and nucleophilic. Protic solvents like
water or alcohols engage in hydrogen bonding with the nucleophile, significantly reducing
its reactivity. Therefore, DMSO enhances the rate of the nucleophilic attack by not
hindering the anion.
7. According to Zaitsev’s rule, which alkene is the major product of an E2 elimination?
A. The least substituted alkene
B. The alkene with the least steric hindrance
C. The most substituted alkene
D. The terminal alkene
Answer: C
Explanation: Zaitsev’s rule states that in an elimination reaction, the most stable alkene is
formed preferentially. Generally, stability increases with the degree of substitution on the
double bond carbons due to hyperconjugation and electronic effects. However, bulky bases
like potassium tert-butoxide may favor the less substituted product, known as the
Hofmann product.
8. What is the hybridization of the oxygen atom in water?
A. sp
B. sp2
C. p
D. sp3
Answer: D
Explanation: The oxygen atom in water is surrounded by two lone pairs and two sigma
bonds to hydrogen. This gives a steric number of four, requiring four hybrid orbitals. The
resulting electron geometry is tetrahedral, which is achieved through sp3 hybridization.
9. Which of the following is a meso compound?
A. (2R, 3R)-2,3-dibromobutane
B. (2R, 3S)-2,3-dibromobutane
C. (2S, 3S)-2,3-dibromobutane
D. (2R, 3R)-2,3-pentanediol
Answer: B
Explanation: A meso compound is an achiral molecule that contains chiral centers but also
possesses an internal plane of symmetry. In (2R, 3S)-2,3-dibromobutane, the two halves of
Q&A with Rationale (CHM2210 Final Exam) | University of Central
Florida
1. Which of the following molecules contains an sp2 hybridized carbon atom?
A. Ethane
B. Ethene
C. Ethyne
D. Methane
Answer: B
Explanation: In ethene, each carbon is double-bonded to another carbon and single-
bonded to two hydrogens. This results in three sigma bonds and one pi bond for each
carbon atom. Consequently, three sp2 hybrid orbitals are formed to accommodate the
trigonal planar geometry required for the sigma framework.
2. Which factor contributes most significantly to the increased acidity of carboxylic acids
compared to alcohols?
A. Inductive effect of the alkyl group
B. Resonance stabilization of the conjugate base
C. Electronegativity of the hydroxyl oxygen
D. Hybridization of the carbon atom
Answer: B
Explanation: Carboxylic acids lose a proton to form a carboxylate anion, which is
stabilized by resonance between two equivalent oxygen atoms. In contrast, the alkoxide ion
formed from an alcohol lacks this resonance stabilization, localizing the negative charge on
a single oxygen. This difference in stability makes the carboxylic acid much more willing to
donate a proton.
3. Identify the relationship between the two molecules: (R)-2-chlorobutane and (S)-2-
chlorobutane.
A. Constitutional isomers
B. Diastereomers
C. Enantiomers
D. Identical compounds
,Answer: C
Explanation: The (R) and (S) designations indicate that these molecules are non-
superimposable mirror images of each other. Because they have the same connectivity but
opposite configurations at every stereocenter, they fit the definition of enantiomers.
Diastereomers would require at least two stereocenters where at least one, but not all,
configurations differ.
4. In a Newman projection, which conformation of butane is the most stable?
A. Gauche
B. Eclipsed
C. Totally eclipsed
D. Anti
Answer: D
Explanation: The anti conformation places the two bulky methyl groups at a dihedral
angle of 180 degrees, minimizing steric repulsion. The gauche conformation involves a 60-
degree angle, which introduces some steric strain known as gauche interaction. Eclipsed
conformations are higher in energy due to both torsional strain and steric hindrance.
5. What is the major product of the reaction between 2-methyl-2-butene and HBr?
A. 2-bromo-2-methylbutane
B. 1-bromo-2-methylbutane
C. 2-bromo-3-methylbutane
D. 2,3-dibromo-2-methylbutane
Answer: A
Explanation: This reaction follows Markovnikov’s rule, where the electrophilic hydrogen
adds to the less substituted carbon to form the most stable carbocation. The 2-methyl-2-
butene substrate forms a tertiary carbocation at the C2 position. The bromide nucleophile
then attacks this tertiary carbocation, resulting in 2-bromo-2-methylbutane.
6. Which solvent is most suitable for an SN2 reaction?
A. Water
B. DMSO (Dimethyl sulfoxide)
C. Methanol
D. Ethanol
Answer: B
, Explanation: SN2 reactions are favored by polar aprotic solvents like DMSO because they
solvate cations well but leave anions relatively ‘naked’ and nucleophilic. Protic solvents like
water or alcohols engage in hydrogen bonding with the nucleophile, significantly reducing
its reactivity. Therefore, DMSO enhances the rate of the nucleophilic attack by not
hindering the anion.
7. According to Zaitsev’s rule, which alkene is the major product of an E2 elimination?
A. The least substituted alkene
B. The alkene with the least steric hindrance
C. The most substituted alkene
D. The terminal alkene
Answer: C
Explanation: Zaitsev’s rule states that in an elimination reaction, the most stable alkene is
formed preferentially. Generally, stability increases with the degree of substitution on the
double bond carbons due to hyperconjugation and electronic effects. However, bulky bases
like potassium tert-butoxide may favor the less substituted product, known as the
Hofmann product.
8. What is the hybridization of the oxygen atom in water?
A. sp
B. sp2
C. p
D. sp3
Answer: D
Explanation: The oxygen atom in water is surrounded by two lone pairs and two sigma
bonds to hydrogen. This gives a steric number of four, requiring four hybrid orbitals. The
resulting electron geometry is tetrahedral, which is achieved through sp3 hybridization.
9. Which of the following is a meso compound?
A. (2R, 3R)-2,3-dibromobutane
B. (2R, 3S)-2,3-dibromobutane
C. (2S, 3S)-2,3-dibromobutane
D. (2R, 3R)-2,3-pentanediol
Answer: B
Explanation: A meso compound is an achiral molecule that contains chiral centers but also
possesses an internal plane of symmetry. In (2R, 3S)-2,3-dibromobutane, the two halves of
