CHM 2210 Final Exam V3 | CHM 2210 Organic Chemistry I | Actual
Q&A with Rationale (CHM2210 Final Exam) | University of Central
Florida
1. What is the hybridization of the central carbon in allene (CH2=C=CH2)?
A. sp
B. sp3
C. sp2
D. p
Answer: A
Explanation: The central carbon in allene is bonded to two carbon atoms via two double
bonds. According to VSEPR theory, a carbon forming two pi bonds must be sp hybridized to
accommodate the linear geometry. The p-orbitals used for the pi bonds are perpendicular
to each other, resulting in the terminal CH2 groups being in different planes.
2. Which of the following represents the most stable conformation of n-butane?
A. Gauche
B. Anti
C. Eclipsed
D. Partially eclipsed
Answer: B
Explanation: The anti conformation places the two large methyl groups 180 degrees apart,
minimizing steric strain. This staggered arrangement represents the global energy
minimum for n-butane. Gauche conformations have a higher energy due to steric
interaction between the methyl groups at 60 degrees.
3. Which acid has the lowest pKa value?
A. Ethanol
B. Acetic acid
C. Trichloroacetic acid
D. Phenol
Answer: C
,Explanation: Trichloroacetic acid is the strongest acid in the list because of the inductive
effect of the three chlorine atoms. These electronegative atoms stabilize the conjugate base
by pulling electron density away from the carboxylate group. A more stable conjugate base
corresponds to a stronger acid and thus a lower pKa value.
4. What is the IUPAC name for (CH3)2CHCH2CH(CH3)2?
A. Isoheptane
B. 2,3-dimethylpentane
C. 2,4-dimethylpentane
D. 2,4-dimethylhexane
Answer: C
Explanation: The longest continuous carbon chain consists of five carbons, making the
parent name pentane. There are two methyl substituents located at the 2nd and 4th
positions when numbered to give the lowest possible locants. Therefore, the correct name
is 2,4-dimethylpentane.
5. Which cyclohexane conformation is generally the most stable?
A. Boat
B. Twist-boat
C. Chair
D. Half-chair
Answer: C
Explanation: The chair conformation is the most stable because it minimizes both
torsional strain and steric strain. In the chair form, all C-H bonds are staggered, and there
are no eclipsed interactions. Other forms like the boat or half-chair suffer from significant
torsional strain and flagpole interactions.
6. A molecule with a non-superimposable mirror image is known as:
A. Achiral
B. Meso
C. Chiral
D. Constitutional
Answer: C
Explanation: Chirality is the geometric property of a molecule that makes it non-
superimposable on its mirror image. Such molecules typically contain at least one
, stereocenter, such as a carbon with four different groups attached. The relationship
between a chiral molecule and its mirror image defines a pair of enantiomers.
7. Which mechanism involves a carbocation intermediate?
A. SN2
B. SN1
C. E2
D. Finkelstein
Answer: B
Explanation: The SN1 mechanism occurs in two steps, the first being the slow dissociation
of the leaving group to form a carbocation. This intermediate is planar and can be attacked
by a nucleophile from either side, often leading to racemization. Stability of this
carbocation determines the rate of the reaction.
8. What is the product of the reaction between propene and HBr in the presence of
peroxides?
A. 2-bromopropane
B. 1,2-dibromopropane
C. 1-bromopropane
D. Propan-2-ol
Answer: C
Explanation: The presence of peroxides initiates a radical mechanism rather than the
standard ionic pathway. This leads to anti-Markovnikov addition, where the bromine atom
adds to the less substituted carbon. Consequently, 1-bromopropane is the major product
instead of the Markovnikov product, 2-bromopropane.
9. Which reagent converts an alkene to a cis-diol?
A. H2, Pd/C
B. mCPBA then H3O+
C. O3, DMS
D. OsO4, NMO
Answer: D
Explanation: Osmium tetroxide (OsO4) reacts with alkenes via a concerted syn-addition to
form a cyclic osmate ester. Subsequent hydrolysis or treatment with NMO yields a cis-1,2-
diol. This differs from anti-dihydroxylation, which is achieved through epoxide opening.
Q&A with Rationale (CHM2210 Final Exam) | University of Central
Florida
1. What is the hybridization of the central carbon in allene (CH2=C=CH2)?
A. sp
B. sp3
C. sp2
D. p
Answer: A
Explanation: The central carbon in allene is bonded to two carbon atoms via two double
bonds. According to VSEPR theory, a carbon forming two pi bonds must be sp hybridized to
accommodate the linear geometry. The p-orbitals used for the pi bonds are perpendicular
to each other, resulting in the terminal CH2 groups being in different planes.
2. Which of the following represents the most stable conformation of n-butane?
A. Gauche
B. Anti
C. Eclipsed
D. Partially eclipsed
Answer: B
Explanation: The anti conformation places the two large methyl groups 180 degrees apart,
minimizing steric strain. This staggered arrangement represents the global energy
minimum for n-butane. Gauche conformations have a higher energy due to steric
interaction between the methyl groups at 60 degrees.
3. Which acid has the lowest pKa value?
A. Ethanol
B. Acetic acid
C. Trichloroacetic acid
D. Phenol
Answer: C
,Explanation: Trichloroacetic acid is the strongest acid in the list because of the inductive
effect of the three chlorine atoms. These electronegative atoms stabilize the conjugate base
by pulling electron density away from the carboxylate group. A more stable conjugate base
corresponds to a stronger acid and thus a lower pKa value.
4. What is the IUPAC name for (CH3)2CHCH2CH(CH3)2?
A. Isoheptane
B. 2,3-dimethylpentane
C. 2,4-dimethylpentane
D. 2,4-dimethylhexane
Answer: C
Explanation: The longest continuous carbon chain consists of five carbons, making the
parent name pentane. There are two methyl substituents located at the 2nd and 4th
positions when numbered to give the lowest possible locants. Therefore, the correct name
is 2,4-dimethylpentane.
5. Which cyclohexane conformation is generally the most stable?
A. Boat
B. Twist-boat
C. Chair
D. Half-chair
Answer: C
Explanation: The chair conformation is the most stable because it minimizes both
torsional strain and steric strain. In the chair form, all C-H bonds are staggered, and there
are no eclipsed interactions. Other forms like the boat or half-chair suffer from significant
torsional strain and flagpole interactions.
6. A molecule with a non-superimposable mirror image is known as:
A. Achiral
B. Meso
C. Chiral
D. Constitutional
Answer: C
Explanation: Chirality is the geometric property of a molecule that makes it non-
superimposable on its mirror image. Such molecules typically contain at least one
, stereocenter, such as a carbon with four different groups attached. The relationship
between a chiral molecule and its mirror image defines a pair of enantiomers.
7. Which mechanism involves a carbocation intermediate?
A. SN2
B. SN1
C. E2
D. Finkelstein
Answer: B
Explanation: The SN1 mechanism occurs in two steps, the first being the slow dissociation
of the leaving group to form a carbocation. This intermediate is planar and can be attacked
by a nucleophile from either side, often leading to racemization. Stability of this
carbocation determines the rate of the reaction.
8. What is the product of the reaction between propene and HBr in the presence of
peroxides?
A. 2-bromopropane
B. 1,2-dibromopropane
C. 1-bromopropane
D. Propan-2-ol
Answer: C
Explanation: The presence of peroxides initiates a radical mechanism rather than the
standard ionic pathway. This leads to anti-Markovnikov addition, where the bromine atom
adds to the less substituted carbon. Consequently, 1-bromopropane is the major product
instead of the Markovnikov product, 2-bromopropane.
9. Which reagent converts an alkene to a cis-diol?
A. H2, Pd/C
B. mCPBA then H3O+
C. O3, DMS
D. OsO4, NMO
Answer: D
Explanation: Osmium tetroxide (OsO4) reacts with alkenes via a concerted syn-addition to
form a cyclic osmate ester. Subsequent hydrolysis or treatment with NMO yields a cis-1,2-
diol. This differs from anti-dihydroxylation, which is achieved through epoxide opening.
