CHM 2210 Final Exam V2 | CHM 2210 Organic Chemistry I | Actual
Q&A with Rationale (CHM2210 Final Exam) | University of Central
Florida
1. What is the hybridization of the oxygen atom in dimethyl ether (CH3OCH3)?
A. sp
B. sp2
C. p
D. sp3
Answer: D
Explanation: In dimethyl ether, the central oxygen atom is bonded to two carbon atoms
and possesses two lone pairs of electrons. This results in a total of four electron domains
around the oxygen atom, which corresponds to a tetrahedral electron geometry.
Consequently, the oxygen atom must be sp3 hybridized to accommodate these four
domains while maintaining its VSEPR-predicted shape.
2. Which of the following molecules has a formal charge of +1 on the central nitrogen atom?
A. Methylamine (CH3NH2)
B. Ammonia (NH3)
C. Nitromethane (CH3NO2)
D. Hydrazine (N2H4)
Answer: C
Explanation: In nitromethane, the nitrogen atom is bonded to one methyl group and two
oxygen atoms, with one oxygen double-bonded and the other single-bonded. Applying the
formal charge formula (Valence - [Lone Pair Electrons + 0.5 * Bonding Electrons]) gives 5 -
(0 + 4) = +1. This positive charge is balanced by the negative charge on the single-bonded
oxygen atom to make the molecule neutral overall.
3. Rank the following compounds in order of increasing acidity: Ethanol, Phenol, Acetic Acid,
Trifluoroacetic Acid.
A. Ethanol < Acetic Acid < Phenol < Trifluoroacetic Acid
B. Ethanol < Phenol < Acetic Acid < Trifluoroacetic Acid
C. Phenol < Ethanol < Acetic Acid < Trifluoroacetic Acid
D. Trifluoroacetic Acid < Acetic Acid < Phenol < Ethanol
,Answer: B
Explanation: Ethanol is the least acidic because its conjugate base is an alkoxide with no
resonance stabilization. Phenol is more acidic due to the resonance stabilization of the
phenoxide ion within the aromatic ring. Trifluoroacetic acid is the most acidic because of
the strong inductive electron-withdrawing effect of the three fluorine atoms which
stabilizes the carboxylate group.
4. According to the IUPAC nomenclature, what is the correct name for the compound
(CH3)2CHCH2CH2Br?
A. 1-bromo-4-methylbutane
B. 4-bromo-2-methylbutane
C. Isobutyl bromide
D. 1-bromo-3-methylbutane
Answer: D
Explanation: The longest carbon chain containing the bromine substituent consists of four
carbons, making the parent alkane ‘butane’. Numbering begins from the end closest to the
first substituent, which is the bromine atom at position 1. Therefore, the methyl group is
located at carbon 3, leading to the name 1-bromo-3-methylbutane.
5. In the most stable conformation of n-butane, what is the dihedral angle between the two
methyl groups?
A. 0 degrees
B. 180 degrees
C. 120 degrees
D. 60 degrees
Answer: B
Explanation: The most stable conformation of n-butane is the anti-staggered
conformation. In this arrangement, the two large methyl groups are positioned as far apart
as possible to minimize steric strain. This results in a dihedral angle of 180 degrees, which
corresponds to the global energy minimum for the molecule.
6. Which of the following describes a ‘Meso’ compound?
A. A molecule with chiral centers and an internal plane of symmetry.
B. A molecule with no chiral centers that is optically active.
C. A mixture of equal amounts of two enantiomers.
D. A molecule that rotates plane-polarized light to the right.
, Answer: A
Explanation: Meso compounds contain two or more stereocenters but are achiral overall
due to an internal plane of symmetry. This symmetry causes the optical rotation of one half
of the molecule to cancel out the rotation of the other half. As a result, meso compounds do
not rotate plane-polarized light and are optically inactive.
7. What is the relationship between (2R,3R)-2,3-dibromobutane and (2S,3S)-2,3-
dibromobutane?
A. Identical
B. Diastereomers
C. Constitutional Isomers
D. Enantiomers
Answer: D
Explanation: Enantiomers are non-superimposable mirror images of each other,
characterized by having the opposite configuration at every chiral center. Since the
configurations are (2R,3R) and (2S,3S), every center has been inverted. These two
molecules will have identical physical properties except for the direction in which they
rotate plane-polarized light.
8. Which solvent is most favorable for an SN2 reaction?
A. DMSO (Dimethyl sulfoxide)
B. Methanol
C. Water
D. Acetic Acid
Answer: A
Explanation: SN2 reactions are favored by polar aprotic solvents like DMSO, DMF, or
acetone. These solvents dissolve the ionic nucleophile but do not form strong hydrogen
bonds with it, leaving the nucleophile ‘naked’ and more reactive. Polar protic solvents like
water or methanol solvate the nucleophile too strongly, which significantly hinders its
ability to attack the electrophile.
9. Which of the following alkyl halides will react fastest in an SN1 reaction?
A. tert-Butyl bromide
B. Ethyl bromide
C. Isopropyl bromide
D. Methyl bromide
Q&A with Rationale (CHM2210 Final Exam) | University of Central
Florida
1. What is the hybridization of the oxygen atom in dimethyl ether (CH3OCH3)?
A. sp
B. sp2
C. p
D. sp3
Answer: D
Explanation: In dimethyl ether, the central oxygen atom is bonded to two carbon atoms
and possesses two lone pairs of electrons. This results in a total of four electron domains
around the oxygen atom, which corresponds to a tetrahedral electron geometry.
Consequently, the oxygen atom must be sp3 hybridized to accommodate these four
domains while maintaining its VSEPR-predicted shape.
2. Which of the following molecules has a formal charge of +1 on the central nitrogen atom?
A. Methylamine (CH3NH2)
B. Ammonia (NH3)
C. Nitromethane (CH3NO2)
D. Hydrazine (N2H4)
Answer: C
Explanation: In nitromethane, the nitrogen atom is bonded to one methyl group and two
oxygen atoms, with one oxygen double-bonded and the other single-bonded. Applying the
formal charge formula (Valence - [Lone Pair Electrons + 0.5 * Bonding Electrons]) gives 5 -
(0 + 4) = +1. This positive charge is balanced by the negative charge on the single-bonded
oxygen atom to make the molecule neutral overall.
3. Rank the following compounds in order of increasing acidity: Ethanol, Phenol, Acetic Acid,
Trifluoroacetic Acid.
A. Ethanol < Acetic Acid < Phenol < Trifluoroacetic Acid
B. Ethanol < Phenol < Acetic Acid < Trifluoroacetic Acid
C. Phenol < Ethanol < Acetic Acid < Trifluoroacetic Acid
D. Trifluoroacetic Acid < Acetic Acid < Phenol < Ethanol
,Answer: B
Explanation: Ethanol is the least acidic because its conjugate base is an alkoxide with no
resonance stabilization. Phenol is more acidic due to the resonance stabilization of the
phenoxide ion within the aromatic ring. Trifluoroacetic acid is the most acidic because of
the strong inductive electron-withdrawing effect of the three fluorine atoms which
stabilizes the carboxylate group.
4. According to the IUPAC nomenclature, what is the correct name for the compound
(CH3)2CHCH2CH2Br?
A. 1-bromo-4-methylbutane
B. 4-bromo-2-methylbutane
C. Isobutyl bromide
D. 1-bromo-3-methylbutane
Answer: D
Explanation: The longest carbon chain containing the bromine substituent consists of four
carbons, making the parent alkane ‘butane’. Numbering begins from the end closest to the
first substituent, which is the bromine atom at position 1. Therefore, the methyl group is
located at carbon 3, leading to the name 1-bromo-3-methylbutane.
5. In the most stable conformation of n-butane, what is the dihedral angle between the two
methyl groups?
A. 0 degrees
B. 180 degrees
C. 120 degrees
D. 60 degrees
Answer: B
Explanation: The most stable conformation of n-butane is the anti-staggered
conformation. In this arrangement, the two large methyl groups are positioned as far apart
as possible to minimize steric strain. This results in a dihedral angle of 180 degrees, which
corresponds to the global energy minimum for the molecule.
6. Which of the following describes a ‘Meso’ compound?
A. A molecule with chiral centers and an internal plane of symmetry.
B. A molecule with no chiral centers that is optically active.
C. A mixture of equal amounts of two enantiomers.
D. A molecule that rotates plane-polarized light to the right.
, Answer: A
Explanation: Meso compounds contain two or more stereocenters but are achiral overall
due to an internal plane of symmetry. This symmetry causes the optical rotation of one half
of the molecule to cancel out the rotation of the other half. As a result, meso compounds do
not rotate plane-polarized light and are optically inactive.
7. What is the relationship between (2R,3R)-2,3-dibromobutane and (2S,3S)-2,3-
dibromobutane?
A. Identical
B. Diastereomers
C. Constitutional Isomers
D. Enantiomers
Answer: D
Explanation: Enantiomers are non-superimposable mirror images of each other,
characterized by having the opposite configuration at every chiral center. Since the
configurations are (2R,3R) and (2S,3S), every center has been inverted. These two
molecules will have identical physical properties except for the direction in which they
rotate plane-polarized light.
8. Which solvent is most favorable for an SN2 reaction?
A. DMSO (Dimethyl sulfoxide)
B. Methanol
C. Water
D. Acetic Acid
Answer: A
Explanation: SN2 reactions are favored by polar aprotic solvents like DMSO, DMF, or
acetone. These solvents dissolve the ionic nucleophile but do not form strong hydrogen
bonds with it, leaving the nucleophile ‘naked’ and more reactive. Polar protic solvents like
water or methanol solvate the nucleophile too strongly, which significantly hinders its
ability to attack the electrophile.
9. Which of the following alkyl halides will react fastest in an SN1 reaction?
A. tert-Butyl bromide
B. Ethyl bromide
C. Isopropyl bromide
D. Methyl bromide
