ORGANIC CHEMISTRY ACS FINAL EXAM
MEGA PACK HIGH-YIELD QUESTIONS
SOLUTIONS AND MASTER RATIONALES
Question 1
An unknown optically active compound with the molecular
formula C6L12O undergoes treatment with dilute aqueous acid to
yield a pair of diastereomeric diols. When the original compound
is subjected to ozonolysis under reductive conditions (Zn/H2O), it
yields acetaldehyde and a four-carbon chiral aldehyde. Which of
the following starting materials best matches this experimental
profile?
A) (E)-4-methylhex-2-ene
B) (Z)-3-methylpent-2-ene
C) (S)-4-methylpent-1-ene
D) (R)-3-methylhex-1-ene
CorreCt Answer: A) (E)-4-methylhex-2-ene
rAtionAle: Ozonolysis of (E)-4-methylhex-2-ene under
reductive conditions cleaves the double bond to produce
acetaldehyde (a two-carbon aldehyde) and 2-methylbutanal (a
four-carbon aldehyde containing a chiral center at C2). Acid-
catalyzed dihydroxylation or hydration processes on an alkene
with an existing chiral center will generate a new stereocenter,
resulting in a mixture of diastereomeric products rather than
enantiomers or meso compounds.
Question 2
Consider the nucleophilic substitution reaction of (R)-2-
,bromobutane with sodium azide (NaN3) in a polar aprotic solvent
such as dimethyl sulfoxide (DMSO). If the concentration of the
nucleophile is doubled while the concentration of the alkyl halide
is kept constant, what occurs to the overall rate of the reaction
and the stereochemical configuration of the resulting product?
A) The reaction rate doubles, and the product configuration
undergoes complete retention to yield (R)-2-azidobutane.
B) The reaction rate quadruples, and the product configuration
undergoes complete inversion to yield (S)-2-azidobutane.
C) The reaction rate doubles, and the product configuration
undergoes complete inversion to yield (S)-2-azidobutane.
D) The reaction rate remains unchanged, and the product
undergoes racemization to yield an equal mixture of enantiomers.
CorreCt Answer: C) The reaction rate doubles,
and the product configuration undergoes complete
inversion to yield (S)-2-azidobutane.
rAtionAle: The reaction conditions describe a classic SN2
pathway: a secondary alkyl halide, a strong nucleophile (azide),
and a polar aprotic solvent (DMSO). The rate law for a
bimolecular substitution is Rate = k[alkyl halide][nucleophile],
meaning doubling the concentration of the azide nucleophile will
directly double the reaction rate. Mechanistically, the nucleophile
attacks from the backside of the carbon-bromine bond, causing a
concerted inversion of configuration at the stereocenter,
changing the (R) reactant into the (S) product.
Question 3
A research chemist attempts to synthesize an ether by reacting
potassium tert-butoxide with 2-bromo-2-methylpropane in
,ethanol at elevated temperatures. Instead of the anticipated
substitution product, the primary isolated organic compound is
found to be a volatile gas that rapidly decolorizes a solution of
bromine in dichloromethane. Which statement correctly identifies
the major product and explains the kinetic outcome of this
reaction?
A) The major product is ethyl tert-butyl ether, formed via an SN2
mechanism because the nucleophile is unhindered.
B) The major product is 2-methylpropene, formed via an E2
mechanism because both the substrate and the base are sterically
hindered.
C) The major product is 2-methylbutene, formed via an E1
mechanism because the solvent stabilizes the carbocation
intermediate.
D) The major product is di-tert-butyl ether, formed via an SN1
mechanism due to the high stability of the tertiary carbocation.
CorreCt Answer: B) The major product is 2-
methylpropene, formed via an E2 mechanism because
both the substrate and the base are sterically hindered.
rAtionAle: The combination of a tertiary alkyl halide (2-
bromo-2-methylpropane) and a strong, sterically hindered bulk
base (potassium tert-butoxide) completely shuts down
substitution pathways. Backside attack is blocked by the bulky
methyl groups on both the substrate and the base. Instead, the
base abstracts a beta-hydrogen in a concerted E2 elimination
process, yielding 2-methylpropene (isobutylene) as the major
product. This alkene decolorizes bromine water due to
electrophilic addition across the newly formed pi bond.
, Question 4
When treated with a mixture of concentrated nitric acid and
concentrated sulfuric acid, benzene undergoes electrophilic
aromatic substitution to form nitrobenzene. If the reaction is
monitored thermochemically, the formation of the arenium ion
(sigma complex) is identified as the rate-determining step. What
is the precise role of concentrated sulfuric acid in this mechanistic
sequence?
A) It acts as a protic solvent that stabilizes the anionic leaving
groups through hydrogen bonding.
B) It acts as a Lewis acid catalyst that directly coordinates to the
benzene ring to increase its nucleophilicity.
C) It acts as a strong Bronsted-Lowry acid to protonate nitric acid,
facilitating the loss of water to generate the electrophilic
nitronium ion.
D) It acts as a reducing agent that prevents the over-oxidation of
the aromatic ring during the course of the reaction.
CorreCt Answer: C) It acts as a strong Bronsted-
Lowry acid to protonate nitric acid, facilitating the loss
of water to generate the electrophilic nitronium ion.
rAtionAle: Nitric acid alone is not electrophilic enough to
disrupt the aromatic stability of a benzene ring. Sulfuric acid,
being a stronger acid than nitric acid, protonates the hydroxyl
group of nitric acid. This transforms the hydroxyl group into an
excellent leaving group (water). The subsequent loss of water
generates the highly reactive nitronium ion (NO2+), which is the
actual electrophile attacked by the aromatic ring in the rate-
determining step.
MEGA PACK HIGH-YIELD QUESTIONS
SOLUTIONS AND MASTER RATIONALES
Question 1
An unknown optically active compound with the molecular
formula C6L12O undergoes treatment with dilute aqueous acid to
yield a pair of diastereomeric diols. When the original compound
is subjected to ozonolysis under reductive conditions (Zn/H2O), it
yields acetaldehyde and a four-carbon chiral aldehyde. Which of
the following starting materials best matches this experimental
profile?
A) (E)-4-methylhex-2-ene
B) (Z)-3-methylpent-2-ene
C) (S)-4-methylpent-1-ene
D) (R)-3-methylhex-1-ene
CorreCt Answer: A) (E)-4-methylhex-2-ene
rAtionAle: Ozonolysis of (E)-4-methylhex-2-ene under
reductive conditions cleaves the double bond to produce
acetaldehyde (a two-carbon aldehyde) and 2-methylbutanal (a
four-carbon aldehyde containing a chiral center at C2). Acid-
catalyzed dihydroxylation or hydration processes on an alkene
with an existing chiral center will generate a new stereocenter,
resulting in a mixture of diastereomeric products rather than
enantiomers or meso compounds.
Question 2
Consider the nucleophilic substitution reaction of (R)-2-
,bromobutane with sodium azide (NaN3) in a polar aprotic solvent
such as dimethyl sulfoxide (DMSO). If the concentration of the
nucleophile is doubled while the concentration of the alkyl halide
is kept constant, what occurs to the overall rate of the reaction
and the stereochemical configuration of the resulting product?
A) The reaction rate doubles, and the product configuration
undergoes complete retention to yield (R)-2-azidobutane.
B) The reaction rate quadruples, and the product configuration
undergoes complete inversion to yield (S)-2-azidobutane.
C) The reaction rate doubles, and the product configuration
undergoes complete inversion to yield (S)-2-azidobutane.
D) The reaction rate remains unchanged, and the product
undergoes racemization to yield an equal mixture of enantiomers.
CorreCt Answer: C) The reaction rate doubles,
and the product configuration undergoes complete
inversion to yield (S)-2-azidobutane.
rAtionAle: The reaction conditions describe a classic SN2
pathway: a secondary alkyl halide, a strong nucleophile (azide),
and a polar aprotic solvent (DMSO). The rate law for a
bimolecular substitution is Rate = k[alkyl halide][nucleophile],
meaning doubling the concentration of the azide nucleophile will
directly double the reaction rate. Mechanistically, the nucleophile
attacks from the backside of the carbon-bromine bond, causing a
concerted inversion of configuration at the stereocenter,
changing the (R) reactant into the (S) product.
Question 3
A research chemist attempts to synthesize an ether by reacting
potassium tert-butoxide with 2-bromo-2-methylpropane in
,ethanol at elevated temperatures. Instead of the anticipated
substitution product, the primary isolated organic compound is
found to be a volatile gas that rapidly decolorizes a solution of
bromine in dichloromethane. Which statement correctly identifies
the major product and explains the kinetic outcome of this
reaction?
A) The major product is ethyl tert-butyl ether, formed via an SN2
mechanism because the nucleophile is unhindered.
B) The major product is 2-methylpropene, formed via an E2
mechanism because both the substrate and the base are sterically
hindered.
C) The major product is 2-methylbutene, formed via an E1
mechanism because the solvent stabilizes the carbocation
intermediate.
D) The major product is di-tert-butyl ether, formed via an SN1
mechanism due to the high stability of the tertiary carbocation.
CorreCt Answer: B) The major product is 2-
methylpropene, formed via an E2 mechanism because
both the substrate and the base are sterically hindered.
rAtionAle: The combination of a tertiary alkyl halide (2-
bromo-2-methylpropane) and a strong, sterically hindered bulk
base (potassium tert-butoxide) completely shuts down
substitution pathways. Backside attack is blocked by the bulky
methyl groups on both the substrate and the base. Instead, the
base abstracts a beta-hydrogen in a concerted E2 elimination
process, yielding 2-methylpropene (isobutylene) as the major
product. This alkene decolorizes bromine water due to
electrophilic addition across the newly formed pi bond.
, Question 4
When treated with a mixture of concentrated nitric acid and
concentrated sulfuric acid, benzene undergoes electrophilic
aromatic substitution to form nitrobenzene. If the reaction is
monitored thermochemically, the formation of the arenium ion
(sigma complex) is identified as the rate-determining step. What
is the precise role of concentrated sulfuric acid in this mechanistic
sequence?
A) It acts as a protic solvent that stabilizes the anionic leaving
groups through hydrogen bonding.
B) It acts as a Lewis acid catalyst that directly coordinates to the
benzene ring to increase its nucleophilicity.
C) It acts as a strong Bronsted-Lowry acid to protonate nitric acid,
facilitating the loss of water to generate the electrophilic
nitronium ion.
D) It acts as a reducing agent that prevents the over-oxidation of
the aromatic ring during the course of the reaction.
CorreCt Answer: C) It acts as a strong Bronsted-
Lowry acid to protonate nitric acid, facilitating the loss
of water to generate the electrophilic nitronium ion.
rAtionAle: Nitric acid alone is not electrophilic enough to
disrupt the aromatic stability of a benzene ring. Sulfuric acid,
being a stronger acid than nitric acid, protonates the hydroxyl
group of nitric acid. This transforms the hydroxyl group into an
excellent leaving group (water). The subsequent loss of water
generates the highly reactive nitronium ion (NO2+), which is the
actual electrophile attacked by the aromatic ring in the rate-
determining step.
