III MAXE • 012 MEHC
University of Michigan
M College of Literature, Science, and the Arts
ARTES • SCIENTIA • VERITAS
EST. 1817
CHEM 210 — Organic Chemistry
E X A M I N AT I O N I I I • A L K E N E & A L KY N E R E A C T I O N M E C H A N I S M S
INSTITUTION University of Michigan COURSE CODE CHEM 210
PROGRAM B.S. in Chemistry / Pre-Health ACADEMIC YEAR
EXAM TITLE Structure & Reactivity II — Exam III TOTAL QUESTIONS 15 Questions
COURSE TITLE Organic Chemistry 210 FORMAT Multiple Choice — Select the Single Best
Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each reaction mechanism question.
▸ Pay close attention to regiochemistry (Markovnikov vs. anti-Markovnikov) and stereochemistry (syn vs. anti addition).
▸ Reagent combinations, intermediates, and product outcomes are all testable content.
▸ Correct answers and detailed mechanistic rationales appear below each question.
▸ A periodic table and pKa table are not provided; memorize relevant values.
SECTION I — DIHYDROXYLATION, EPOXIDATION, HALOGENATION & Questions 1 –
HYDRATION 15
1. A student treats an alkene with OsO4 followed by Na2SO3 / H2O. What is the stereochemical outcome of this
dihydroxylation?
A. Anti addition of two –OH groups
B. Syn addition of two –OH groups
C. Markovnikov addition of one –OH group only
D. Anti-Markovnikov addition of one –OH group only
CORRECT ANSWER B — Syn addition of two –OH groups
RATIONALE OsO4 (osmium tetroxide) forms a cyclic osmate ester intermediate with the alkene, delivering both oxygen
atoms to the same face of the double bond. Subsequent reductive workup with Na2SO3 / H2O hydrolyzes the
osmate ester to release the vicinal diol with syn stereochemistry. KMnO4 under cold, basic conditions also
gives syn dihydroxylation via a cyclic manganate ester, though hot acidic KMnO4 leads to oxidative cleavage
instead.
, 2. Which reagent system is used to convert an alkene into an epoxide, and which double bond reacts preferentially in
a polyunsaturated system?
A. OsO4 / Na2SO3; least substituted double bond reacts first
B. RCO3H or mCPBA; most substituted double bond reacts first
C. BH3 / H2O2, NaOH; most substituted double bond reacts first
D. H2 / Pd-C; least substituted double bond reacts first
CORRECT ANSWER B — RCO3H or mCPBA; most substituted double bond reacts first
RATIONALE Peroxycarboxylic acids (RCO3H) and meta-chloroperoxybenzoic acid (mCPBA) are electrophilic epoxidizing
agents that transfer an oxygen atom to the alkene π bond. The reaction is concerted and stereospecific (syn
addition of the O atom). In a polyunsaturated system, the most electron-rich (most substituted) double bond
reacts preferentially because it is the most nucleophilic. This is the opposite selectivity of hydroboration,
which favors the least hindered double bond for steric reasons.
3. Treatment of an alkene with Br2 in H2O yields a halohydrin. Which statement correctly describes the regiochemistry
and stereochemistry of this transformation?
A. Anti addition; –OH adds to the less substituted carbon; Markovnikov orientation
B. Syn addition; –OH adds to the more substituted carbon; anti-Markovnikov orientation
C. Anti addition; –OH adds to the more substituted carbon; Markovnikov orientation
D. Syn addition; –OH adds to the less substituted carbon; Markovnikov orientation
CORRECT ANSWER C — Anti addition; –OH adds to the more substituted carbon; Markovnikov orientation
RATIONALE Halohydrin formation proceeds through a bridged halonium ion (epoxy ion) intermediate. The nucleophile
(H2O) attacks the more substituted carbon of the halonium ion because that carbon bears greater partial
positive charge (Markovnikov orientation). Since the nucleophile attacks from the face opposite the bridging
halogen, the overall addition is anti. The resulting product has the –OH on the more substituted carbon and
the halogen on the less substituted carbon, added to opposite faces of the original double bond.
4. An alkene is treated with Cl2 in an inert solvent (CCl4). What is the stereochemical outcome and key intermediate of
this halogenation reaction?
A. Syn addition via a carbocation intermediate
B. Anti addition via a halonium ion (epoxy ion) intermediate
C. Syn addition via a halonium ion intermediate
D. Anti addition via a carbocation intermediate
CORRECT ANSWER B — Anti addition via a halonium ion (epoxy ion) intermediate
RATIONALE Halogenation of alkenes with Cl2 or Br2 proceeds through a bridged halonium ion intermediate (the "epoxy
ion" referenced in the notes). The electrophilic halogen forms a three-membered ring with the two alkene
carbons. The halide counterion then attacks from the face opposite the bridging halogen (backside attack),
resulting in anti addition of the two halogen atoms. This stereospecific anti addition can be demonstrated by
the formation of a meso compound from cis-2-butene and a racemic mixture from trans-2-butene.
University of Michigan
M College of Literature, Science, and the Arts
ARTES • SCIENTIA • VERITAS
EST. 1817
CHEM 210 — Organic Chemistry
E X A M I N AT I O N I I I • A L K E N E & A L KY N E R E A C T I O N M E C H A N I S M S
INSTITUTION University of Michigan COURSE CODE CHEM 210
PROGRAM B.S. in Chemistry / Pre-Health ACADEMIC YEAR
EXAM TITLE Structure & Reactivity II — Exam III TOTAL QUESTIONS 15 Questions
COURSE TITLE Organic Chemistry 210 FORMAT Multiple Choice — Select the Single Best
Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each reaction mechanism question.
▸ Pay close attention to regiochemistry (Markovnikov vs. anti-Markovnikov) and stereochemistry (syn vs. anti addition).
▸ Reagent combinations, intermediates, and product outcomes are all testable content.
▸ Correct answers and detailed mechanistic rationales appear below each question.
▸ A periodic table and pKa table are not provided; memorize relevant values.
SECTION I — DIHYDROXYLATION, EPOXIDATION, HALOGENATION & Questions 1 –
HYDRATION 15
1. A student treats an alkene with OsO4 followed by Na2SO3 / H2O. What is the stereochemical outcome of this
dihydroxylation?
A. Anti addition of two –OH groups
B. Syn addition of two –OH groups
C. Markovnikov addition of one –OH group only
D. Anti-Markovnikov addition of one –OH group only
CORRECT ANSWER B — Syn addition of two –OH groups
RATIONALE OsO4 (osmium tetroxide) forms a cyclic osmate ester intermediate with the alkene, delivering both oxygen
atoms to the same face of the double bond. Subsequent reductive workup with Na2SO3 / H2O hydrolyzes the
osmate ester to release the vicinal diol with syn stereochemistry. KMnO4 under cold, basic conditions also
gives syn dihydroxylation via a cyclic manganate ester, though hot acidic KMnO4 leads to oxidative cleavage
instead.
, 2. Which reagent system is used to convert an alkene into an epoxide, and which double bond reacts preferentially in
a polyunsaturated system?
A. OsO4 / Na2SO3; least substituted double bond reacts first
B. RCO3H or mCPBA; most substituted double bond reacts first
C. BH3 / H2O2, NaOH; most substituted double bond reacts first
D. H2 / Pd-C; least substituted double bond reacts first
CORRECT ANSWER B — RCO3H or mCPBA; most substituted double bond reacts first
RATIONALE Peroxycarboxylic acids (RCO3H) and meta-chloroperoxybenzoic acid (mCPBA) are electrophilic epoxidizing
agents that transfer an oxygen atom to the alkene π bond. The reaction is concerted and stereospecific (syn
addition of the O atom). In a polyunsaturated system, the most electron-rich (most substituted) double bond
reacts preferentially because it is the most nucleophilic. This is the opposite selectivity of hydroboration,
which favors the least hindered double bond for steric reasons.
3. Treatment of an alkene with Br2 in H2O yields a halohydrin. Which statement correctly describes the regiochemistry
and stereochemistry of this transformation?
A. Anti addition; –OH adds to the less substituted carbon; Markovnikov orientation
B. Syn addition; –OH adds to the more substituted carbon; anti-Markovnikov orientation
C. Anti addition; –OH adds to the more substituted carbon; Markovnikov orientation
D. Syn addition; –OH adds to the less substituted carbon; Markovnikov orientation
CORRECT ANSWER C — Anti addition; –OH adds to the more substituted carbon; Markovnikov orientation
RATIONALE Halohydrin formation proceeds through a bridged halonium ion (epoxy ion) intermediate. The nucleophile
(H2O) attacks the more substituted carbon of the halonium ion because that carbon bears greater partial
positive charge (Markovnikov orientation). Since the nucleophile attacks from the face opposite the bridging
halogen, the overall addition is anti. The resulting product has the –OH on the more substituted carbon and
the halogen on the less substituted carbon, added to opposite faces of the original double bond.
4. An alkene is treated with Cl2 in an inert solvent (CCl4). What is the stereochemical outcome and key intermediate of
this halogenation reaction?
A. Syn addition via a carbocation intermediate
B. Anti addition via a halonium ion (epoxy ion) intermediate
C. Syn addition via a halonium ion intermediate
D. Anti addition via a carbocation intermediate
CORRECT ANSWER B — Anti addition via a halonium ion (epoxy ion) intermediate
RATIONALE Halogenation of alkenes with Cl2 or Br2 proceeds through a bridged halonium ion intermediate (the "epoxy
ion" referenced in the notes). The electrophilic halogen forms a three-membered ring with the two alkene
carbons. The halide counterion then attacks from the face opposite the bridging halogen (backside attack),
resulting in anti addition of the two halogen atoms. This stereospecific anti addition can be demonstrated by
the formation of a meso compound from cis-2-butene and a racemic mixture from trans-2-butene.
