3 MAXE 012 MEHC
Portland State University
● ●
PSU Department of Chemistry — CHEM 210 Organic Chemistry I
LET KNOWLEDGE SERVE THE CITY
CHEM 210
CHEM 210 — Examination 3
ORGANIC CHEMISTRY I: ALKENE/ALKYNE ADDITION REACTIONS, STEREOCHEMISTRY & REAGENTS | 2026/2027
INSTITUTION Portland State University (PSU) COURSE CODE CHEM 210 — Organic Chemistry I
PROGRAM Undergraduate Chemistry / Pre-Professional ACADEMIC YEAR
EXAM TITLE Examination 3 — Reaction Mechanisms & Reagents TOTAL QUESTIONS 25 Questions
SUBJECT AREAS Addition Reactions, Stereochemistry, Reagents, FORMAT Multiple Choice — Select the Single Best Answer
Mechanisms
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each multiple-choice question.
▸ Topics include: electrophilic addition, hydroboration, epoxidation, dihydroxylation, ozonolysis, hydrogenation, halogenation, and elimination reactions.
▸ Know the stereochemistry (syn vs. anti), regiochemistry (Markovnikov vs. anti-Markovnikov), and stereochemical outcome (racemic vs. non-racemic) for each
reaction.
▸ Memorize reagents for each transformation: BH₃/H₂O₂/NaOH for hydroboration, McPBA for epoxidation, KMnO₄/OsO₄ for dihydroxylation, O₃ for ozonolysis, H₂/Pt for
hydrogenation, and X₂ for halogenation.
▸ Correct answers and detailed rationales appear below each question for exam review purposes.
SECTION I — ELECTROPHILIC ADDITION, HYDROBORATION & EPOXIDATION Questions 1 – 10
1. What are the stereochemistry and regiochemistry of standard electrophilic addition (e.g., H₂O/H₂SO₄ to an alkene)?
A. Anti stereochemistry, Anti-Markovnikov regiochemistry
B. Syn AND Anti stereochemistry (racemic mixture), MARKOVNIKOV regiochemistry (H adds to less substituted carbon)
C. Syn only, Markovnikov regiochemistry
D. Anti only, Anti-Markovnikov regiochemistry
CORRECT ANSWER B — Electrophilic addition gives a MIXTURE of syn and anti (racemic), with MARKOVNIKOV regiochemistry
RATIONALE Electrophilic addition (e.g., H₂O/H₂SO₄, HX) proceeds through a carbocation intermediate. The carbocation is trigonal planar (sp2), so the
nucleophile can attack from either face → produces BOTH syn and anti addition (racemic mixture if a chiral center is formed). Regiochemistry is
MARKOVNIKOV: H⁺ adds to the LESS substituted carbon, forming the MORE stable carbocation intermediate. The nucleophile (H₂O, Br⁻) then attacks
the more substituted carbon. This is the classic mechanism: protonation → carbocation → nucleophilic attack. The syn/anti mixture distinguishes
this from concerted additions like hydroboration (syn only).
2. What reagents are used for standard electrophilic hydration of an alkene?
A. 1) BH₃, 2) H₂O₂, NaOH
B. H₂O / H₂SO₄ — water adds across the double bond with Markovnikov regiochemistry
C. McPBA, CH₂Cl₂
D. H₂ / Pt
CORRECT ANSWER B — H₂O / H₂SO₄ (acid-catalyzed hydration); produces an alcohol with Markovnikov regiochemistry
RATIONALE Acid-catalyzed hydration (H₂O/H₂SO₄) adds H and OH across the double bond. Mechanism: (1) Protonation of the alkene by H₃O⁺ to form the most
stable carbocation (Markovnikov). (2) Water attacks the carbocation. (3) Deprotonation yields the alcohol. Option A is hydroboration-oxidation
(anti-Markovnikov). Option C is epoxidation (forms an epoxide). Option D is hydrogenation (adds H₂, removes π bond). The key: H₂O/H₂SO₄ =
Markovnikov alcohol; BH₃ then H₂O₂/NaOH = anti-Markovnikov alcohol.
3. What are the stereochemistry and regiochemistry of hydroboration-oxidation?
A. Anti stereochemistry, Markovnikov regiochemistry
B. SYN stereochemistry (H and OH add to same face), ANTI-MARKOVNIKOV regiochemistry (OH on LESS substituted carbon)
C. Syn AND Anti mixture, Markovnikov regiochemistry
D. Anti only, Anti-Markovnikov regiochemistry
CORRECT ANSWER B — Hydroboration: SYN addition of H and OH; ANTI-MARKOVNIKOV (OH adds to LESS substituted carbon)
RATIONALE Hydroboration-oxidation is a two-step process: (1) BH₃ adds to the alkene in a concerted syn fashion — both B and H add to the SAME face. Boron
adds to the LESS substituted carbon (anti-Markovnikov) due to steric factors. (2) Oxidation with H₂O₂/NaOH replaces B with OH with RETENTION of
configuration. The result: net SYN addition of H and OH, with OH on the LESS substituted carbon. This is complementary to acid-catalyzed
hydration (Markovnikov, mixed syn/anti). Hydroboration produces NO carbocation — the reaction is concerted and stereospecific.
, 4. What reagents are required for hydroboration-oxidation of an alkene to an alcohol?
A. H₂O / H₂SO₄
B. 1) BH₃, 2) H₂O₂, NaOH — borane followed by basic hydrogen peroxide oxidation
C. KMnO₄, cold
D. O₃, then Zn
CORRECT ANSWER B — 1) BH₃ (or B₂H₆), 2) H₂O₂, NaOH; the two-step sequence gives anti-Markovnikov syn addition of H and OH
RATIONALE Hydroboration uses borane (BH₃, typically as BH₃·THF complex or B₂H₆). Step 1: BH₃ adds syn across the double bond (B to less substituted C, H to
more substituted C). Step 2: Oxidation with basic H₂O₂ replaces the C-B bond with C-OH with retention of configuration. The net result is anti-
Markovnikov hydration with syn stereochemistry. Option A gives Markovnikov alcohol. Option C (KMnO₄, cold) gives syn dihydroxylation (vicinal
diol). Option D cleaves the double bond (ozonolysis). Hydroboration is one of the most important synthetic methods for anti-Markovnikov alcohol
formation.
5. What is epoxidation and what are its stereochemical characteristics?
A. Addition of two OH groups with anti stereochemistry
B. Formation of a three-membered epoxide ring (C-O-C); SYN stereochemistry — the oxygen adds to the same face of the alkene
C. Cleavage of the double bond to form carbonyl compounds
D. Addition of H₂ across the double bond
CORRECT ANSWER B — Epoxidation forms a three-membered cyclic ether (epoxide/oxirane) with SYN stereochemistry; the O adds to one face
RATIONALE Epoxidation converts an alkene to an epoxide (oxirane) — a three-membered ring containing oxygen. Reagent: McPBA (meta-chloroperoxybenzoic
acid) or other peroxyacids (RCO₃H). The reaction is CONCERTED — the peroxyacid delivers the oxygen atom to the π bond in a single step, adding to
the SAME face (syn addition). If the alkene has substituents, the epoxide retains the relative stereochemistry. Epoxides are valuable synthetic
intermediates because they can be opened by nucleophiles. Option A describes dihydroxylation with OsO₄ (syn) or epoxide opening (anti). Option C
describes ozonolysis. Option D describes hydrogenation.
6. What is the standard reagent for epoxidation of an alkene?
A. KMnO₄
B. McPBA (meta-chloroperoxybenzoic acid, RCO₃H) — a peroxyacid that delivers oxygen to the double bond
C. OsO₄
D. O₃
CORRECT ANSWER B — McPBA (meta-chloroperoxybenzoic acid) is the most common peroxyacid for epoxidation; it is electrophilic and delivers O to the alkene
RATIONALE Peroxyacids (RCO₃H) are the reagents of choice for epoxidation. McPBA (MCPBA) is most common — it is crystalline, stable, and easy to handle. The
mechanism: the peroxyacid's electrophilic oxygen attacks the π bond in a concerted syn addition, transferring the oxygen and forming a carboxylic
acid byproduct. Option A (KMnO₄) gives dihydroxylation (vicinal diol) or oxidative cleavage depending on conditions. Option C (OsO₄) gives syn
dihydroxylation. Option D (O₃) cleaves the double bond (ozonolysis). Each reagent has a distinct outcome — knowing which reagent does what is
essential for synthesis problems.
7. What is dihydroxylation and what stereochemistry does it produce?
A. Anti addition of two OH groups to form a trans diol
B. SYN addition of two OH groups to form a cis vicinal diol (both OH on the same face)
C. Markovnikov addition of H and OH
D. Cleavage of the double bond
CORRECT ANSWER B — Dihydroxylation with KMnO₄ (cold) or OsO₄ adds two OH groups with SYN stereochemistry (cis vicinal diol)
RATIONALE Syn dihydroxylation converts an alkene to a vicinal diol (glycol) with both OH groups on the SAME face. Reagents: OsO₄ (catalytic with NMO co-
oxidant) or cold, dilute KMnO₄. Mechanism (OsO₄): concerted [3+2] cycloaddition forms an osmate ester, then hydrolysis releases the syn diol. Anti
dihydroxylation is achieved by epoxidation followed by acid-catalyzed ring opening (trans diol). The syn vs. anti distinction is critical: syn
dihydroxylation via OsO₄ gives the cis diol from a cyclic alkene; anti dihydroxylation via epoxide opening gives the trans diol. Both are
stereospecific.
8. What reagents are used for syn dihydroxylation of an alkene? (Select all that apply)
A. KMnO₄ (cold, dilute) — potassium permanganate
B. OsO₄ (osmium tetroxide, catalytic with NMO) — the most reliable syn dihydroxylation reagent
C. McPBA then H₃O⁺
D. O₃ then Zn
CORRECT ANSWER A, B — Both cold KMnO₄ and OsO₄ give SYN dihydroxylation (cis vicinal diol)
RATIONALE Syn dihydroxylation: (A) KMnO₄ (cold, dilute, basic) forms a cyclic manganate ester that hydrolyzes to the cis diol. Drawback: over-oxidation can
occur. (B) OsO₄ (catalytic) with NMO (N-methylmorpholine N-oxide) as co-oxidant gives clean syn dihydroxylation — the gold standard. Option C
(McPBA then H₃O⁺) gives ANTI dihydroxylation — epoxidation followed by trans ring opening. Option D (O₃ then Zn) is reductive ozonolysis —
CLEAVES the double bond to carbonyls, not dihydroxylation. The syn vs. anti distinction is a classic exam question.
Portland State University
● ●
PSU Department of Chemistry — CHEM 210 Organic Chemistry I
LET KNOWLEDGE SERVE THE CITY
CHEM 210
CHEM 210 — Examination 3
ORGANIC CHEMISTRY I: ALKENE/ALKYNE ADDITION REACTIONS, STEREOCHEMISTRY & REAGENTS | 2026/2027
INSTITUTION Portland State University (PSU) COURSE CODE CHEM 210 — Organic Chemistry I
PROGRAM Undergraduate Chemistry / Pre-Professional ACADEMIC YEAR
EXAM TITLE Examination 3 — Reaction Mechanisms & Reagents TOTAL QUESTIONS 25 Questions
SUBJECT AREAS Addition Reactions, Stereochemistry, Reagents, FORMAT Multiple Choice — Select the Single Best Answer
Mechanisms
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each multiple-choice question.
▸ Topics include: electrophilic addition, hydroboration, epoxidation, dihydroxylation, ozonolysis, hydrogenation, halogenation, and elimination reactions.
▸ Know the stereochemistry (syn vs. anti), regiochemistry (Markovnikov vs. anti-Markovnikov), and stereochemical outcome (racemic vs. non-racemic) for each
reaction.
▸ Memorize reagents for each transformation: BH₃/H₂O₂/NaOH for hydroboration, McPBA for epoxidation, KMnO₄/OsO₄ for dihydroxylation, O₃ for ozonolysis, H₂/Pt for
hydrogenation, and X₂ for halogenation.
▸ Correct answers and detailed rationales appear below each question for exam review purposes.
SECTION I — ELECTROPHILIC ADDITION, HYDROBORATION & EPOXIDATION Questions 1 – 10
1. What are the stereochemistry and regiochemistry of standard electrophilic addition (e.g., H₂O/H₂SO₄ to an alkene)?
A. Anti stereochemistry, Anti-Markovnikov regiochemistry
B. Syn AND Anti stereochemistry (racemic mixture), MARKOVNIKOV regiochemistry (H adds to less substituted carbon)
C. Syn only, Markovnikov regiochemistry
D. Anti only, Anti-Markovnikov regiochemistry
CORRECT ANSWER B — Electrophilic addition gives a MIXTURE of syn and anti (racemic), with MARKOVNIKOV regiochemistry
RATIONALE Electrophilic addition (e.g., H₂O/H₂SO₄, HX) proceeds through a carbocation intermediate. The carbocation is trigonal planar (sp2), so the
nucleophile can attack from either face → produces BOTH syn and anti addition (racemic mixture if a chiral center is formed). Regiochemistry is
MARKOVNIKOV: H⁺ adds to the LESS substituted carbon, forming the MORE stable carbocation intermediate. The nucleophile (H₂O, Br⁻) then attacks
the more substituted carbon. This is the classic mechanism: protonation → carbocation → nucleophilic attack. The syn/anti mixture distinguishes
this from concerted additions like hydroboration (syn only).
2. What reagents are used for standard electrophilic hydration of an alkene?
A. 1) BH₃, 2) H₂O₂, NaOH
B. H₂O / H₂SO₄ — water adds across the double bond with Markovnikov regiochemistry
C. McPBA, CH₂Cl₂
D. H₂ / Pt
CORRECT ANSWER B — H₂O / H₂SO₄ (acid-catalyzed hydration); produces an alcohol with Markovnikov regiochemistry
RATIONALE Acid-catalyzed hydration (H₂O/H₂SO₄) adds H and OH across the double bond. Mechanism: (1) Protonation of the alkene by H₃O⁺ to form the most
stable carbocation (Markovnikov). (2) Water attacks the carbocation. (3) Deprotonation yields the alcohol. Option A is hydroboration-oxidation
(anti-Markovnikov). Option C is epoxidation (forms an epoxide). Option D is hydrogenation (adds H₂, removes π bond). The key: H₂O/H₂SO₄ =
Markovnikov alcohol; BH₃ then H₂O₂/NaOH = anti-Markovnikov alcohol.
3. What are the stereochemistry and regiochemistry of hydroboration-oxidation?
A. Anti stereochemistry, Markovnikov regiochemistry
B. SYN stereochemistry (H and OH add to same face), ANTI-MARKOVNIKOV regiochemistry (OH on LESS substituted carbon)
C. Syn AND Anti mixture, Markovnikov regiochemistry
D. Anti only, Anti-Markovnikov regiochemistry
CORRECT ANSWER B — Hydroboration: SYN addition of H and OH; ANTI-MARKOVNIKOV (OH adds to LESS substituted carbon)
RATIONALE Hydroboration-oxidation is a two-step process: (1) BH₃ adds to the alkene in a concerted syn fashion — both B and H add to the SAME face. Boron
adds to the LESS substituted carbon (anti-Markovnikov) due to steric factors. (2) Oxidation with H₂O₂/NaOH replaces B with OH with RETENTION of
configuration. The result: net SYN addition of H and OH, with OH on the LESS substituted carbon. This is complementary to acid-catalyzed
hydration (Markovnikov, mixed syn/anti). Hydroboration produces NO carbocation — the reaction is concerted and stereospecific.
, 4. What reagents are required for hydroboration-oxidation of an alkene to an alcohol?
A. H₂O / H₂SO₄
B. 1) BH₃, 2) H₂O₂, NaOH — borane followed by basic hydrogen peroxide oxidation
C. KMnO₄, cold
D. O₃, then Zn
CORRECT ANSWER B — 1) BH₃ (or B₂H₆), 2) H₂O₂, NaOH; the two-step sequence gives anti-Markovnikov syn addition of H and OH
RATIONALE Hydroboration uses borane (BH₃, typically as BH₃·THF complex or B₂H₆). Step 1: BH₃ adds syn across the double bond (B to less substituted C, H to
more substituted C). Step 2: Oxidation with basic H₂O₂ replaces the C-B bond with C-OH with retention of configuration. The net result is anti-
Markovnikov hydration with syn stereochemistry. Option A gives Markovnikov alcohol. Option C (KMnO₄, cold) gives syn dihydroxylation (vicinal
diol). Option D cleaves the double bond (ozonolysis). Hydroboration is one of the most important synthetic methods for anti-Markovnikov alcohol
formation.
5. What is epoxidation and what are its stereochemical characteristics?
A. Addition of two OH groups with anti stereochemistry
B. Formation of a three-membered epoxide ring (C-O-C); SYN stereochemistry — the oxygen adds to the same face of the alkene
C. Cleavage of the double bond to form carbonyl compounds
D. Addition of H₂ across the double bond
CORRECT ANSWER B — Epoxidation forms a three-membered cyclic ether (epoxide/oxirane) with SYN stereochemistry; the O adds to one face
RATIONALE Epoxidation converts an alkene to an epoxide (oxirane) — a three-membered ring containing oxygen. Reagent: McPBA (meta-chloroperoxybenzoic
acid) or other peroxyacids (RCO₃H). The reaction is CONCERTED — the peroxyacid delivers the oxygen atom to the π bond in a single step, adding to
the SAME face (syn addition). If the alkene has substituents, the epoxide retains the relative stereochemistry. Epoxides are valuable synthetic
intermediates because they can be opened by nucleophiles. Option A describes dihydroxylation with OsO₄ (syn) or epoxide opening (anti). Option C
describes ozonolysis. Option D describes hydrogenation.
6. What is the standard reagent for epoxidation of an alkene?
A. KMnO₄
B. McPBA (meta-chloroperoxybenzoic acid, RCO₃H) — a peroxyacid that delivers oxygen to the double bond
C. OsO₄
D. O₃
CORRECT ANSWER B — McPBA (meta-chloroperoxybenzoic acid) is the most common peroxyacid for epoxidation; it is electrophilic and delivers O to the alkene
RATIONALE Peroxyacids (RCO₃H) are the reagents of choice for epoxidation. McPBA (MCPBA) is most common — it is crystalline, stable, and easy to handle. The
mechanism: the peroxyacid's electrophilic oxygen attacks the π bond in a concerted syn addition, transferring the oxygen and forming a carboxylic
acid byproduct. Option A (KMnO₄) gives dihydroxylation (vicinal diol) or oxidative cleavage depending on conditions. Option C (OsO₄) gives syn
dihydroxylation. Option D (O₃) cleaves the double bond (ozonolysis). Each reagent has a distinct outcome — knowing which reagent does what is
essential for synthesis problems.
7. What is dihydroxylation and what stereochemistry does it produce?
A. Anti addition of two OH groups to form a trans diol
B. SYN addition of two OH groups to form a cis vicinal diol (both OH on the same face)
C. Markovnikov addition of H and OH
D. Cleavage of the double bond
CORRECT ANSWER B — Dihydroxylation with KMnO₄ (cold) or OsO₄ adds two OH groups with SYN stereochemistry (cis vicinal diol)
RATIONALE Syn dihydroxylation converts an alkene to a vicinal diol (glycol) with both OH groups on the SAME face. Reagents: OsO₄ (catalytic with NMO co-
oxidant) or cold, dilute KMnO₄. Mechanism (OsO₄): concerted [3+2] cycloaddition forms an osmate ester, then hydrolysis releases the syn diol. Anti
dihydroxylation is achieved by epoxidation followed by acid-catalyzed ring opening (trans diol). The syn vs. anti distinction is critical: syn
dihydroxylation via OsO₄ gives the cis diol from a cyclic alkene; anti dihydroxylation via epoxide opening gives the trans diol. Both are
stereospecific.
8. What reagents are used for syn dihydroxylation of an alkene? (Select all that apply)
A. KMnO₄ (cold, dilute) — potassium permanganate
B. OsO₄ (osmium tetroxide, catalytic with NMO) — the most reliable syn dihydroxylation reagent
C. McPBA then H₃O⁺
D. O₃ then Zn
CORRECT ANSWER A, B — Both cold KMnO₄ and OsO₄ give SYN dihydroxylation (cis vicinal diol)
RATIONALE Syn dihydroxylation: (A) KMnO₄ (cold, dilute, basic) forms a cyclic manganate ester that hydrolyzes to the cis diol. Drawback: over-oxidation can
occur. (B) OsO₄ (catalytic) with NMO (N-methylmorpholine N-oxide) as co-oxidant gives clean syn dihydroxylation — the gold standard. Option C
(McPBA then H₃O⁺) gives ANTI dihydroxylation — epoxidation followed by trans ring opening. Option D (O₃ then Zn) is reductive ozonolysis —
CLEAVES the double bond to carbonyls, not dihydroxylation. The syn vs. anti distinction is a classic exam question.
