CHEM 222 | REACTION MECHANISMS | 2026–2027
CHEM 222 Reaction Mechanisms
Practice Questions
2026–2027 | 110 Original Questions, Answers & Detailed Rationales
Mechanism Strategy and High-Yield Review
1. Identify the electrophile: Locate a polarized bond, carbonyl carbon, activated aromatic electrophile,
or positively charged intermediate.
2. Identify the nucleophile or base: Ask whether the reagent donates an electron pair to carbon or
abstracts a proton.
3. Check the substrate: Steric hindrance, resonance, and carbocation stability govern competing
pathways.
4. Track electron flow: Use curved arrows from electron pairs or bonds to the atoms receiving electrons.
5. Confirm regiochemistry and stereochemistry: Check Markovnikov versus anti-Markovnikov
outcomes, syn/anti addition, inversion, and antiperiplanar elimination.
6. Check the workup: Acid/base workup, tautomerization, and oxidation or reduction steps determine
isolated product
Independent original educational resource • Page 1
,CHEM 222 | REACTION MECHANISMS | 2026–2027
Substitution and elimination
Question 1. Under conventional textbook conditions, which mechanism best describes 1-
bromobutane with NaI in acetone?
A. E1
B. SN2
C. E2
D. SN1
Correct Answer: B. SN2
Detailed Rationale: Iodide attacks the primary carbon from the backside as bromide leaves. Key
principle: primary unhindered substrate and polar aprotic solvent.
Why the Other Options Are Incorrect: A — E1 involves a different bond-making or bond-breaking
pathway; here, iodide attacks the primary carbon from the backside as bromide leaves. C — E2
involves a different bond-making or bond-breaking pathway; here, iodide attacks the primary
carbon from the backside as bromide leaves. D — SN1 involves a different bond-making or bond-
breaking pathway; here, iodide attacks the primary carbon from the backside as bromide leaves.
Question 2. For 2-bromobutane with NaOEt in hot ethanol, what is the key bond-making or bond-
breaking event?
A. bulky base removes a beta hydrogen as bromide leaves
B. cyanide attacks carbon bearing bromine as bromide leaves
C. ethoxide removes a beta hydrogen while bromide departs
D. ionization gives a tertiary carbocation before water attacks
Correct Answer: C. ethoxide removes a beta hydrogen while bromide departs
Detailed Rationale: The mechanism is E2: ethoxide removes a beta hydrogen while bromide
departs. This follows because strong base and heat favor elimination.
Why the Other Options Are Incorrect: A — This describes a different transformation; the specified
reagents follow E2, so ethoxide removes a beta hydrogen while bromide departs. B — This describes
a different transformation; the specified reagents follow E2, so ethoxide removes a beta hydrogen
while bromide departs. D — This describes a different transformation; the specified reagents follow
E2, so ethoxide removes a beta hydrogen while bromide departs.
Independent original educational resource • Page 2
,CHEM 222 | REACTION MECHANISMS | 2026–2027
Question 3. What is the expected principal product of tert-butyl bromide in water?
A. 2-methylpropene
B. butanenitrile
C. butene isomers
D. tert-butanol after deprotonation
Correct Answer: D. tert-butanol after deprotonation
Detailed Rationale: Ionization gives a tertiary carbocation before water attacks; therefore the
principal product is tert-butanol after deprotonation. Key principle: tertiary carbocation and polar
protic solvent.
Why the Other Options Are Incorrect: A — That product corresponds to a different substrate or
reagent sequence; the stated conditions give tert-butanol after deprotonation by SN1. B — That
product corresponds to a different substrate or reagent sequence; the stated conditions give tert-
butanol after deprotonation by SN1. C — That product corresponds to a different substrate or
reagent sequence; the stated conditions give tert-butanol after deprotonation by SN1.
Question 4. Under conventional textbook conditions, which mechanism best describes tert-butyl
bromide with KOtBu?
A. E2
B. SN2
C. SN1
D. E1
Correct Answer: A. E2
Detailed Rationale: Bulky base removes a beta hydrogen as bromide leaves. Key principle: strong
sterically hindered base.
Why the Other Options Are Incorrect: B — SN2 involves a different bond-making or bond-breaking
pathway; here, bulky base removes a beta hydrogen as bromide leaves. C — SN1 involves a
different bond-making or bond-breaking pathway; here, bulky base removes a beta hydrogen as
bromide leaves. D — E1 involves a different bond-making or bond-breaking pathway; here, bulky
base removes a beta hydrogen as bromide leaves.
Independent original educational resource • Page 3
, CHEM 222 | REACTION MECHANISMS | 2026–2027
Question 5. For 1-bromopropane with NaCN in DMSO, what is the key bond-making or bond-
breaking event?
A. base removes an axial beta hydrogen anti to axial leaving group
B. cyanide attacks carbon bearing bromine as bromide leaves
C. methoxide abstracts an antiperiplanar beta hydrogen while bromide leaves
D. azide displaces bromide by backside attack
Correct Answer: B. cyanide attacks carbon bearing bromine as bromide leaves
Detailed Rationale: The mechanism is SN2: cyanide attacks carbon bearing bromine as bromide
leaves. This follows because primary substrate and good carbon nucleophile.
Why the Other Options Are Incorrect: A — This describes a different transformation; the specified
reagents follow SN2, so cyanide attacks carbon bearing bromine as bromide leaves. C — This
describes a different transformation; the specified reagents follow SN2, so cyanide attacks carbon
bearing bromine as bromide leaves. D — This describes a different transformation; the specified
reagents follow SN2, so cyanide attacks carbon bearing bromine as bromide leaves.
Question 6. What is the expected principal product of 2-bromobutane with NaOCH3 in DMSO?
A. cyclohexene
B. 2-methoxy-2-methylbutane after deprotonation
C. butene isomers
D. benzyl azide
Correct Answer: C. butene isomers
Detailed Rationale: Methoxide abstracts an antiperiplanar beta hydrogen while bromide leaves;
therefore the principal product is butene isomers. Key principle: secondary halide and strong base
compete with substitution.
Why the Other Options Are Incorrect: A — That product corresponds to a different substrate or
reagent sequence; the stated conditions give butene isomers by E2. B — That product corresponds
to a different substrate or reagent sequence; the stated conditions give butene isomers by E2. D —
That product corresponds to a different substrate or reagent sequence; the stated conditions give
butene isomers by E2.
Independent original educational resource • Page 4
CHEM 222 Reaction Mechanisms
Practice Questions
2026–2027 | 110 Original Questions, Answers & Detailed Rationales
Mechanism Strategy and High-Yield Review
1. Identify the electrophile: Locate a polarized bond, carbonyl carbon, activated aromatic electrophile,
or positively charged intermediate.
2. Identify the nucleophile or base: Ask whether the reagent donates an electron pair to carbon or
abstracts a proton.
3. Check the substrate: Steric hindrance, resonance, and carbocation stability govern competing
pathways.
4. Track electron flow: Use curved arrows from electron pairs or bonds to the atoms receiving electrons.
5. Confirm regiochemistry and stereochemistry: Check Markovnikov versus anti-Markovnikov
outcomes, syn/anti addition, inversion, and antiperiplanar elimination.
6. Check the workup: Acid/base workup, tautomerization, and oxidation or reduction steps determine
isolated product
Independent original educational resource • Page 1
,CHEM 222 | REACTION MECHANISMS | 2026–2027
Substitution and elimination
Question 1. Under conventional textbook conditions, which mechanism best describes 1-
bromobutane with NaI in acetone?
A. E1
B. SN2
C. E2
D. SN1
Correct Answer: B. SN2
Detailed Rationale: Iodide attacks the primary carbon from the backside as bromide leaves. Key
principle: primary unhindered substrate and polar aprotic solvent.
Why the Other Options Are Incorrect: A — E1 involves a different bond-making or bond-breaking
pathway; here, iodide attacks the primary carbon from the backside as bromide leaves. C — E2
involves a different bond-making or bond-breaking pathway; here, iodide attacks the primary
carbon from the backside as bromide leaves. D — SN1 involves a different bond-making or bond-
breaking pathway; here, iodide attacks the primary carbon from the backside as bromide leaves.
Question 2. For 2-bromobutane with NaOEt in hot ethanol, what is the key bond-making or bond-
breaking event?
A. bulky base removes a beta hydrogen as bromide leaves
B. cyanide attacks carbon bearing bromine as bromide leaves
C. ethoxide removes a beta hydrogen while bromide departs
D. ionization gives a tertiary carbocation before water attacks
Correct Answer: C. ethoxide removes a beta hydrogen while bromide departs
Detailed Rationale: The mechanism is E2: ethoxide removes a beta hydrogen while bromide
departs. This follows because strong base and heat favor elimination.
Why the Other Options Are Incorrect: A — This describes a different transformation; the specified
reagents follow E2, so ethoxide removes a beta hydrogen while bromide departs. B — This describes
a different transformation; the specified reagents follow E2, so ethoxide removes a beta hydrogen
while bromide departs. D — This describes a different transformation; the specified reagents follow
E2, so ethoxide removes a beta hydrogen while bromide departs.
Independent original educational resource • Page 2
,CHEM 222 | REACTION MECHANISMS | 2026–2027
Question 3. What is the expected principal product of tert-butyl bromide in water?
A. 2-methylpropene
B. butanenitrile
C. butene isomers
D. tert-butanol after deprotonation
Correct Answer: D. tert-butanol after deprotonation
Detailed Rationale: Ionization gives a tertiary carbocation before water attacks; therefore the
principal product is tert-butanol after deprotonation. Key principle: tertiary carbocation and polar
protic solvent.
Why the Other Options Are Incorrect: A — That product corresponds to a different substrate or
reagent sequence; the stated conditions give tert-butanol after deprotonation by SN1. B — That
product corresponds to a different substrate or reagent sequence; the stated conditions give tert-
butanol after deprotonation by SN1. C — That product corresponds to a different substrate or
reagent sequence; the stated conditions give tert-butanol after deprotonation by SN1.
Question 4. Under conventional textbook conditions, which mechanism best describes tert-butyl
bromide with KOtBu?
A. E2
B. SN2
C. SN1
D. E1
Correct Answer: A. E2
Detailed Rationale: Bulky base removes a beta hydrogen as bromide leaves. Key principle: strong
sterically hindered base.
Why the Other Options Are Incorrect: B — SN2 involves a different bond-making or bond-breaking
pathway; here, bulky base removes a beta hydrogen as bromide leaves. C — SN1 involves a
different bond-making or bond-breaking pathway; here, bulky base removes a beta hydrogen as
bromide leaves. D — E1 involves a different bond-making or bond-breaking pathway; here, bulky
base removes a beta hydrogen as bromide leaves.
Independent original educational resource • Page 3
, CHEM 222 | REACTION MECHANISMS | 2026–2027
Question 5. For 1-bromopropane with NaCN in DMSO, what is the key bond-making or bond-
breaking event?
A. base removes an axial beta hydrogen anti to axial leaving group
B. cyanide attacks carbon bearing bromine as bromide leaves
C. methoxide abstracts an antiperiplanar beta hydrogen while bromide leaves
D. azide displaces bromide by backside attack
Correct Answer: B. cyanide attacks carbon bearing bromine as bromide leaves
Detailed Rationale: The mechanism is SN2: cyanide attacks carbon bearing bromine as bromide
leaves. This follows because primary substrate and good carbon nucleophile.
Why the Other Options Are Incorrect: A — This describes a different transformation; the specified
reagents follow SN2, so cyanide attacks carbon bearing bromine as bromide leaves. C — This
describes a different transformation; the specified reagents follow SN2, so cyanide attacks carbon
bearing bromine as bromide leaves. D — This describes a different transformation; the specified
reagents follow SN2, so cyanide attacks carbon bearing bromine as bromide leaves.
Question 6. What is the expected principal product of 2-bromobutane with NaOCH3 in DMSO?
A. cyclohexene
B. 2-methoxy-2-methylbutane after deprotonation
C. butene isomers
D. benzyl azide
Correct Answer: C. butene isomers
Detailed Rationale: Methoxide abstracts an antiperiplanar beta hydrogen while bromide leaves;
therefore the principal product is butene isomers. Key principle: secondary halide and strong base
compete with substitution.
Why the Other Options Are Incorrect: A — That product corresponds to a different substrate or
reagent sequence; the stated conditions give butene isomers by E2. B — That product corresponds
to a different substrate or reagent sequence; the stated conditions give butene isomers by E2. D —
That product corresponds to a different substrate or reagent sequence; the stated conditions give
butene isomers by E2.
Independent original educational resource • Page 4
