CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
CHEM 222 Organic Chemistry II
Final Practice Exam 2026–2027 | 125 Questions & Worked Solutions
Independent educational practice material, not an official university exam. Questions 1–110 are adapted
from the companion comprehensive study guide; questions 111–125 add numerical and interpretive
problems. Exact course scope varies by institution.
Instructions
Answer all 125 questions before reviewing the worked solutions. For reagent problems, identify the
reacting functional group, mechanism, regioselectivity and stereochemistry. For calculations, show the
formula, substitution, and result.
Quick Reference
Degree of unsaturation: DBE = (2C + 2 + N − H − X)/2; oxygen and sulfur are omitted.
Percent yield: Actual yield ÷ theoretical yield × 100%.
Chromatography: Rf = distance traveled by compound ÷ distance traveled by solvent front.
First-order half-life: t½ = 0.693/k.
Aromaticity: Cyclic, planar, fully conjugated, with 4n+2 π electrons.
NMR: Integration counts relative hydrogens; simple splitting follows n+1 for n equivalent neighbors.
Part 1 — Organic chemistry review
Question 1. What is the hybridization of a carbonyl carbon?
A. sp
B. sp3d
C. sp2
D. sp3
Correct Answer: C. sp2
Original educational review | Page 1
,CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: The carbonyl carbon has three
electron domains and trigonal-planar geometry.
Why other options are incorrect: A (sp): This alternative does not match the key functional-group
behavior or definition established in the worked solution. B (sp3d): This alternative does not match
the key functional-group behavior or definition established in the worked solution. D (sp3): This
alternative does not match the key functional-group behavior or definition established in the
worked solution.
Question 2. Which factor best stabilizes a carboxylate conjugate base?
A. Resonance delocalization across two oxygens
B. A localized carbon-centered charge
C. A tetrahedral carbonyl carbon
D. Loss of oxygen electronegativity
Correct Answer: A. Resonance delocalization across two oxygens
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Carboxylate negative charge is
delocalized over two oxygen atoms.
Why other options are incorrect: B (A localized carbon-centered charge): This alternative does not
match the key functional-group behavior or definition established in the worked solution. C (A
tetrahedral carbonyl carbon): This alternative does not match the key functional-group behavior or
definition established in the worked solution. D (Loss of oxygen electronegativity): This alternative
does not match the key functional-group behavior or definition established in the worked solution.
Question 3. Which compound is generally the strongest acid?
A. An alkane
B. An ether
C. A carboxylic acid
D. An alcohol
Correct Answer: C. A carboxylic acid
Original educational review | Page 2
,CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: A resonance-stabilized
carboxylate forms upon deprotonation.
Why other options are incorrect: A (An alkane): This alternative does not match the key functional-
group behavior or definition established in the worked solution. B (An ether): This alternative does
not match the key functional-group behavior or definition established in the worked solution. D (An
alcohol): This alternative does not match the key functional-group behavior or definition
established in the worked solution.
Question 4. Which orbital overlap forms the pi bond in an alkene?
A. Side-by-side p-orbital overlap
B. End-on sp3 overlap
C. Hydrogen 1s–1s overlap
D. Side-by-side s orbitals
Correct Answer: A. Side-by-side p-orbital overlap
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Parallel p orbitals overlap
laterally to form a pi bond.
Why other options are incorrect: B (End-on sp3 overlap): This alternative does not match the key
functional-group behavior or definition established in the worked solution. C (Hydrogen 1s–1s
overlap): This alternative does not match the key functional-group behavior or definition
established in the worked solution. D (Side-by-side s orbitals): This alternative does not match the
key functional-group behavior or definition established in the worked solution.
Question 5. What is the formal charge on oxygen with three single bonds and one lone pair?
A. −1
B. +2
C. +1
D. 0
Correct Answer: C. +1
Original educational review | Page 3
, CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Oxygen has six valence electrons;
one lone pair and three bonds assign five electrons, giving +1.
Why other options are incorrect: A (−1): This alternative does not match the key functional-group
behavior or definition established in the worked solution. B (+2): This alternative does not match the
key functional-group behavior or definition established in the worked solution. D (0): This
alternative does not match the key functional-group behavior or definition established in the
worked solution.
Question 6. Which species is typically the best leaving group?
A. Iodide
B. Fluoride
C. Hydroxide
D. Amide anion
Correct Answer: A. Iodide
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Iodide is relatively stable after
departure and is a weak base.
Why other options are incorrect: B (Fluoride): This alternative does not match the key functional-
group behavior or definition established in the worked solution. C (Hydroxide): This alternative does
not match the key functional-group behavior or definition established in the worked solution. D
(Amide anion): This alternative does not match the key functional-group behavior or definition
established in the worked solution.
Question 7. Which feature makes an amide nitrogen less basic than an amine nitrogen?
A. Absence of nitrogen lone pairs
B. A formal negative charge on nitrogen
C. Resonance donation into the carbonyl
D. Extra sp3 carbon bonds
Correct Answer: C. Resonance donation into the carbonyl
Original educational review | Page 4
CHEM 222 Organic Chemistry II
Final Practice Exam 2026–2027 | 125 Questions & Worked Solutions
Independent educational practice material, not an official university exam. Questions 1–110 are adapted
from the companion comprehensive study guide; questions 111–125 add numerical and interpretive
problems. Exact course scope varies by institution.
Instructions
Answer all 125 questions before reviewing the worked solutions. For reagent problems, identify the
reacting functional group, mechanism, regioselectivity and stereochemistry. For calculations, show the
formula, substitution, and result.
Quick Reference
Degree of unsaturation: DBE = (2C + 2 + N − H − X)/2; oxygen and sulfur are omitted.
Percent yield: Actual yield ÷ theoretical yield × 100%.
Chromatography: Rf = distance traveled by compound ÷ distance traveled by solvent front.
First-order half-life: t½ = 0.693/k.
Aromaticity: Cyclic, planar, fully conjugated, with 4n+2 π electrons.
NMR: Integration counts relative hydrogens; simple splitting follows n+1 for n equivalent neighbors.
Part 1 — Organic chemistry review
Question 1. What is the hybridization of a carbonyl carbon?
A. sp
B. sp3d
C. sp2
D. sp3
Correct Answer: C. sp2
Original educational review | Page 1
,CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: The carbonyl carbon has three
electron domains and trigonal-planar geometry.
Why other options are incorrect: A (sp): This alternative does not match the key functional-group
behavior or definition established in the worked solution. B (sp3d): This alternative does not match
the key functional-group behavior or definition established in the worked solution. D (sp3): This
alternative does not match the key functional-group behavior or definition established in the
worked solution.
Question 2. Which factor best stabilizes a carboxylate conjugate base?
A. Resonance delocalization across two oxygens
B. A localized carbon-centered charge
C. A tetrahedral carbonyl carbon
D. Loss of oxygen electronegativity
Correct Answer: A. Resonance delocalization across two oxygens
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Carboxylate negative charge is
delocalized over two oxygen atoms.
Why other options are incorrect: B (A localized carbon-centered charge): This alternative does not
match the key functional-group behavior or definition established in the worked solution. C (A
tetrahedral carbonyl carbon): This alternative does not match the key functional-group behavior or
definition established in the worked solution. D (Loss of oxygen electronegativity): This alternative
does not match the key functional-group behavior or definition established in the worked solution.
Question 3. Which compound is generally the strongest acid?
A. An alkane
B. An ether
C. A carboxylic acid
D. An alcohol
Correct Answer: C. A carboxylic acid
Original educational review | Page 2
,CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: A resonance-stabilized
carboxylate forms upon deprotonation.
Why other options are incorrect: A (An alkane): This alternative does not match the key functional-
group behavior or definition established in the worked solution. B (An ether): This alternative does
not match the key functional-group behavior or definition established in the worked solution. D (An
alcohol): This alternative does not match the key functional-group behavior or definition
established in the worked solution.
Question 4. Which orbital overlap forms the pi bond in an alkene?
A. Side-by-side p-orbital overlap
B. End-on sp3 overlap
C. Hydrogen 1s–1s overlap
D. Side-by-side s orbitals
Correct Answer: A. Side-by-side p-orbital overlap
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Parallel p orbitals overlap
laterally to form a pi bond.
Why other options are incorrect: B (End-on sp3 overlap): This alternative does not match the key
functional-group behavior or definition established in the worked solution. C (Hydrogen 1s–1s
overlap): This alternative does not match the key functional-group behavior or definition
established in the worked solution. D (Side-by-side s orbitals): This alternative does not match the
key functional-group behavior or definition established in the worked solution.
Question 5. What is the formal charge on oxygen with three single bonds and one lone pair?
A. −1
B. +2
C. +1
D. 0
Correct Answer: C. +1
Original educational review | Page 3
, CHEM 222 | ORGANIC CHEMISTRY II | FINAL PRACTICE EXAM
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Oxygen has six valence electrons;
one lone pair and three bonds assign five electrons, giving +1.
Why other options are incorrect: A (−1): This alternative does not match the key functional-group
behavior or definition established in the worked solution. B (+2): This alternative does not match the
key functional-group behavior or definition established in the worked solution. D (0): This
alternative does not match the key functional-group behavior or definition established in the
worked solution.
Question 6. Which species is typically the best leaving group?
A. Iodide
B. Fluoride
C. Hydroxide
D. Amide anion
Correct Answer: A. Iodide
Worked Solution: Step 1: Identify the functional group or chemical principle in the stem. Step 2:
Compare the available structures, reagents, or outcomes. Step 3: Iodide is relatively stable after
departure and is a weak base.
Why other options are incorrect: B (Fluoride): This alternative does not match the key functional-
group behavior or definition established in the worked solution. C (Hydroxide): This alternative does
not match the key functional-group behavior or definition established in the worked solution. D
(Amide anion): This alternative does not match the key functional-group behavior or definition
established in the worked solution.
Question 7. Which feature makes an amide nitrogen less basic than an amine nitrogen?
A. Absence of nitrogen lone pairs
B. A formal negative charge on nitrogen
C. Resonance donation into the carbonyl
D. Extra sp3 carbon bonds
Correct Answer: C. Resonance donation into the carbonyl
Original educational review | Page 4
