University of Michigan
LANIF • 012 MEHC
M College of Literature, Science, and the Arts
ARTES • SCIENTIA • VERITAS
EST. 1817
CHEM 210 — Organic Chemistry
F I N A L E X A M I N AT I O N • ST R U C T U R E , R E A C T I V I TY & 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 — Final Exam TOTAL QUESTIONS 41 Questions
COURSE TITLE Organic Chemistry 210 FORMAT Multiple Choice — Select the Single Best
Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each multiple-choice question.
▸ Questions cover SOPBAR calculations, resonance, hybridization, nomenclature, conformations, stereochemistry, spectroscopy,
acid-base theory, and reaction mechanisms.
▸ Pay close attention to regiochemistry (Markovnikov vs. anti-Markovnikov), stereochemistry (syn vs. anti, cis vs. trans, R/S, E/Z),
and mechanistic intermediates.
▸ Correct answers and detailed rationales appear below each question for comprehensive final exam review.
▸ A periodic table is not provided; key IR frequencies, pKa trends, and relative reactivity values should be known from memory.
SECTION I — FUNDAMENTALS: STRUCTURE, SPECTROSCOPY, Questions 1
STEREOCHEMISTRY & MECHANISMS – 41
1. SOPBAR is used to find which of the following?
A. Number of chiral centers
B. Degrees of unsaturation
C. pKa of an acid
D. Chemical shift in NMR
CORRECT ANSWER B — Degrees of unsaturation
RATIONALE SOPBAR is a mnemonic for the formula used to calculate degrees of unsaturation (also called index of
hydrogen deficiency, IHD). The formula is ((2C+2) − H − X + N)/2, where C = number of carbons, H = hydrogens,
X = halogens (counted as H equivalents), and N = nitrogen (adds one to the hydrogen count). Each degree of
unsaturation corresponds to one π bond or one ring. A double bond = 1 degree, a triple bond = 2 degrees, a
ring = 1 degree. SOPBAR stands for the elements accounted for in the formula (S = subtract, O and P are not
counted, B = both halogens and nitrogens, A = add, R = result divided by 2).
, 2. What is the correct formula for calculating degrees of unsaturation from a molecular formula?
A. ((2C+2) + H − X − N)/2
B. ((2C+2) − H − X + N)/2
C. ((2C−2) + H + X − N)/2
D. ((C+2) − H − X + N)/2
CORRECT ANSWER B — ((2C+2) − H − X + N)/2
RATIONALE The formula ((2C+2) − H − X + N)/2 derives from the fact that a fully saturated acyclic hydrocarbon has the
formula CₙH₂ₙ₊₂. Each degree of unsaturation (π bond or ring) reduces the hydrogen count by 2. Halogens (X)
replace hydrogens one-for-one, so they are subtracted like hydrogens. Nitrogen, being trivalent, adds one
extra hydrogen to the saturated formula for each N present, so it is added in the numerator. Oxygen (divalent)
does not affect the hydrogen count and is ignored. The result is divided by 2 because each unsaturation
removes 2 hydrogens from the saturated reference.
3. Which of the following is NOT a valid resonance rule?
A. Do not break single bonds
B. Do not exceed the octet for second-row elements
C. Keep the net charge the same across all resonance forms
D. Maximize formal charge separation for greater stability
CORRECT ANSWER D — Maximize formal charge separation for greater stability
RATIONALE Resonance rules require minimizing formal charge separation, not maximizing it. The most stable resonance
contributor has full octets, minimal formal charges, and negative charges on the most electronegative atoms.
The three core rules are: (1) never break single (sigma) bonds — only π bonds and lone pairs move; (2) never
exceed the octet for second-row elements (C, N, O, F); (3) the net charge must remain constant across all
resonance forms. Additional guidelines state that structures with complete octets are more stable than those
with electron-deficient atoms, and structures with charge separation are less stable than neutral ones.
4. Which resonance pattern involves a lone pair adjacent to a π bond?
A. Pattern 1: Lone pair next to π bond
B. Pattern 2: π bond next to positive charge
C. Pattern 3: Double bond between atoms of differing electronegativity
D. All of the above are resonance patterns, but A describes Pattern 1 specifically
CORRECT ANSWER D — All of the above are resonance patterns, but A describes Pattern 1 specifically
RATIONALE All three are recognized resonance patterns. Pattern 1: A lone pair adjacent to a π bond (e.g., enolate anion,
allylic lone pair) can delocalize into the π system. Pattern 2: A π bond adjacent to a positive charge (e.g., allylic
carbocation) allows the π electrons to shift toward the positive charge. Pattern 3: A double bond between
atoms of differing electronegativity (e.g., C=O carbonyl) has a significant resonance contributor with charge
separation (C⁺–O⁻). These three patterns cover the vast majority of resonance situations encountered in
organic chemistry. Recognizing them allows rapid prediction of resonance-stabilized intermediates and
products.
LANIF • 012 MEHC
M College of Literature, Science, and the Arts
ARTES • SCIENTIA • VERITAS
EST. 1817
CHEM 210 — Organic Chemistry
F I N A L E X A M I N AT I O N • ST R U C T U R E , R E A C T I V I TY & 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 — Final Exam TOTAL QUESTIONS 41 Questions
COURSE TITLE Organic Chemistry 210 FORMAT Multiple Choice — Select the Single Best
Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each multiple-choice question.
▸ Questions cover SOPBAR calculations, resonance, hybridization, nomenclature, conformations, stereochemistry, spectroscopy,
acid-base theory, and reaction mechanisms.
▸ Pay close attention to regiochemistry (Markovnikov vs. anti-Markovnikov), stereochemistry (syn vs. anti, cis vs. trans, R/S, E/Z),
and mechanistic intermediates.
▸ Correct answers and detailed rationales appear below each question for comprehensive final exam review.
▸ A periodic table is not provided; key IR frequencies, pKa trends, and relative reactivity values should be known from memory.
SECTION I — FUNDAMENTALS: STRUCTURE, SPECTROSCOPY, Questions 1
STEREOCHEMISTRY & MECHANISMS – 41
1. SOPBAR is used to find which of the following?
A. Number of chiral centers
B. Degrees of unsaturation
C. pKa of an acid
D. Chemical shift in NMR
CORRECT ANSWER B — Degrees of unsaturation
RATIONALE SOPBAR is a mnemonic for the formula used to calculate degrees of unsaturation (also called index of
hydrogen deficiency, IHD). The formula is ((2C+2) − H − X + N)/2, where C = number of carbons, H = hydrogens,
X = halogens (counted as H equivalents), and N = nitrogen (adds one to the hydrogen count). Each degree of
unsaturation corresponds to one π bond or one ring. A double bond = 1 degree, a triple bond = 2 degrees, a
ring = 1 degree. SOPBAR stands for the elements accounted for in the formula (S = subtract, O and P are not
counted, B = both halogens and nitrogens, A = add, R = result divided by 2).
, 2. What is the correct formula for calculating degrees of unsaturation from a molecular formula?
A. ((2C+2) + H − X − N)/2
B. ((2C+2) − H − X + N)/2
C. ((2C−2) + H + X − N)/2
D. ((C+2) − H − X + N)/2
CORRECT ANSWER B — ((2C+2) − H − X + N)/2
RATIONALE The formula ((2C+2) − H − X + N)/2 derives from the fact that a fully saturated acyclic hydrocarbon has the
formula CₙH₂ₙ₊₂. Each degree of unsaturation (π bond or ring) reduces the hydrogen count by 2. Halogens (X)
replace hydrogens one-for-one, so they are subtracted like hydrogens. Nitrogen, being trivalent, adds one
extra hydrogen to the saturated formula for each N present, so it is added in the numerator. Oxygen (divalent)
does not affect the hydrogen count and is ignored. The result is divided by 2 because each unsaturation
removes 2 hydrogens from the saturated reference.
3. Which of the following is NOT a valid resonance rule?
A. Do not break single bonds
B. Do not exceed the octet for second-row elements
C. Keep the net charge the same across all resonance forms
D. Maximize formal charge separation for greater stability
CORRECT ANSWER D — Maximize formal charge separation for greater stability
RATIONALE Resonance rules require minimizing formal charge separation, not maximizing it. The most stable resonance
contributor has full octets, minimal formal charges, and negative charges on the most electronegative atoms.
The three core rules are: (1) never break single (sigma) bonds — only π bonds and lone pairs move; (2) never
exceed the octet for second-row elements (C, N, O, F); (3) the net charge must remain constant across all
resonance forms. Additional guidelines state that structures with complete octets are more stable than those
with electron-deficient atoms, and structures with charge separation are less stable than neutral ones.
4. Which resonance pattern involves a lone pair adjacent to a π bond?
A. Pattern 1: Lone pair next to π bond
B. Pattern 2: π bond next to positive charge
C. Pattern 3: Double bond between atoms of differing electronegativity
D. All of the above are resonance patterns, but A describes Pattern 1 specifically
CORRECT ANSWER D — All of the above are resonance patterns, but A describes Pattern 1 specifically
RATIONALE All three are recognized resonance patterns. Pattern 1: A lone pair adjacent to a π bond (e.g., enolate anion,
allylic lone pair) can delocalize into the π system. Pattern 2: A π bond adjacent to a positive charge (e.g., allylic
carbocation) allows the π electrons to shift toward the positive charge. Pattern 3: A double bond between
atoms of differing electronegativity (e.g., C=O carbonyl) has a significant resonance contributor with charge
separation (C⁺–O⁻). These three patterns cover the vast majority of resonance situations encountered in
organic chemistry. Recognizing them allows rapid prediction of resonance-stabilized intermediates and
products.
