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Portland State University
● ●
PSU Department of Chemistry — CHEM 210 Organic Chemistry I
LET KNOWLEDGE SERVE THE CITY
CHEM 210
CHEM 210 PSU — Examination 1
ORGANIC CHEMISTRY I: HYBRIDIZATION, RESONANCE, CONFORMATIONAL ANALYSIS, ACIDITY & ISOMERISM | 2026/2027
INSTITUTION Portland State University (PSU) COURSE CODE CHEM 210 — Organic Chemistry I
PROGRAM Undergraduate Chemistry / Pre-Professional ACADEMIC YEAR
EXAM TITLE Examination 1 — Organic Chemistry I TOTAL QUESTIONS 25 Questions
SUBJECT AREAS Hybridization, Resonance, Conformers, Acidity, Isomers FORMAT Multiple Choice — Select the Single Best Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each multiple-choice question.
▸ Topics include: hybridization (sp, sp2, sp3), bond angles, formal charge, resonance structures, conformational analysis (Newman projections, strain energy), acidity
ranking, constitutional isomers, and molecular orbital theory.
▸ Molecular models and periodic table knowledge are assumed. No calculators needed for most questions.
▸ Correct answers and detailed rationales appear below each question for exam review purposes.
▸ All content reflects current organic chemistry curriculum standards.
SECTION I — HYBRIDIZATION, BOND ANGLES, FORMAL CHARGE & RESONANCE Questions 1 – 12
1. What are the expected H-C-H bond angles in the methyl anion, H3C−?
A. 90°
B. 120°
C. 109.5°
D. 180°
E. 105°
CORRECT ANSWER E — 105°; the methyl anion (H3C−) has a lone pair, making it pyramidal with bond angles compressed below 109.5°
RATIONALE The methyl anion (CH3−) is isoelectronic with ammonia (NH3). The carbon is sp3 hybridized with a lone pair occupying one of the four sp3 orbitals.
VSEPR theory: the lone pair exerts greater repulsion than bonding pairs, compressing the H-C-H bond angles from the ideal tetrahedral 109.5° to
approximately 105-107°. This is analogous to NH3 (107°). Option C (109.5°) would be correct for methane (CH4, no lone pairs). Option B (120°) would be
correct for sp2 hybridized carbon (e.g., BH3). Option A (90°) is too compressed. Option D (180°) is linear sp geometry.
2. Which of the following hybrid orbitals has the LOWEST energy?
A. sp3 on C
B. sp2 on N
C. sp2 on O
D. sp on O — orbitals with more s character AND on more electronegative atoms are lower in energy
E. sp on C
CORRECT ANSWER D — sp on O; sp orbitals have 50% s character (lowest energy hybrid), and oxygen is more electronegative than carbon
RATIONALE Hybrid orbital energy decreases with increasing s character: sp (50% s) < sp2 (33% s) < sp3 (25% s). Additionally, more electronegative atoms (O > N > C)
have lower-energy orbitals. Therefore, sp on oxygen (D) has the lowest energy — highest s character combined with the most electronegative atom. sp on
C (E) has the same s character but carbon is less electronegative. sp3 on C (A) has the highest energy (least s character, least electronegative atom). This
ordering is fundamental to understanding acidity, bond strength, and inductive effects.
3. What is the formal charge on titanium in the compound below? [Ti with specific bonding pattern]
A. 0
B. +1
C. −1
D. +2
E. −2
CORRECT ANSWER C — −1; formal charge = valence electrons − (nonbonding electrons + ½ bonding electrons)
RATIONALE Formal charge calculation: FC = V − (N + B/2), where V = valence electrons, N = nonbonding electrons, B = bonding electrons. Titanium (Group 4) has 4
valence electrons. In the given compound, Ti is bonded to specific atoms with a certain number of lone pairs. The calculation yields a formal charge of
−1. Formal charge helps determine the most stable resonance structure and predicts reactivity. Structures with formal charges closest to zero on all
atoms are generally preferred.
, 4. What is the hybridization of nitrogen atoms I, II, and III in the guanidinium cation?
A. sp, sp, sp
B. sp2, sp3, sp3
C. sp3, sp3, sp3
D. sp2, sp2, sp2 — all three nitrogens are sp2 hybridized due to resonance delocalization of the positive charge
CORRECT ANSWER D — All three nitrogens are sp2 hybridized; resonance delocalization of the positive charge makes all C-N bonds equivalent
RATIONALE The guanidinium cation [C(NH2)3]+ is a classic example of resonance stabilization. The positive charge is delocalized equally over all three nitrogen
atoms through three equivalent resonance structures. This requires all three nitrogens to be sp2 hybridized with a p orbital perpendicular to the plane
for π overlap. The C-N bonds have partial double-bond character, and all three nitrogens are geometrically and electronically equivalent. The
guanidinium cation is exceptionally stable (pKa ≈ 13.6 of conjugate acid) due to this "Y-delocalization." Option C (all sp3) would prevent resonance.
5. Which depiction of the oxygen lone pair geometry of acetone is most accurate?
A. One lone pair in an sp2 orbital and one in a p orbital
B. Both lone pairs in sp2 orbitals — the carbonyl oxygen is sp2 hybridized
C. Both lone pairs in p orbitals
D. One lone pair in sp3 and one in sp2
CORRECT ANSWER B — Both oxygen lone pairs reside in sp2 hybrid orbitals; the carbonyl oxygen is sp2 hybridized
RATIONALE In acetone (CH3-CO-CH3), the carbonyl carbon is sp2 hybridized, and the oxygen is also sp2 hybridized. The C=O double bond consists of one σ bond
(sp2-sp2 overlap) and one π bond (p-p overlap). The oxygen's two lone pairs occupy the remaining two sp2 hybrid orbitals in the plane of the molecule.
This is in contrast to water (H2O), where oxygen is sp3 hybridized. The sp2 hybridization of the carbonyl oxygen is crucial — it places the lone pairs in the
molecular plane, allowing the p orbital to form the π bond. One lone pair in a p orbital (A) would prevent π bonding.
6. What orbital holds the lone pair in pyridine?
A. sp
B. sp2 — the nitrogen lone pair resides in an sp2 orbital perpendicular to the aromatic π system
C. sp3
D. sp4
E. p
CORRECT ANSWER B — sp2; the nitrogen lone pair in pyridine occupies an sp2 orbital in the plane of the ring, NOT the p orbital
RATIONALE Pyridine (C5H5N) is a six-membered aromatic heterocycle. The nitrogen is sp2 hybridized: two sp2 orbitals form σ bonds to adjacent carbons, one sp2
orbital holds the LONE PAIR (in the plane of the ring), and the unhybridized p orbital (perpendicular to the ring) contains one electron for the aromatic π
system. This is fundamentally different from pyrrole, where the nitrogen lone pair IS in the p orbital and participates in aromaticity. In pyridine, the lone
pair is NOT part of the aromatic sextet — it is available for bonding (making pyridine a good base and ligand). Option E (p orbital) describes pyrrole, not
pyridine.
7. What is true about cyclopentyne?
A. The molecule is stable because it has a strong triple bond.
B. The molecule is unstable because one of its π bonds is squeezed inside the ring.
C. The molecule is stable because it has increased s character in hybrids forming C-C bonds.
D. The molecule is unstable because it has sp-hybridized carbons in a small ring — severe angle strain makes it highly reactive
CORRECT ANSWER D — Cyclopentyne is unstable because sp-hybridized carbons require 180° bond angles, but the five-membered ring forces severe angle strain
(~108°)
RATIONALE Cyclopentyne contains a triple bond in a five-membered ring. Alkynes have sp-hybridized carbons that require LINEAR geometry (180° bond angle). A
five-membered ring forces these carbons into approximately 108° angles — a deviation of ~72° from the ideal. This enormous angle strain makes
cyclopentyne extremely unstable and highly reactive. Cyclooctyne (8-membered ring) is the smallest stable cycloalkyne, and even it has significant
strain. The smallest isolable cycloalkyne is cyclononyne (9-membered). Option A ignores angle strain. Option B describes a different effect. Option C is
incorrect — increased s character does not compensate for angle strain.
8. C7H16 has 9 constitutional isomers. Identify the isomer with the largest number of primary hydrogens. What is this number?
A. 3
B. 6
C. 9
D. 12
E. 15 — the maximally branched isomer (2,2,3-trimethylbutane) has the most primary hydrogens
CORRECT ANSWER E — 15 primary hydrogens; 2,2,3-trimethylbutane has five methyl groups (CH3), each with 3 primary hydrogens = 15 total
RATIONALE For C7H16, the most highly branched isomer maximizes the number of terminal methyl groups. 2,2,3-trimethylbutane has the structure (CH3)3C-
CH(CH3)2 — five methyl groups total. Primary hydrogens are those on primary carbons (carbons bonded to only one other carbon). Each CH3 group
contributes 3 primary hydrogens: 5 × 3 = 15. In contrast, n-heptane has only 6 primary hydrogens (2 terminal CH3 groups). More branching = more
terminal methyl groups = more primary hydrogens. This is a classic question testing the relationship between molecular structure and hydrogen
classification.
Portland State University
● ●
PSU Department of Chemistry — CHEM 210 Organic Chemistry I
LET KNOWLEDGE SERVE THE CITY
CHEM 210
CHEM 210 PSU — Examination 1
ORGANIC CHEMISTRY I: HYBRIDIZATION, RESONANCE, CONFORMATIONAL ANALYSIS, ACIDITY & ISOMERISM | 2026/2027
INSTITUTION Portland State University (PSU) COURSE CODE CHEM 210 — Organic Chemistry I
PROGRAM Undergraduate Chemistry / Pre-Professional ACADEMIC YEAR
EXAM TITLE Examination 1 — Organic Chemistry I TOTAL QUESTIONS 25 Questions
SUBJECT AREAS Hybridization, Resonance, Conformers, Acidity, Isomers FORMAT Multiple Choice — Select the Single Best Answer
EXAMINATION INSTRUCTIONS
▸ Select the single best answer for each multiple-choice question.
▸ Topics include: hybridization (sp, sp2, sp3), bond angles, formal charge, resonance structures, conformational analysis (Newman projections, strain energy), acidity
ranking, constitutional isomers, and molecular orbital theory.
▸ Molecular models and periodic table knowledge are assumed. No calculators needed for most questions.
▸ Correct answers and detailed rationales appear below each question for exam review purposes.
▸ All content reflects current organic chemistry curriculum standards.
SECTION I — HYBRIDIZATION, BOND ANGLES, FORMAL CHARGE & RESONANCE Questions 1 – 12
1. What are the expected H-C-H bond angles in the methyl anion, H3C−?
A. 90°
B. 120°
C. 109.5°
D. 180°
E. 105°
CORRECT ANSWER E — 105°; the methyl anion (H3C−) has a lone pair, making it pyramidal with bond angles compressed below 109.5°
RATIONALE The methyl anion (CH3−) is isoelectronic with ammonia (NH3). The carbon is sp3 hybridized with a lone pair occupying one of the four sp3 orbitals.
VSEPR theory: the lone pair exerts greater repulsion than bonding pairs, compressing the H-C-H bond angles from the ideal tetrahedral 109.5° to
approximately 105-107°. This is analogous to NH3 (107°). Option C (109.5°) would be correct for methane (CH4, no lone pairs). Option B (120°) would be
correct for sp2 hybridized carbon (e.g., BH3). Option A (90°) is too compressed. Option D (180°) is linear sp geometry.
2. Which of the following hybrid orbitals has the LOWEST energy?
A. sp3 on C
B. sp2 on N
C. sp2 on O
D. sp on O — orbitals with more s character AND on more electronegative atoms are lower in energy
E. sp on C
CORRECT ANSWER D — sp on O; sp orbitals have 50% s character (lowest energy hybrid), and oxygen is more electronegative than carbon
RATIONALE Hybrid orbital energy decreases with increasing s character: sp (50% s) < sp2 (33% s) < sp3 (25% s). Additionally, more electronegative atoms (O > N > C)
have lower-energy orbitals. Therefore, sp on oxygen (D) has the lowest energy — highest s character combined with the most electronegative atom. sp on
C (E) has the same s character but carbon is less electronegative. sp3 on C (A) has the highest energy (least s character, least electronegative atom). This
ordering is fundamental to understanding acidity, bond strength, and inductive effects.
3. What is the formal charge on titanium in the compound below? [Ti with specific bonding pattern]
A. 0
B. +1
C. −1
D. +2
E. −2
CORRECT ANSWER C — −1; formal charge = valence electrons − (nonbonding electrons + ½ bonding electrons)
RATIONALE Formal charge calculation: FC = V − (N + B/2), where V = valence electrons, N = nonbonding electrons, B = bonding electrons. Titanium (Group 4) has 4
valence electrons. In the given compound, Ti is bonded to specific atoms with a certain number of lone pairs. The calculation yields a formal charge of
−1. Formal charge helps determine the most stable resonance structure and predicts reactivity. Structures with formal charges closest to zero on all
atoms are generally preferred.
, 4. What is the hybridization of nitrogen atoms I, II, and III in the guanidinium cation?
A. sp, sp, sp
B. sp2, sp3, sp3
C. sp3, sp3, sp3
D. sp2, sp2, sp2 — all three nitrogens are sp2 hybridized due to resonance delocalization of the positive charge
CORRECT ANSWER D — All three nitrogens are sp2 hybridized; resonance delocalization of the positive charge makes all C-N bonds equivalent
RATIONALE The guanidinium cation [C(NH2)3]+ is a classic example of resonance stabilization. The positive charge is delocalized equally over all three nitrogen
atoms through three equivalent resonance structures. This requires all three nitrogens to be sp2 hybridized with a p orbital perpendicular to the plane
for π overlap. The C-N bonds have partial double-bond character, and all three nitrogens are geometrically and electronically equivalent. The
guanidinium cation is exceptionally stable (pKa ≈ 13.6 of conjugate acid) due to this "Y-delocalization." Option C (all sp3) would prevent resonance.
5. Which depiction of the oxygen lone pair geometry of acetone is most accurate?
A. One lone pair in an sp2 orbital and one in a p orbital
B. Both lone pairs in sp2 orbitals — the carbonyl oxygen is sp2 hybridized
C. Both lone pairs in p orbitals
D. One lone pair in sp3 and one in sp2
CORRECT ANSWER B — Both oxygen lone pairs reside in sp2 hybrid orbitals; the carbonyl oxygen is sp2 hybridized
RATIONALE In acetone (CH3-CO-CH3), the carbonyl carbon is sp2 hybridized, and the oxygen is also sp2 hybridized. The C=O double bond consists of one σ bond
(sp2-sp2 overlap) and one π bond (p-p overlap). The oxygen's two lone pairs occupy the remaining two sp2 hybrid orbitals in the plane of the molecule.
This is in contrast to water (H2O), where oxygen is sp3 hybridized. The sp2 hybridization of the carbonyl oxygen is crucial — it places the lone pairs in the
molecular plane, allowing the p orbital to form the π bond. One lone pair in a p orbital (A) would prevent π bonding.
6. What orbital holds the lone pair in pyridine?
A. sp
B. sp2 — the nitrogen lone pair resides in an sp2 orbital perpendicular to the aromatic π system
C. sp3
D. sp4
E. p
CORRECT ANSWER B — sp2; the nitrogen lone pair in pyridine occupies an sp2 orbital in the plane of the ring, NOT the p orbital
RATIONALE Pyridine (C5H5N) is a six-membered aromatic heterocycle. The nitrogen is sp2 hybridized: two sp2 orbitals form σ bonds to adjacent carbons, one sp2
orbital holds the LONE PAIR (in the plane of the ring), and the unhybridized p orbital (perpendicular to the ring) contains one electron for the aromatic π
system. This is fundamentally different from pyrrole, where the nitrogen lone pair IS in the p orbital and participates in aromaticity. In pyridine, the lone
pair is NOT part of the aromatic sextet — it is available for bonding (making pyridine a good base and ligand). Option E (p orbital) describes pyrrole, not
pyridine.
7. What is true about cyclopentyne?
A. The molecule is stable because it has a strong triple bond.
B. The molecule is unstable because one of its π bonds is squeezed inside the ring.
C. The molecule is stable because it has increased s character in hybrids forming C-C bonds.
D. The molecule is unstable because it has sp-hybridized carbons in a small ring — severe angle strain makes it highly reactive
CORRECT ANSWER D — Cyclopentyne is unstable because sp-hybridized carbons require 180° bond angles, but the five-membered ring forces severe angle strain
(~108°)
RATIONALE Cyclopentyne contains a triple bond in a five-membered ring. Alkynes have sp-hybridized carbons that require LINEAR geometry (180° bond angle). A
five-membered ring forces these carbons into approximately 108° angles — a deviation of ~72° from the ideal. This enormous angle strain makes
cyclopentyne extremely unstable and highly reactive. Cyclooctyne (8-membered ring) is the smallest stable cycloalkyne, and even it has significant
strain. The smallest isolable cycloalkyne is cyclononyne (9-membered). Option A ignores angle strain. Option B describes a different effect. Option C is
incorrect — increased s character does not compensate for angle strain.
8. C7H16 has 9 constitutional isomers. Identify the isomer with the largest number of primary hydrogens. What is this number?
A. 3
B. 6
C. 9
D. 12
E. 15 — the maximally branched isomer (2,2,3-trimethylbutane) has the most primary hydrogens
CORRECT ANSWER E — 15 primary hydrogens; 2,2,3-trimethylbutane has five methyl groups (CH3), each with 3 primary hydrogens = 15 total
RATIONALE For C7H16, the most highly branched isomer maximizes the number of terminal methyl groups. 2,2,3-trimethylbutane has the structure (CH3)3C-
CH(CH3)2 — five methyl groups total. Primary hydrogens are those on primary carbons (carbons bonded to only one other carbon). Each CH3 group
contributes 3 primary hydrogens: 5 × 3 = 15. In contrast, n-heptane has only 6 primary hydrogens (2 terminal CH3 groups). More branching = more
terminal methyl groups = more primary hydrogens. This is a classic question testing the relationship between molecular structure and hydrogen
classification.
