TEST BANK FOR ORGANIC CHEMISTRY PRINCIPLES AND MECHANISMS 3RD ED.ITION (JOEL M. KARTY)
,TABLE OF CONTENTS
PART I — FOUNDATIONS OF ORGANIC CHEMISTRY
Chapter 1
Atomic and Molecular Structure
Chapter 2
Three-Dimensional Geometry, Intermolecular Interactions, and Physical Properties
Chapter 3
Valence Bond Theory and Molecular Orbital Theory
Chapter 4
Isomerism I — Conformers and Constitutional Isomers
Chapter 5
Isomerism II — Chirality, Enantiomers, and Diastereomers
Chapter 6
Proton Transfer Reactions
Chapter 7
Elementary Steps in Organic Mechanisms
Chapter 8
An Introduction to Multistep Mechanisms — SN1 and E1 Reactions and Their Comparisons to SN2 and E2
Reactions
Chapter 9
Competition among SN2, SN1, E2, and E1 Reactions
PART II — ORGANIC SYNTHESIS & ALKENE CHEMISTRY
Chapter 10
Organic Synthesis I — Nucleophilic Substitution and Elimination Reactions and Functional Group
Transformations
Chapter 11
Organic Synthesis II — Reactions That Alter the Carbon Skeleton, and Designing Multistep Syntheses
Chapter 12
,Electrophilic Addition to Nonpolar π Bonds I — Addition of a Brønsted Acid
Chapter 13
Electrophilic Addition to Nonpolar π Bonds II
Chapter 14
Conjugation and Aromaticity
PART III — SPECTROSCOPY & STRUCTURE DETERMINATION
Chapter 15
Structure Determination I — Mass Spectrometry
Chapter 16
Structure Determination II — Infrared Spectroscopy and Ultraviolet–Visible Spectroscopy
Chapter 17
Structure Determination III — Nuclear Magnetic Resonance Spectroscopy
PART IV — CARBONYL CHEMISTRY & ADVANCED SYNTHESIS
Chapter 18
Nucleophilic Addition to Polar π Bonds I — Reagents That Are Strongly Nucleophilic
Chapter 19
Nucleophilic Addition to Polar π Bonds II — Reagents That Are Weakly Nucleophilic or Non-nucleophilic, and
Acid and Base Catalysis
Chapter 20
Redox Reactions; Organometallic Reagents and Their Reactions
Chapter 21
Organic Synthesis III — Intermediate Topics in Synthesis Design
Chapter 22
Nucleophilic Addition–Elimination Reactions I — Reagents That Are Strongly Nucleophilic
Chapter 23
Nucleophilic Addition–Elimination Reactions II — Reagents That Are Weakly Nucleophilic or Non-
nucleophilic
,PART V — AROMATIC CHEMISTRY & PERICYCLIC REACTIONS
Chapter 24
Aromatic Substitution I — Electrophilic Aromatic Substitution on Benzene and Useful Accompanying
Reactions
Chapter 25
Aromatic Substitution II — Reactions of Substituted Benzenes and Other Rings
Chapter 26
The Diels–Alder Reaction, Syn Dihydroxylation, and Oxidative Cleavage
Chapter 27
Reactions Involving Radicals
PART VI — POLYMER & BIOLOGICAL ORGANIC CHEMISTRY
Chapter 28
Polymers
Chapter 29
Biomolecules I — An Overview of the Four Major Classes of Biomolecules
Chapter 30
Biomolecules II — Representative Biochemical Processes Involving Biomolecules
Exam Bank Features
• Comprehensive Coverage of All 30 Chapters
• Premium Board-Style Mechanism Challenges
• Original Multiple-Choice Questions
• Detailed Answer Rationales
• Why the Other Options Are Incorrect
• Mechanism Insights
• High-Yield Organic Chemistry Exam Tips
• Ideal for Undergraduate Organic Chemistry, ACS Preparation, and Comprehensive Course Review
,Chapter 1 — Atomic and Molecular Structure
Mechanism Challenge 1
A synthetic organic chemist is designing a new reaction that requires a carbon-carbon bond capable of
withstanding high temperatures without readily breaking. Which bond would provide the greatest bond
strength under these conditions?
A. Carbon-carbon single bond (C–C)
B. Carbon-carbon double bond (C=C)
C. Carbon-carbon triple bond (C≡C)
D. Carbon-hydrogen bond (C–H)
Correct Answer
C. Carbon-carbon triple bond (C≡C)
Comprehensive Rationale
A carbon-carbon triple bond is the strongest covalent bond commonly encountered between two carbon
atoms. It consists of one σ bond and two π bonds, producing the highest bond order (3). Greater bond order
results in stronger orbital overlap, shorter bond length, and a higher bond dissociation energy than either a
double or single bond. Consequently, alkynes possess shorter and stronger carbon-carbon bonds than alkenes
or alkanes.
Why the Other Options Are Incorrect
A. A carbon-carbon single bond contains only one σ bond and therefore has the lowest bond order and the
weakest carbon-carbon bond.
B. A carbon-carbon double bond is stronger than a single bond but weaker than a triple bond because it
contains one σ bond and one π bond.
D. Although many carbon-hydrogen bonds are relatively strong, the question specifically compares carbon-
carbon bonds.
Mechanism Insight
Bond order directly influences molecular stability. As bond order increases from one to three, bond length
decreases while bond strength increases. These trends are fundamental for understanding reaction
mechanisms and predicting which bonds are most resistant to cleavage.
Exam Tip
Remember the relationship:
Bond Strength: C≡C > C=C > C–C
,Bond Length: C–C > C=C > C≡C
Mechanism Challenge 2
A graduate student compares methane (CH₄), ethene (C₂H₄), and ethyne (C₂H₂) while studying carbon
hybridization. Which statement correctly explains why the carbon atoms in ethyne hold their bonding
electrons closer to the nucleus than those in methane?
A. sp-hybridized orbitals contain a greater percentage of s-character than sp³-hybridized orbitals.
B. sp³-hybridized orbitals contain more s-character than sp orbitals.
C. Triple bonds contain only π bonds, which pull electrons closer to the nucleus.
D. Carbon atoms in alkynes have a greater nuclear charge than carbon atoms in alkanes.
Correct Answer
A. sp-hybridized orbitals contain a greater percentage of s-character than sp³-hybridized orbitals.
Comprehensive Rationale
An sp-hybridized carbon contains 50% s-character, whereas an sp³-hybridized carbon contains only 25% s-
character. Because s orbitals are closer to the nucleus than p orbitals, electrons in orbitals with greater s-
character experience stronger nuclear attraction. This produces shorter, stronger bonds and contributes to the
greater acidity of terminal alkynes.
Why the Other Options Are Incorrect
B. The relationship is reversed; sp³ orbitals have the lowest percentage of s-character.
C. A triple bond consists of one σ bond and two π bonds, not only π bonds.
D. Every carbon atom has the same nuclear charge (six protons). Hybridization—not nuclear charge—accounts
for the difference.
Mechanism Insight
The percentage of s-character influences several important properties, including bond length, bond strength,
electronegativity, and acidity. These trends are frequently applied when comparing reaction intermediates and
predicting organic reactivity.
Exam Tip
s-character trend:
sp (50%) > sp² (33%) > sp³ (25%)
More s-character means stronger, shorter bonds and greater acidity of attached hydrogens.
,Mechanism Challenge 3
An instructor asks students to identify the primary reason carbon is capable of forming millions of stable
organic compounds. Which property of carbon best explains this extraordinary structural diversity?
A. Carbon readily forms strong covalent bonds with itself, allowing the formation of long chains, rings, and
branched structures.
B. Carbon always forms ionic compounds with other elements.
C. Carbon possesses the highest electronegativity in the periodic table.
D. Carbon can accommodate more than eight electrons in its valence shell.
Correct Answer
A. Carbon readily forms strong covalent bonds with itself, allowing the formation of long chains, rings, and
branched structures.
Comprehensive Rationale
Carbon's ability to catenate—form stable covalent bonds with other carbon atoms—is the foundation of
organic chemistry. Combined with its tetravalency and ability to form single, double, and triple bonds, carbon
can generate an enormous variety of molecular architectures, including linear, branched, cyclic, and aromatic
compounds.
Why the Other Options Are Incorrect
B. Organic compounds are dominated by covalent rather than ionic bonding.
C. Fluorine, not carbon, is the most electronegative element.
D. Carbon obeys the octet rule and does not normally expand its valence shell.
Mechanism Insight
Carbon's unique bonding capabilities explain why it serves as the backbone of nearly all biologically important
molecules and synthetic organic compounds.
Exam Tip
When asked why carbon is the central element of organic chemistry, remember the combination of
tetravalency, catenation, and the ability to form single, double, and triple bonds.
Mechanism Challenge 4
While constructing the Lewis structure of carbon dioxide (CO₂), a student notices that the central carbon atom
forms two double bonds. Why is this arrangement favored?
A. It allows every atom to achieve a complete valence shell while minimizing formal charges.
B. Carbon cannot form single bonds with oxygen.
,C. Oxygen prefers to have six bonding electrons instead of eight.
D. Carbon expands its valence shell beyond eight electrons.
Correct Answer
A. It allows every atom to achieve a complete valence shell while minimizing formal charges.
Comprehensive Rationale
The most stable Lewis structure for carbon dioxide contains two carbon-oxygen double bonds. This
arrangement gives carbon and both oxygen atoms complete octets while producing formal charges of zero on
all atoms. Lewis structures that minimize formal charges generally represent the most stable electron
distribution.
Why the Other Options Are Incorrect
B. Carbon can form single bonds with oxygen in many compounds, such as alcohols and ethers.
C. Oxygen typically seeks an octet of eight valence electrons, not six.
D. Carbon is a second-period element and cannot expand its valence shell beyond an octet under normal
conditions.
Mechanism Insight
Evaluating formal charges is one of the most reliable methods for selecting the most stable Lewis structure
before predicting molecular reactivity.
Exam Tip
When comparing Lewis structures, prioritize those that:
• Give atoms complete octets whenever possible.
• Minimize formal charges.
• Place negative formal charges on more electronegative atoms.
Mechanism Challenge 5
A researcher compares carbon (C), nitrogen (N), oxygen (O), and fluorine (F) across the second period of the
periodic table. Which trend correctly describes how electronegativity changes from carbon to fluorine?
A. It decreases because atomic size increases.
B. It remains essentially constant across the period.
C. It increases because effective nuclear charge increases while atomic radius decreases.
D. It first decreases and then increases.
, Correct Answer
C. It increases because effective nuclear charge increases while atomic radius decreases.
Comprehensive Rationale
Moving from left to right across a period, the number of protons increases while shielding changes very little.
As a result, the effective nuclear charge experienced by valence electrons increases, causing atoms to attract
shared electrons more strongly. Consequently, electronegativity increases across the period, with fluorine
being the most electronegative element.
Why the Other Options Are Incorrect
A. Atomic radius decreases—not increases—across a period.
B. Electronegativity changes significantly across the second period.
D. The trend is a steady increase rather than an irregular pattern.
Mechanism Insight
Electronegativity determines bond polarity, influences reaction mechanisms, and helps predict nucleophilic
and electrophilic behavior.
Exam Tip
Across a period:
• Electronegativity ↑
• Ionization energy ↑
• Atomic radius ↓
These periodic trends are frequently tested together.
Mechanism Challenge 6
During the analysis of hydrogen fluoride (HF), a student concludes that the bond between hydrogen and
fluorine is highly polar. What is the primary reason for this observation?
A. Fluorine attracts the shared bonding electrons much more strongly than hydrogen.
B. Hydrogen contributes two valence electrons to the bond.
C. Fluorine forms ionic bonds with every element.
D. The electrons are shared equally between hydrogen and fluorine.
Correct Answer
A. Fluorine attracts the shared bonding electrons much more strongly than hydrogen.
, Comprehensive Rationale
Fluorine has the highest electronegativity of any element, allowing it to attract shared bonding electrons more
strongly than hydrogen. This unequal sharing creates a polar covalent bond with a partial negative charge (δ–)
on fluorine and a partial positive charge (δ+) on hydrogen.
Bond polarity influences physical properties such as boiling point, intermolecular forces, and chemical
reactivity.
Why the Other Options Are Incorrect
B. Hydrogen contributes one valence electron, not two.
C. Fluorine forms many covalent compounds, including HF.
D. Equal sharing would produce a nonpolar covalent bond.
Mechanism Insight
Bond polarity determines electron-rich and electron-poor regions within molecules, allowing chemists to
predict sites of nucleophilic and electrophilic attack.
Exam Tip
Greater differences in electronegativity produce greater bond polarity, but not every polar bond is ionic.
Mechanism Challenge 7
An undergraduate is comparing sodium chloride (NaCl) with methane (CH₄). Which statement correctly
distinguishes the bonding found in these two compounds?
A. Sodium chloride contains ionic bonding, whereas methane contains covalent bonding.
B. Both compounds contain only ionic bonds.
C. Both compounds contain only covalent bonds.
D. Methane contains ionic bonds because carbon is more electronegative than hydrogen.
Correct Answer
A. Sodium chloride contains ionic bonding, whereas methane contains covalent bonding.
Comprehensive Rationale
Sodium chloride forms when sodium transfers an electron to chlorine, producing oppositely charged ions held
together by electrostatic attraction. Methane, in contrast, forms when carbon and hydrogen share electrons
through covalent bonds. Understanding the distinction between electron transfer and electron sharing is
fundamental to predicting the properties of chemical compounds.
Why the Other Options Are Incorrect
,TABLE OF CONTENTS
PART I — FOUNDATIONS OF ORGANIC CHEMISTRY
Chapter 1
Atomic and Molecular Structure
Chapter 2
Three-Dimensional Geometry, Intermolecular Interactions, and Physical Properties
Chapter 3
Valence Bond Theory and Molecular Orbital Theory
Chapter 4
Isomerism I — Conformers and Constitutional Isomers
Chapter 5
Isomerism II — Chirality, Enantiomers, and Diastereomers
Chapter 6
Proton Transfer Reactions
Chapter 7
Elementary Steps in Organic Mechanisms
Chapter 8
An Introduction to Multistep Mechanisms — SN1 and E1 Reactions and Their Comparisons to SN2 and E2
Reactions
Chapter 9
Competition among SN2, SN1, E2, and E1 Reactions
PART II — ORGANIC SYNTHESIS & ALKENE CHEMISTRY
Chapter 10
Organic Synthesis I — Nucleophilic Substitution and Elimination Reactions and Functional Group
Transformations
Chapter 11
Organic Synthesis II — Reactions That Alter the Carbon Skeleton, and Designing Multistep Syntheses
Chapter 12
,Electrophilic Addition to Nonpolar π Bonds I — Addition of a Brønsted Acid
Chapter 13
Electrophilic Addition to Nonpolar π Bonds II
Chapter 14
Conjugation and Aromaticity
PART III — SPECTROSCOPY & STRUCTURE DETERMINATION
Chapter 15
Structure Determination I — Mass Spectrometry
Chapter 16
Structure Determination II — Infrared Spectroscopy and Ultraviolet–Visible Spectroscopy
Chapter 17
Structure Determination III — Nuclear Magnetic Resonance Spectroscopy
PART IV — CARBONYL CHEMISTRY & ADVANCED SYNTHESIS
Chapter 18
Nucleophilic Addition to Polar π Bonds I — Reagents That Are Strongly Nucleophilic
Chapter 19
Nucleophilic Addition to Polar π Bonds II — Reagents That Are Weakly Nucleophilic or Non-nucleophilic, and
Acid and Base Catalysis
Chapter 20
Redox Reactions; Organometallic Reagents and Their Reactions
Chapter 21
Organic Synthesis III — Intermediate Topics in Synthesis Design
Chapter 22
Nucleophilic Addition–Elimination Reactions I — Reagents That Are Strongly Nucleophilic
Chapter 23
Nucleophilic Addition–Elimination Reactions II — Reagents That Are Weakly Nucleophilic or Non-
nucleophilic
,PART V — AROMATIC CHEMISTRY & PERICYCLIC REACTIONS
Chapter 24
Aromatic Substitution I — Electrophilic Aromatic Substitution on Benzene and Useful Accompanying
Reactions
Chapter 25
Aromatic Substitution II — Reactions of Substituted Benzenes and Other Rings
Chapter 26
The Diels–Alder Reaction, Syn Dihydroxylation, and Oxidative Cleavage
Chapter 27
Reactions Involving Radicals
PART VI — POLYMER & BIOLOGICAL ORGANIC CHEMISTRY
Chapter 28
Polymers
Chapter 29
Biomolecules I — An Overview of the Four Major Classes of Biomolecules
Chapter 30
Biomolecules II — Representative Biochemical Processes Involving Biomolecules
Exam Bank Features
• Comprehensive Coverage of All 30 Chapters
• Premium Board-Style Mechanism Challenges
• Original Multiple-Choice Questions
• Detailed Answer Rationales
• Why the Other Options Are Incorrect
• Mechanism Insights
• High-Yield Organic Chemistry Exam Tips
• Ideal for Undergraduate Organic Chemistry, ACS Preparation, and Comprehensive Course Review
,Chapter 1 — Atomic and Molecular Structure
Mechanism Challenge 1
A synthetic organic chemist is designing a new reaction that requires a carbon-carbon bond capable of
withstanding high temperatures without readily breaking. Which bond would provide the greatest bond
strength under these conditions?
A. Carbon-carbon single bond (C–C)
B. Carbon-carbon double bond (C=C)
C. Carbon-carbon triple bond (C≡C)
D. Carbon-hydrogen bond (C–H)
Correct Answer
C. Carbon-carbon triple bond (C≡C)
Comprehensive Rationale
A carbon-carbon triple bond is the strongest covalent bond commonly encountered between two carbon
atoms. It consists of one σ bond and two π bonds, producing the highest bond order (3). Greater bond order
results in stronger orbital overlap, shorter bond length, and a higher bond dissociation energy than either a
double or single bond. Consequently, alkynes possess shorter and stronger carbon-carbon bonds than alkenes
or alkanes.
Why the Other Options Are Incorrect
A. A carbon-carbon single bond contains only one σ bond and therefore has the lowest bond order and the
weakest carbon-carbon bond.
B. A carbon-carbon double bond is stronger than a single bond but weaker than a triple bond because it
contains one σ bond and one π bond.
D. Although many carbon-hydrogen bonds are relatively strong, the question specifically compares carbon-
carbon bonds.
Mechanism Insight
Bond order directly influences molecular stability. As bond order increases from one to three, bond length
decreases while bond strength increases. These trends are fundamental for understanding reaction
mechanisms and predicting which bonds are most resistant to cleavage.
Exam Tip
Remember the relationship:
Bond Strength: C≡C > C=C > C–C
,Bond Length: C–C > C=C > C≡C
Mechanism Challenge 2
A graduate student compares methane (CH₄), ethene (C₂H₄), and ethyne (C₂H₂) while studying carbon
hybridization. Which statement correctly explains why the carbon atoms in ethyne hold their bonding
electrons closer to the nucleus than those in methane?
A. sp-hybridized orbitals contain a greater percentage of s-character than sp³-hybridized orbitals.
B. sp³-hybridized orbitals contain more s-character than sp orbitals.
C. Triple bonds contain only π bonds, which pull electrons closer to the nucleus.
D. Carbon atoms in alkynes have a greater nuclear charge than carbon atoms in alkanes.
Correct Answer
A. sp-hybridized orbitals contain a greater percentage of s-character than sp³-hybridized orbitals.
Comprehensive Rationale
An sp-hybridized carbon contains 50% s-character, whereas an sp³-hybridized carbon contains only 25% s-
character. Because s orbitals are closer to the nucleus than p orbitals, electrons in orbitals with greater s-
character experience stronger nuclear attraction. This produces shorter, stronger bonds and contributes to the
greater acidity of terminal alkynes.
Why the Other Options Are Incorrect
B. The relationship is reversed; sp³ orbitals have the lowest percentage of s-character.
C. A triple bond consists of one σ bond and two π bonds, not only π bonds.
D. Every carbon atom has the same nuclear charge (six protons). Hybridization—not nuclear charge—accounts
for the difference.
Mechanism Insight
The percentage of s-character influences several important properties, including bond length, bond strength,
electronegativity, and acidity. These trends are frequently applied when comparing reaction intermediates and
predicting organic reactivity.
Exam Tip
s-character trend:
sp (50%) > sp² (33%) > sp³ (25%)
More s-character means stronger, shorter bonds and greater acidity of attached hydrogens.
,Mechanism Challenge 3
An instructor asks students to identify the primary reason carbon is capable of forming millions of stable
organic compounds. Which property of carbon best explains this extraordinary structural diversity?
A. Carbon readily forms strong covalent bonds with itself, allowing the formation of long chains, rings, and
branched structures.
B. Carbon always forms ionic compounds with other elements.
C. Carbon possesses the highest electronegativity in the periodic table.
D. Carbon can accommodate more than eight electrons in its valence shell.
Correct Answer
A. Carbon readily forms strong covalent bonds with itself, allowing the formation of long chains, rings, and
branched structures.
Comprehensive Rationale
Carbon's ability to catenate—form stable covalent bonds with other carbon atoms—is the foundation of
organic chemistry. Combined with its tetravalency and ability to form single, double, and triple bonds, carbon
can generate an enormous variety of molecular architectures, including linear, branched, cyclic, and aromatic
compounds.
Why the Other Options Are Incorrect
B. Organic compounds are dominated by covalent rather than ionic bonding.
C. Fluorine, not carbon, is the most electronegative element.
D. Carbon obeys the octet rule and does not normally expand its valence shell.
Mechanism Insight
Carbon's unique bonding capabilities explain why it serves as the backbone of nearly all biologically important
molecules and synthetic organic compounds.
Exam Tip
When asked why carbon is the central element of organic chemistry, remember the combination of
tetravalency, catenation, and the ability to form single, double, and triple bonds.
Mechanism Challenge 4
While constructing the Lewis structure of carbon dioxide (CO₂), a student notices that the central carbon atom
forms two double bonds. Why is this arrangement favored?
A. It allows every atom to achieve a complete valence shell while minimizing formal charges.
B. Carbon cannot form single bonds with oxygen.
,C. Oxygen prefers to have six bonding electrons instead of eight.
D. Carbon expands its valence shell beyond eight electrons.
Correct Answer
A. It allows every atom to achieve a complete valence shell while minimizing formal charges.
Comprehensive Rationale
The most stable Lewis structure for carbon dioxide contains two carbon-oxygen double bonds. This
arrangement gives carbon and both oxygen atoms complete octets while producing formal charges of zero on
all atoms. Lewis structures that minimize formal charges generally represent the most stable electron
distribution.
Why the Other Options Are Incorrect
B. Carbon can form single bonds with oxygen in many compounds, such as alcohols and ethers.
C. Oxygen typically seeks an octet of eight valence electrons, not six.
D. Carbon is a second-period element and cannot expand its valence shell beyond an octet under normal
conditions.
Mechanism Insight
Evaluating formal charges is one of the most reliable methods for selecting the most stable Lewis structure
before predicting molecular reactivity.
Exam Tip
When comparing Lewis structures, prioritize those that:
• Give atoms complete octets whenever possible.
• Minimize formal charges.
• Place negative formal charges on more electronegative atoms.
Mechanism Challenge 5
A researcher compares carbon (C), nitrogen (N), oxygen (O), and fluorine (F) across the second period of the
periodic table. Which trend correctly describes how electronegativity changes from carbon to fluorine?
A. It decreases because atomic size increases.
B. It remains essentially constant across the period.
C. It increases because effective nuclear charge increases while atomic radius decreases.
D. It first decreases and then increases.
, Correct Answer
C. It increases because effective nuclear charge increases while atomic radius decreases.
Comprehensive Rationale
Moving from left to right across a period, the number of protons increases while shielding changes very little.
As a result, the effective nuclear charge experienced by valence electrons increases, causing atoms to attract
shared electrons more strongly. Consequently, electronegativity increases across the period, with fluorine
being the most electronegative element.
Why the Other Options Are Incorrect
A. Atomic radius decreases—not increases—across a period.
B. Electronegativity changes significantly across the second period.
D. The trend is a steady increase rather than an irregular pattern.
Mechanism Insight
Electronegativity determines bond polarity, influences reaction mechanisms, and helps predict nucleophilic
and electrophilic behavior.
Exam Tip
Across a period:
• Electronegativity ↑
• Ionization energy ↑
• Atomic radius ↓
These periodic trends are frequently tested together.
Mechanism Challenge 6
During the analysis of hydrogen fluoride (HF), a student concludes that the bond between hydrogen and
fluorine is highly polar. What is the primary reason for this observation?
A. Fluorine attracts the shared bonding electrons much more strongly than hydrogen.
B. Hydrogen contributes two valence electrons to the bond.
C. Fluorine forms ionic bonds with every element.
D. The electrons are shared equally between hydrogen and fluorine.
Correct Answer
A. Fluorine attracts the shared bonding electrons much more strongly than hydrogen.
, Comprehensive Rationale
Fluorine has the highest electronegativity of any element, allowing it to attract shared bonding electrons more
strongly than hydrogen. This unequal sharing creates a polar covalent bond with a partial negative charge (δ–)
on fluorine and a partial positive charge (δ+) on hydrogen.
Bond polarity influences physical properties such as boiling point, intermolecular forces, and chemical
reactivity.
Why the Other Options Are Incorrect
B. Hydrogen contributes one valence electron, not two.
C. Fluorine forms many covalent compounds, including HF.
D. Equal sharing would produce a nonpolar covalent bond.
Mechanism Insight
Bond polarity determines electron-rich and electron-poor regions within molecules, allowing chemists to
predict sites of nucleophilic and electrophilic attack.
Exam Tip
Greater differences in electronegativity produce greater bond polarity, but not every polar bond is ionic.
Mechanism Challenge 7
An undergraduate is comparing sodium chloride (NaCl) with methane (CH₄). Which statement correctly
distinguishes the bonding found in these two compounds?
A. Sodium chloride contains ionic bonding, whereas methane contains covalent bonding.
B. Both compounds contain only ionic bonds.
C. Both compounds contain only covalent bonds.
D. Methane contains ionic bonds because carbon is more electronegative than hydrogen.
Correct Answer
A. Sodium chloride contains ionic bonding, whereas methane contains covalent bonding.
Comprehensive Rationale
Sodium chloride forms when sodium transfers an electron to chlorine, producing oppositely charged ions held
together by electrostatic attraction. Methane, in contrast, forms when carbon and hydrogen share electrons
through covalent bonds. Understanding the distinction between electron transfer and electron sharing is
fundamental to predicting the properties of chemical compounds.
Why the Other Options Are Incorrect
