Apply Electron Sharing Reactions — Apply Reactivity 3.3—Electron sharing reactions in problem-solving and explanations - IB Chemistry

SECTION 1

Historical Context & Motivation

For centuries, chemists understood that substances combine to form new materials, but the mechanism by which atoms actually bond remained a mystery. The discovery of the electron in 1897 opened the door to understanding how atoms interact at the subatomic level. Early models of bonding focused on the complete transfer of electrons between atoms—what we now call ionic bonding—but this model could not explain the stable molecules formed between nonmetal atoms like H₂ or O₂. A new idea was needed: one in which atoms share electrons rather than transfer them.

1897 • Discovery of the Electron

J.J. Thomson discovers the electron using cathode ray experiments, proving atoms contain negatively charged particles that could participate in bonding.

1916 • Lewis Electron Pair Model

Gilbert N. Lewis proposes that atoms can achieve stability by sharing electron pairs, introducing the concept of the covalent bond and the dot-structure notation still used today.

1927 • Quantum Mechanical Bonding Theory

Walter Heitler and Fritz London apply quantum mechanics to the hydrogen molecule, showing mathematically why shared electrons stabilize the bond.

1931 • Hybridization & Directed Bonds

Linus Pauling introduces orbital hybridization (sp, sp², sp³), explaining the three-dimensional geometry of covalent molecules such as methane and water.

1965 • Woodward–Hoffmann Rules

Robert B. Woodward and Roald Hoffmann show that orbital symmetry governs which electron-sharing reactions are allowed, deepening mechanistic chemistry.

The central question driving this topic is deceptively simple: How do we predict and explain what happens when atoms share electrons? In IB Chemistry Reactivity 3.3, you will apply electron-sharing principles to real reactions—drawing mechanisms, predicting products, and explaining why certain pathways are favored over others.

SECTION 2

Core Principles of Electron Sharing

Electron sharing reactions—commonly called covalent bonding reactions—are governed by a small set of foundational ideas. Understanding these principles lets you predict products, draw reaction mechanisms with curly arrows, and explain why reactions proceed the way they do. Below are the core concepts you need.

Nucleophiles & Electrophiles

A nucleophile is an electron-rich species that donates a pair of electrons. An electrophile is electron-poor and accepts them. Every electron-sharing mechanism starts by identifying these two roles.

Curly Arrow Notation

A full curly arrow (⟶) represents the movement of an electron pair from nucleophile to electrophile. A half-headed (fishhook) arrow shows movement of a single electron, used in radical mechanisms.

Bond Formation & Bond Breaking

In heterolytic fission, one atom keeps both bonding electrons. In homolytic fission, each atom takes one electron. The type of fission determines whether the pathway is ionic or radical.

Sigma (σ) and Pi (π) Bonds

A sigma bond forms from head-on orbital overlap and allows free rotation. A pi bond forms from sideways overlap and restricts rotation, making double and triple bonds more reactive at the π component.

Electronegativity & Bond Polarity

When two atoms share electrons unequally due to differing electronegativity, the bond is polar. The resulting δ⁺ and δ⁻ sites determine where nucleophilic or electrophilic attack occurs in a reaction.

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KEY TAKEAWAY

Think of electron-sharing reactions like a handshake. A nucleophile is the hand reaching out (it has electrons to offer), and an electrophile is the hand receiving (it needs electrons). A curly arrow simply traces the path of that handshake—from the donor to the acceptor. Every mechanism you draw is just a sequence of these "handshakes."

SECTION 3

Visualizing Electron Sharing Mechanisms

The diagram below illustrates the nucleophilic substitution mechanism (S_N2) for the reaction between hydroxide ion (OH⁻) and bromomethane (CH₃Br). This is one of the most important electron-sharing reaction types you will encounter. Notice how the curly arrow originates from the lone pair on the nucleophile (OH⁻) and points toward the electrophilic carbon. Simultaneously, the C–Br bond breaks heterolytically, with both electrons going to the leaving group (Br⁻).

The S_N2 mechanism in one step. The cyan curly arrow shows the lone pair on OH⁻ attacking the electrophilic carbon, while the red curly arrow shows the C–Br bonding electrons departing with the bromine, forming Br⁻. The transition state (dashed yellow box) is the highest-energy point where both bonds are partially formed and partially broken.

In the diagram, observe three critical features. First, the curly arrow tail always starts at the electron source—a lone pair or a bond—and the arrowhead always points to the electron sink (the atom accepting the electrons). Second, the partial charges δ⁺ and δ⁻ on the C–Br bond arise because bromine is more electronegative than carbon, pulling shared electron density toward itself. Third, notice that this S_N2 reaction happens in one concerted step—bond making and bond breaking occur simultaneously.

SECTION 4

How Electron Sharing Mechanisms Work

Bond Energies and Reaction Feasibility

Whether an electron-sharing reaction proceeds depends on the balance between energy needed to break existing bonds and energy released when new bonds form. We quantify this with average bond enthalpies. If the total energy released by bond formation exceeds the energy consumed by bond breaking, the reaction is exothermic and tends to be favoured.

ENTHALPY CHANGE FROM BOND ENTHALPIES

ΔH = Σ (bonds broken) − Σ (bonds formed)

ΔH = enthalpy change of reaction (kJ mol⁻¹). A negative value means the reaction releases energy (exothermic). Each sum uses average bond enthalpy values from the IB Data Booklet.

Types of Electron Sharing Mechanisms in IB Chemistry

IB Chemistry focuses on several key mechanism types. Nucleophilic substitution (S_N1 and S_N2) involves a nucleophile replacing a leaving group on a carbon atom. Electrophilic addition occurs when an electrophile adds across a C=C double bond, breaking the π bond. Free-radical substitution involves homolytic fission and single-electron transfer in three stages: initiation, propagation, and termination.

GENERAL NUCLEOPHILIC SUBSTITUTION

Nu:⁻ + R−LG → R−Nu + LG⁻

Nu:⁻ = nucleophile with a lone pair; R = organic group (often an alkyl group); LG = leaving group. The colon represents the lone pair that forms the new bond.

ELECTROPHILIC ADDITION TO ALKENES

C=C + E⁺ → C−C−E (carbocation intermediate) → final product

E⁺ = electrophile (e.g., H⁺ from HBr). The π electrons of the double bond attack E⁺. A carbocation intermediate forms, which is then attacked by the remaining nucleophilic part (e.g., Br⁻).

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IB Exam Tip

When drawing mechanisms, always use full curly arrows (double-headed) for heterolytic mechanisms and half-headed (fishhook) arrows for radical mechanisms. The IB mark scheme awards separate marks for correct arrow placement, so practice this carefully.

SECTION 5

Classifying Electron Sharing Reactions

Different electron-sharing reactions follow different patterns. The diagram below compares the three main mechanism types you need for IB Chemistry, showing the flow of electrons at each stage. Recognizing the pattern quickly is the key to applying these mechanisms in exam questions.

Comparison of three electron-sharing mechanism types assessed in IB Chemistry. The left column shows nucleophilic substitution (S_N2), the center shows electrophilic addition to alkenes, and the right shows free-radical substitution. Note the different arrow styles: full curly arrows for heterolytic mechanisms and half-headed arrows for radical mechanisms.

Feature	Nucleophilic Substitution	Electrophilic Addition	Free-Radical Substitution
Substrate	Haloalkane (R−X)	Alkene (C=C)	Alkane (C−H)
Reagent	Nucleophile (OH⁻, NH₃, CN⁻)	Electrophile (HBr, Br₂)	Halogen (Cl₂, Br₂) + UV
Arrow type	Full curly (pair)	Full curly (pair)	Half-headed (single e⁻)
Fission type	Heterolytic	Heterolytic	Homolytic
Conditions	Warm aqueous or reflux	Room temperature	UV light required

Summary of three key electron-sharing mechanism types for IB Chemistry

SECTION 6

Worked Example: Electrophilic Addition of HBr to Propene

Let us work through a complete IB-style problem. We will draw the mechanism for the electrophilic addition of HBr to propene (CH₃CH=CH₂), predict the major product using Markovnikov's rule, and calculate the enthalpy change using bond enthalpies.

Electrophilic Addition of HBr to Propene

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Step 1 — Identify the Substrate and Reagent

The substrate is propene (CH₃CH=CH₂), which contains a C=C double bond. The reagent is HBr, a strong acid that provides H⁺ as the electrophile and Br⁻ as the nucleophile.

Electrophile = H⁺; Nucleophile = Br⁻; Substrate = alkene π bond
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Step 2 — Draw the First Curly Arrow (π Bond Attacks H⁺)

Draw a full curly arrow from the C=C π bond to the H atom of HBr. This represents the electron-rich π bond acting as the nucleophile, attacking the electrophilic H. Simultaneously, draw a second curly arrow from the H–Br bond to Br, showing heterolytic fission. The H bonds to one of the carbons, and Br⁻ is released.

A carbocation intermediate forms on the carbon that did NOT bond to H.
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Step 3 — Apply Markovnikov's Rule to Determine Which Carbon Gets H⁺

Markovnikov's rule states that H⁺ adds to the carbon with more existing hydrogen atoms. In propene, the terminal CH₂ already has two H atoms, so H⁺ adds there, forming CH₃. This places the positive charge on the central carbon (a secondary carbocation), which is more stable than a primary one.

Carbocation forms at C2: CH₃C⁺HCH₃
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Step 4 — Draw the Second Curly Arrow (Br⁻ Attacks C⁺)

Now draw a curly arrow from the lone pair on Br⁻ to the positively charged carbon. This completes the addition—both H and Br have added across the former double bond.

Major product = 2-bromopropane (CH₃CHBrCH₃)
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Step 5 — Verify Using Bond Enthalpies (Optional Calculation)

Bonds broken: C=C (614 kJ mol⁻¹) + H–Br (366 kJ mol⁻¹) = 980 kJ mol⁻¹ total. Bonds formed: C–C (346 kJ mol⁻¹) + C–H (414 kJ mol⁻¹) + C–Br (285 kJ mol⁻¹) = 1045 kJ mol⁻¹ total. ΔH = 980 − 1045 = −65 kJ mol⁻¹.

ΔH ≈ −65 kJ mol⁻¹ (exothermic, confirming the reaction is energetically favourable)

SECTION 7

Strengths & Limitations of the Mechanism Model

Curly-arrow mechanisms are powerful tools, but like all models in science they have both strengths and limitations. Understanding these will help you use them wisely on exams and know when a more sophisticated approach is needed.

Strengths	Limitations
Clearly show which bonds break and which form in each step	Do not show the 3D arrangement of the transition state in detail
Predict products and regiochemistry (e.g., Markovnikov's rule)	Cannot predict reaction rates without additional kinetic data
Distinguish between heterolytic and homolytic pathways using arrow types	Oversimplify electron movement—real electrons exist in delocalized orbitals
Work consistently across organic, inorganic, and biochemical reactions	Average bond enthalpies give only approximate ΔH values, not exact ones

Strengths and limitations of curly-arrow mechanism models

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PERSPECTIVE

Curly-arrow mechanisms are like a road map for a car journey. The map shows which roads connect which cities (which bonds form and break), and it can tell you the general direction (electron flow from nucleophile to electrophile). However, the map does not tell you how fast traffic is moving or what the weather is like along the way. For that level of detail, chemists turn to computational chemistry and transition-state theory.

SECTION 8

Connection to Advanced Theory

At the IB level, you learn to draw mechanisms using curly arrows and predict products. At the university level, these concepts deepen significantly. Molecular orbital (MO) theory replaces the simple overlap picture with a full quantum-mechanical treatment. Frontier molecular orbital theory explains reactivity by looking at the HOMO (highest occupied molecular orbital) of the nucleophile and the LUMO (lowest unoccupied molecular orbital) of the electrophile. Understanding where these overlap in space and energy predicts not just whether a reaction occurs, but how fast it occurs and with what stereochemistry.

Concept	IB Level (Reactivity 3.3)	University Level
Electron movement	Curly arrows show pair movement	HOMO–LUMO interactions govern reactivity
Product prediction	Markovnikov's rule (empirical)	Carbocation stability from hyperconjugation and induction
Stereochemistry	Sₙ2 gives inversion (basic)	Detailed orbital symmetry analysis (Woodward–Hoffmann rules)
Energy analysis	Bond enthalpy calculations (approximate)	Potential energy surfaces and computational modeling

IB vs. university-level treatment of electron-sharing reactions

The good news is that the curly-arrow skills you build now remain the universal language of organic chemistry at every level. University professors, pharmaceutical researchers, and materials scientists all use the same arrow notation. By mastering it now, you are building a foundation that will serve you throughout any future chemistry course.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

In a curly-arrow mechanism, the tail of the arrow always starts at the electron source and the head always points to the electron sink. Explain why a nucleophile acts as the electron source and an electrophile acts as the electron sink, using the terms "electron-rich" and "electron-poor" in your answer.

Show answer

A nucleophile is described as electron-rich because it possesses a lone pair or a region of high electron density that it can donate. An electrophile is electron-poor—it has an empty or partially empty orbital, or carries a partial or full positive charge, that can accept electrons. Since electrons flow from regions of high density to regions of low density, the curly arrow naturally starts at the nucleophile (source) and ends at the electrophile (sink). This is analogous to the way water flows downhill—electrons flow from where there are "too many" to where there is a deficit.

Question 2: BASIC CALCULATION

Using the bond enthalpies: C–H = 414 kJ mol⁻¹, Cl–Cl = 242 kJ mol⁻¹, C–Cl = 324 kJ mol⁻¹, and H–Cl = 431 kJ mol⁻¹, calculate the enthalpy change for the free-radical substitution: CH₄ + Cl₂ → CH₃Cl + HCl.

Show answer

Bonds broken: one C–H (414 kJ mol⁻¹) + one Cl–Cl (242 kJ mol⁻¹) = 656 kJ mol⁻¹. Bonds formed: one C–Cl (324 kJ mol⁻¹) + one H–Cl (431 kJ mol⁻¹) = 755 kJ mol⁻¹. ΔH = 656 − 755 = −99 kJ mol⁻¹. The reaction is exothermic, releasing approximately 99 kJ of energy per mole of methane that reacts. This negative enthalpy change confirms that the products are more stable than the reactants, which is consistent with the reaction being energetically favorable once initiated by UV light.

Question 3: INTERMEDIATE

When but-1-ene (CH₂=CHCH₂CH₃) reacts with HBr, two possible products can form: 1-bromobutane and 2-bromobutane. Using Markovnikov's rule and the concept of carbocation stability, predict and explain which is the major product. Draw the two curly arrows for the first step of the mechanism.

Show answer

According to Markovnikov's rule, H⁺ adds to the carbon of the double bond that already has more hydrogens—that is, C1 (the terminal =CH₂). This places the carbocation on C2, forming a secondary carbocation (CH₃C⁺HCH₂CH₃), which is more stable than the primary alternative at C1. In the first step, the curly arrow starts at the C=C π bond and points to the H of HBr, showing the π electrons attacking the electrophile. A second curly arrow starts at the H–Br bond and points to Br, showing heterolytic fission. Because the secondary carbocation is stabilized by the inductive effect of two alkyl groups, the major product is 2-bromobutane (CH₃CHBrCH₂CH₃).

Question 4: APPLIED

A student performs a nucleophilic substitution reaction between sodium hydroxide (NaOH) and 1-bromobutane (CH₃CH₂CH₂CH₂Br) under reflux conditions. The student expects to produce butan-1-ol. Write the balanced equation, identify the nucleophile and leaving group, draw the mechanism with two curly arrows, and explain why heating under reflux is necessary.

Show answer

Balanced equation: CH₃CH₂CH₂CH₂Br + NaOH → CH₃CH₂CH₂CH₂OH + NaBr. The nucleophile is the hydroxide ion (OH⁻), which has lone pairs available for donation. The leaving group is Br⁻, which departs with the bonding pair from the C–Br bond. In the Sₙ2 mechanism, the first curly arrow goes from the lone pair on OH⁻ to the electrophilic carbon bearing the Br. The second curly arrow goes from the C–Br bond to Br, representing heterolytic fission. Heating under reflux is necessary because the reaction has a significant activation energy; reflux maintains a high temperature to increase the proportion of molecules with enough kinetic energy to overcome the activation barrier, while condensing the vapour back into the flask to prevent loss of volatile organic reactant.

Question 5: CRITICAL THINKING

Free-radical substitution of methane with bromine (CH₄ + Br₂ → CH₃Br + HBr) is much slower than the equivalent reaction with chlorine, even though both proceed by the same radical mechanism. Using the bond enthalpies C–H = 414, H–Br = 366, H–Cl = 431, Br–Br = 193, and Cl–Cl = 242 kJ mol⁻¹, compare the enthalpy changes of the first propagation step for each halogen. Explain how this relates to the difference in reaction rate.

Show answer

First propagation step with Cl: Cl• + CH₄ → HCl + •CH₃. Bonds broken: C–H = 414. Bonds formed: H–Cl = 431. ΔH = 414 − 431 = −17 kJ mol⁻¹ (exothermic). First propagation step with Br: Br• + CH₄ → HBr + •CH₃. Bonds broken: C–H = 414. Bonds formed: H–Br = 366. ΔH = 414 − 366 = +48 kJ mol⁻¹ (endothermic). The bromine propagation step is endothermic, meaning it has a higher activation energy than the exothermic chlorine step. According to the Arrhenius relationship, a higher activation energy means a smaller fraction of molecular collisions have sufficient energy to react at a given temperature, leading to a slower rate. This illustrates how bond enthalpy data, combined with mechanistic reasoning, can explain observable differences in reaction rates—even when the overall mechanism type is identical.

SUMMARY

Summary — Applying Electron Sharing Reactions

Summary

Electron sharing reactions lie at the heart of covalent chemistry. Every mechanism begins by identifying the nucleophile (electron-rich, donates a pair) and the electrophile (electron-poor, accepts a pair). Curly arrows trace the movement of electron pairs from source to sink, with full-headed arrows for heterolytic mechanisms and half-headed arrows for homolytic (radical) mechanisms. The three core mechanism types for IB Chemistry are nucleophilic substitution (Sₙ1 and Sₙ2), electrophilic addition to alkenes, and free-radical substitution of alkanes.

To apply these ideas in problem-solving, first classify the substrate and reagent, then draw arrows showing electron flow, and finally use bond enthalpy calculations (ΔH = Σ bonds broken − Σ bonds formed) to verify energy changes. Markovnikov's rule predicts the major product of electrophilic addition by placing the positive charge on the more substituted (more stable) carbon. Electronegativity differences create the δ⁺ and δ⁻ sites that determine where nucleophilic or electrophilic attack occurs. Mastering these principles gives you a systematic approach to any reaction mechanism question on the IB exam.