Understand Electron-Pair Sharing Reactions — Understand Reactivity 3.4—Electron-pair sharing reactions - IB Chemistry

SECTION 1

Historical Context & Motivation

For centuries, chemists could describe what formed in a chemical reaction, but they struggled to explain how bonds actually broke and reformed at the atomic level. The idea that reactions proceed through the directed movement of electron pairs was a breakthrough that transformed chemistry from a descriptive science into one that could predict and design new reactions. Understanding the history of this concept helps us appreciate why electron-pair sharing reactions are central to modern organic and inorganic chemistry.

1916 • Lewis's Electron-Pair Bond

Gilbert N. Lewis proposed that a covalent bond consists of a shared pair of electrons between two atoms. This simple yet powerful idea became the foundation for all electron-pair sharing models.

1923 • Lewis Acids and Bases

Lewis expanded his theory by defining an acid as an electron-pair acceptor and a base as an electron-pair donor, broadening the concept of acid-base chemistry far beyond proton transfer.

1933 • Ingold's Nucleophile & Electrophile

Christopher Ingold introduced the terms nucleophile ("nucleus-loving") and electrophile ("electron-loving") to describe how reagents interact with electron-rich and electron-poor sites.

1941 • Curly Arrow Notation

Sir Robert Robinson popularized the use of curly arrows (curved arrows) to show the movement of electron pairs in reaction mechanisms, giving chemists a universal visual language for tracking bond making and breaking.

1960s • Frontier Molecular Orbital Theory

Kenichi Fukui developed frontier molecular orbital (FMO) theory, explaining electron-pair sharing reactions in terms of HOMO-LUMO interactions. This work earned him the Nobel Prize in 1981.

These milestones reveal a central question in chemistry: when two molecules collide, what determines which bonds break and which new bonds form? The answer lies in understanding where electron pairs are located, how they move, and what drives them from one atom to another. This lesson explores the principles behind electron-pair sharing reactions and equips you with the tools to predict and explain these processes.

SECTION 2

Core Principles & Definitions

Electron-pair sharing reactions revolve around the idea that bonds form and break because pairs of electrons move between atoms or molecules. To understand this fully, you need to master a handful of foundational concepts that serve as the vocabulary for reaction mechanisms in IB Chemistry.

Nucleophile

A species that donates an electron pair to form a new covalent bond. Nucleophiles are electron-rich — they have a lone pair or a region of high electron density (e.g., OH⁻, NH₃, CN⁻).

Electrophile

A species that accepts an electron pair to form a new covalent bond. Electrophiles are electron-poor — they carry a partial or full positive charge (e.g., H⁺, carbocations, BF₃).

Curly Arrow (Curved Arrow)

A double-headed arrow that shows the movement of an electron pair from its source (tail) to its destination (head). The tail starts at the electron-rich site; the head points to the electron-poor site.

Heterolytic Bond Breaking

When a covalent bond breaks unevenly so that both electrons go to one atom, producing ions. This contrasts with homolytic fission, where each atom takes one electron.

Lewis Acid–Base Reaction

An electron-pair sharing reaction viewed through the Lewis lens: a Lewis base (nucleophile) donates its electron pair to a Lewis acid (electrophile), forming a coordinate (dative) covalent bond.

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KEY TAKEAWAY

Think of a nucleophile as a person carrying a gift (an electron pair) and an electrophile as someone eagerly waiting to receive it. The curly arrow is like a map showing the gift's journey — it always starts at the giver and ends at the receiver. Every electron-pair sharing reaction is just this exchange: the electron-rich species shares with the electron-poor species, and a new bond is formed.

SECTION 3

Visual Explanation — Curly Arrow Mechanisms

The curly arrow is the single most important tool for communicating how electron pairs move during a reaction. A double-headed curly arrow always represents the movement of two electrons (one pair). The diagram below shows a generic nucleophilic attack on an electrophile, illustrating how a new bond forms and an old bond breaks simultaneously.

The cyan curly arrow shows the nucleophile's lone pair moving toward the electrophilic carbon. The pink arrow shows the C–L bond electrons departing with the leaving group. In the product, the new Nu–C bond has formed and L departs as L⁻.

Notice how each curly arrow accounts for exactly one electron pair. The tail of the cyan arrow originates at the lone pair on Nu⁻, and the arrowhead lands between Nu and C — this is where the new bond forms. Simultaneously, the pink arrow shows the bonding pair in the C–L bond departing entirely with the leaving group L, which becomes L⁻. This coordinated movement of electron pairs is the essence of electron-pair sharing reactions.

SECTION 4

How Electron-Pair Sharing Works — Mechanism Deep Dive

Coordinate (Dative) Covalent Bond Formation

In a standard covalent bond, each atom contributes one electron to the shared pair. In a coordinate covalent bond (also called a dative bond), both electrons come from the same atom — the nucleophile. This is exactly what happens in every electron-pair sharing reaction. The nucleophile provides both electrons, and the electrophile simply accepts them into its empty or partially vacant orbital.

Identifying Nucleophilic and Electrophilic Sites

To predict where electron pairs will move, you need to locate the electron-rich and electron-poor regions in a molecule. Electronegativity differences within bonds create partial charges: the more electronegative atom pulls electron density toward itself (δ⁻), leaving the less electronegative atom electron-poor (δ⁺). For example, in a C–Cl bond, chlorine is more electronegative, so carbon bears a partial positive charge and acts as the electrophilic centre.

Frontier Molecular Orbital Theory — Extension Only (Beyond IB Scope)

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Not Required for IB Assessment — Extension Reading

The following discussion of HOMO–LUMO frontier molecular orbital theory goes beyond the IB Diploma Programme specification for Reactivity 3.4. You are not required to know this material for IB exams. It is included here for interest only, for students who wish to preview university-level ideas. The IB exam will only ask you to draw and interpret curly-arrow mechanisms for SN2, electrophilic addition to alkenes, and nucleophilic addition to carbonyl compounds.

At a deeper level studied at university, electron-pair sharing can be understood using frontier molecular orbitals. The nucleophile's electron pair resides in its Highest Occupied Molecular Orbital (HOMO), while the electrophile has a Lowest Unoccupied Molecular Orbital (LUMO) ready to accept electrons. A reaction occurs when the HOMO of the nucleophile overlaps effectively with the LUMO of the electrophile. The smaller the energy gap between HOMO and LUMO, the easier the reaction. This is a concept explored in university-level organic and physical chemistry courses.

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IB Exam Tip

On the IB exam, you will be expected to draw curly arrows showing electron-pair movement for nucleophilic substitution (S_N2), electrophilic addition to alkenes, and nucleophilic addition to carbonyl compounds. Always start the arrow at the electron source (lone pair or π bond) and end it at the electron-deficient centre.

SECTION 5

Classifying Electron-Pair Sharing Reactions

Electron-pair sharing reactions can be classified by the type of species that initiates the reaction (nucleophile or electrophile) and the structural change that occurs. In IB Chemistry, you will encounter three major categories: nucleophilic substitution, electrophilic addition, and nucleophilic addition. The diagram below compares these three pathways.

Comparison of three electron-pair sharing reaction types. In each case, the electron-rich species (nucleophile or π bond) donates an electron pair to the electron-poor species (electrophile or δ⁺ carbon).

Feature	Nucleophilic Substitution	Electrophilic Addition	Nucleophilic Addition
Substrate	Haloalkane (sp³ C)	Alkene or alkyne (C=C)	Aldehyde or ketone (C=O)
Attacking species	Nucleophile (e.g., OH⁻, CN⁻)	Electrophile (e.g., H⁺, Br⁺)	Nucleophile (e.g., H⁻, CN⁻)
Bond changes	One made, one broken	π bond broken, two σ bonds made	π bond broken, one σ bond made
Leaving group?	Yes	No	No

Summary of three electron-pair sharing reaction types

SECTION 6

Worked Example — S_N2 Reaction of Bromomethane with Hydroxide

Let's work through a complete electron-pair sharing reaction mechanism. We will examine the S_N2 nucleophilic substitution of bromomethane (CH₃Br) with hydroxide ion (OH⁻).

S_N2 Reaction: CH₃Br + OH⁻ → CH₃OH + Br⁻

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Step 1 — Identify the Nucleophile and Electrophile

The hydroxide ion (OH⁻) has three lone pairs on the oxygen atom and carries a negative charge, making it an excellent nucleophile. The carbon in CH₃Br is bonded to the electronegative bromine atom, which pulls electron density away from it. This makes the carbon electrophilic (δ⁺). Bromine is the leaving group because it can stabilize the negative charge after departure.

Nucleophile: OH⁻ | Electrophile: Cδ⁺ in CH₃Br | Leaving group: Br⁻
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Step 2 — Draw the Curly Arrow from Nucleophile to Electrophile

Draw a curly arrow starting from a lone pair on the oxygen of OH⁻ (the tail), pointing to the carbon atom in CH₃Br (the head). This arrow represents the formation of the new O–C bond using the electron pair from oxygen.

Arrow 1: lone pair on O → C atom (bond formation)
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Step 3 — Draw the Curly Arrow for Bond Breaking

Since carbon can only form four bonds, the C–Br bond must break as the new O–C bond forms. Draw a second curly arrow from the C–Br bond (the tail) to the bromine atom (the head). This represents the heterolytic fission of the C–Br bond — both bonding electrons go to bromine.

Arrow 2: C–Br bond → Br atom (bond breaking)
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Step 4 — Write the Products

The nucleophile has replaced the leaving group. The organic product is methanol (CH₃OH), and the inorganic product is bromide ion (Br⁻). In S_N2, this happens in a single, concerted step — bond making and breaking occur simultaneously.

CH₃Br + OH⁻ → CH₃OH + Br⁻
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Step 5 — Verify Conservation

Check that atoms and charges are balanced. Reactant side: one C, three H, one Br, one O, one H, overall charge = −1. Product side: one C, three H, one O, one H, one Br, overall charge = −1. Everything balances, confirming that our curly arrows correctly tracked the electron-pair movement.

Atoms balanced ✓ | Charges balanced ✓ | Mechanism complete ✓

SECTION 6B

Worked Example — Electrophilic Addition of HBr to Ethene

The IB specification requires you to draw complete curly-arrow mechanisms for electrophilic addition to alkenes. This is one of the most important reaction types you will meet in Reactivity 3.4. In electrophilic addition, the electron-rich π bond of an alkene acts as the nucleophile and donates a pair of electrons to an incoming electrophile (such as HBr or Br₂). Let's work through the addition of HBr to ethene (CH₂=CH₂) step by step.

Electrophilic Addition: CH₂=CH₂ + HBr → CH₃CH₂Br

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Step 1 — Identify the Roles of Each Species

In HBr, the H–Br bond is polar. The hydrogen end is electron-poor (δ⁺) and acts as the electrophile. The π bond of ethene has electron density above and below the C=C bond, making the alkene the nucleophile in this reaction. There is no leaving group — instead, both atoms of HBr add across the double bond.

Nucleophile: π bond of C=C | Electrophile: Hδ⁺ in HBr | No leaving group
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Step 2 — Draw the Curly Arrows for Step 1 (π attack on H⁺)

Draw the first curly arrow starting from the C=C π bond (the tail) and pointing to the H atom of HBr (the head). This arrow shows the π electrons attacking the electrophilic hydrogen. At the same time, draw a second curly arrow from the H–Br bond to the bromine atom, showing heterolytic fission of the H–Br bond. Both electrons of the H–Br bond go to Br, which leaves as Br⁻.

Arrow 1: C=C π bond → H of HBr (new C–H bond forms) | Arrow 2: H–Br bond → Br (H–Br breaks, Br⁻ departs)
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Step 3 — Identify the Carbocation Intermediate

After step 1, one carbon of the original double bond has formed a new C–H bond and now has four bonds. The other carbon has lost its share of the π electrons and is left with only three bonds — it carries a positive charge. This species is called a carbocation (CH₃–CH⁺₂). The carbocation intermediate is the electrophilic species in the second step.

Intermediate: CH₃–CH⁺ (ethyl carbocation) + Br⁻
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Step 4 — Draw the Curly Arrow for Step 2 (Br⁻ attacks carbocation)

In the second step, the bromide ion (Br⁻) generated in step 1 now acts as a nucleophile. Draw a curly arrow from a lone pair on Br⁻ (the tail) to the positively charged carbon of the carbocation (the head). This forms the new C–Br bond and gives the final product.

Arrow 3: lone pair on Br⁻ → C⁺ of carbocation (new C–Br bond forms)
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Step 5 — Write the Overall Equation and Verify

The product is bromoethane (CH₃CH₂Br). Both H and Br have added across the original double bond. Check: reactants have 2C, 4H, 1Br; product has 2C, 5H, 1Br — wait, the H from HBr adds too, giving 2C, 5H, 1Br overall. Charges: reactants neutral (HBr and CH₂=CH₂), product neutral (CH₃CH₂Br). ✓

CH₂=CH₂ + HBr → CH₃CH₂Br

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IB Exam Tip — Electrophilic Addition

Key points to remember for IB electrophilic addition: (1) The π bond acts as the nucleophile — always start your first curly arrow from the C=C bond. (2) Three curly arrows are needed in total for addition of HBr: one from π bond to H, one from H–Br bond to Br, and one from Br⁻ lone pair to the carbocation. (3) The same two-step pattern applies to the addition of Br₂ to an alkene, except both steps involve bromine species.

SECTION 6C

Worked Example — Nucleophilic Addition to a Carbonyl Compound

The IB specification also requires curly-arrow mechanisms for nucleophilic addition to carbonyl compounds (aldehydes and ketones). In this reaction type, a nucleophile attacks the electron-poor carbon of a polar C=O bond. A classic IB example is the addition of cyanide ion (CN⁻) to ethanal (CH₃CHO).

Nucleophilic Addition: CH₃CHO + HCN → CH₃CH(OH)CN

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Step 1 — Identify the Nucleophile and Electrophilic Centre

In ethanal, the C=O bond is strongly polarised because oxygen is much more electronegative than carbon. This makes the carbonyl carbon electron-poor (C^δ+) and therefore the electrophilic site. The cyanide ion (CN⁻) has a lone pair on the carbon atom, making it the nucleophile. Unlike substitution, there is no leaving group — the π bond simply breaks and both electrons go to oxygen.

Nucleophile: CN⁻ (lone pair on C of CN⁻) | Electrophile: Cδ⁺ of C=O | No leaving group
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Step 2 — Draw the Two Curly Arrows

Draw the first curly arrow from the lone pair on the carbon of CN⁻ (the tail) to the carbonyl carbon of ethanal (the head). This shows the formation of the new C–CN bond. Simultaneously, draw a second curly arrow from the C=O π bond (the tail) to the oxygen atom (the head). This shows the π bond electrons moving entirely onto oxygen as the double bond breaks down to a single bond, giving oxygen a negative charge (O⁻).

Arrow 1: lone pair on C of CN⁻ → carbonyl C (new C–C bond) | Arrow 2: C=O π bond → O (O gains lone pair, becomes O⁻)
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Step 3 — Identify the Tetrahedral Intermediate

After the nucleophilic attack, the carbonyl carbon has changed from trigonal planar (sp² hybridisation, three groups) to tetrahedral (sp³ hybridisation, four groups). The intermediate is an alkoxide ion — a species with a negatively charged oxygen (–O⁻). The formula of the intermediate is CH₃CH(CN)O⁻.

Intermediate: CH₃CH(CN)O⁻ (alkoxide ion, tetrahedral carbon)
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Step 4 — Protonation to Give the Final Product

In aqueous solution (or with a proton source such as dilute acid), the alkoxide ion is quickly protonated. A water molecule (or H⁺) donates a proton to the O⁻. This gives the final product: a hydroxynitrile (also called a cyanohydrin). The overall equation is CH₃CHO + HCN → CH₃CH(OH)CN.

CH₃CHO + HCN → CH₃CH(OH)CN (2-hydroxypropanenitrile)
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Step 5 — Verify Conservation and Note Key Feature

Verify atom and charge balance. Reactants: CH₃CHO + HCN gives 3C, 4H, 1O, 1N, neutral overall. Product: CH₃CH(OH)CN gives 3C, 4H, 1O, 1N, neutral overall. ✓ The carbonyl carbon has become tetrahedral in the product — a key feature of nucleophilic addition.

Atoms balanced ✓ | Charges balanced ✓ | C=O (trigonal planar) → C–OH (tetrahedral) ✓

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IB Exam Tip — Nucleophilic Addition

Key points to remember for IB nucleophilic addition: (1) The nucleophile always attacks the carbonyl carbon (Cδ⁺), not the oxygen. (2) Two curly arrows are needed in the addition step: one from the nucleophile's lone pair to the carbonyl carbon, and one from the C=O π bond to the oxygen. (3) The product has a tetrahedral carbon where the carbonyl carbon was. (4) This pattern applies to any nucleophile (e.g., H⁻ from NaBH₄ reducing an aldehyde) — the key difference is the identity of the nucleophile.

SECTION 7

Heterolytic vs. Homolytic — Comparing Bond-Breaking Mechanisms

Not all bond-breaking processes involve electron-pair sharing. It's important to distinguish heterolytic fission (where one atom takes both electrons) from homolytic fission (where each atom takes one electron). The type of fission determines whether the reaction proceeds via electron pairs or free radicals.

Feature	Heterolytic Fission	Homolytic Fission
Electron distribution	Both electrons go to one atom	One electron to each atom
Products	Ions (cation + anion)	Free radicals
Arrow notation	Double-headed curly arrow (two electrons)	Single-headed fish-hook arrow (one electron)
Favoured conditions	Polar solvents, polar bonds	UV light, high temperature, non-polar bonds
Example	CH₃Br → CH₃⁺ + Br⁻	Cl₂ → Cl• + Cl•

Comparison of heterolytic and homolytic bond fission

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KEY TAKEAWAY

Think of breaking a bond like splitting a pair of chopsticks. In heterolytic fission, one person keeps both chopsticks (like one atom keeping both electrons) — this creates an imbalance (ions). In homolytic fission, each person takes one chopstick (one electron each) — both walk away with the same thing (radicals). Electron-pair sharing reactions always use the heterolytic pathway because they involve the coordinated movement of complete electron pairs.

SECTION 8

Connection to Advanced Theory — Reaction Kinetics & Stereochemistry

Understanding electron-pair sharing reactions at the mechanism level opens the door to predicting reaction rates and stereochemical outcomes. For instance, an S_N2 reaction always produces inversion of configuration (the nucleophile attacks from the opposite side of the leaving group). The S_N1 mechanism — in which a carbocation intermediate forms before the nucleophile attacks — is not required for IB Reactivity 3.4 and is included in the comparison table below for extension purposes only. You will not be assessed on SN1 in your IB exams.

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Extension Only — SN1 is Beyond IB Scope

The S_N1 mechanism is beyond the scope of the IB Diploma Programme Reactivity 3.4 specification. The IB specification focuses on S_N2 mechanisms. The table below is provided as extension reading for students who are curious about how S_N1 differs, but this content will not appear on your IB exam.

Feature	S_N2 (IB Level)	S_N1 (Extension — Beyond IB Scope)
Number of steps	One (concerted)	Two (via carbocation)
Rate law	Rate = k[Nu⁻][substrate]	Rate = k[substrate]
Stereochemistry	Inversion (Walden inversion)	Racemization (mixture)
Preferred substrate	Primary haloalkane (less steric hindrance)	Tertiary haloalkane (stable carbocation)
Solvent effect	Favoured in aprotic polar solvents	Favoured in protic polar solvents

Comparing SN2 (IB level) and SN1 (extension only) reaction pathways

As you progress beyond IB, you will see how the concepts of steric effects and solvent polarity interweave to give you a complete picture of reactivity. The curly-arrow mechanisms you learn now form the essential foundation for university-level organic chemistry, biochemistry, and even drug design. Every pharmaceutical synthesis involves dozens of carefully planned electron-pair sharing reactions.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why a hydroxide ion (OH⁻) acts as a nucleophile while a hydrogen ion (H⁺) acts as an electrophile. Use the terms 'electron pair', 'donor', and 'acceptor' in your answer.

Show answer

The hydroxide ion has three lone pairs of electrons on the oxygen atom and carries a full negative charge, giving it a high electron density. This makes it an electron-pair donor — a nucleophile. In contrast, the hydrogen ion has an empty 1s orbital and no electrons, so it can only accept an electron pair — making it an electrophile. In their reaction, the lone pair from OH⁻ (nucleophile/donor) forms a coordinate bond with H⁺ (electrophile/acceptor), producing water (H₂O). This is a classic Lewis acid-base reaction.

Question 2: BASIC CALCULATION

In the reaction NH₃ + BF₃ → NH₃BF₃, identify the nucleophile and the electrophile. Draw the curly arrow for this reaction and state what type of bond is formed.

Show answer

NH₃ is the nucleophile (Lewis base) because the nitrogen atom has a lone pair available for donation. BF₃ is the electrophile (Lewis acid) because boron has an empty p orbital and only six electrons in its outer shell — it needs two more to complete its octet. The curly arrow starts at the lone pair on nitrogen and points toward the boron atom. The bond formed is a coordinate (dative) covalent bond, because both electrons in the new N–B bond originate from nitrogen. In the product NH₃BF₃, boron now has an octet and nitrogen carries a formal positive charge.

Question 3: INTERMEDIATE

The reaction of 1-bromopropane (CH₃CH₂CH₂Br) with cyanide ion (CN⁻) proceeds via an S_N2 mechanism. (a) Write the overall equation. (b) Identify the nucleophile, electrophilic centre, and leaving group. (c) Describe the curly arrows needed for the mechanism.

Show answer

(a) CH₃CH₂CH₂Br + CN⁻ → CH₃CH₂CH₂CN + Br⁻. (b) The nucleophile is CN⁻ (it has a lone pair on carbon), the electrophilic centre is the carbon bonded to bromine (Cδ⁺ due to the C–Br polarity), and the leaving group is Br⁻. (c) Two curly arrows are needed. Arrow 1 goes from the lone pair on the carbon of CN⁻ to the electrophilic carbon of the substrate — this forms the new C–C bond. Arrow 2 goes from the C–Br bond to the bromine atom — this represents the heterolytic breaking of the C–Br bond. Both arrows move simultaneously in this concerted S_N2 process. The product, butanenitrile, has the CN group replacing Br.

Question 4: APPLIED

Ethene (CH₂=CH₂) reacts with hydrogen bromide (HBr) via an electrophilic addition mechanism. (a) Explain why the π bond in ethene acts as a nucleophile. (b) Draw the curly arrows for both steps of this reaction. (c) Identify the intermediate formed after step 1.

Show answer

(a) The π bond in ethene consists of electron density located above and below the plane of the C=C bond. This region of high electron density is attractive to electrophiles, so the π electrons effectively act as a nucleophile — they donate to the incoming electrophile. (b) Step 1: A curly arrow goes from the C=C π bond to the H atom of HBr (forming a new C–H bond), while a second curly arrow goes from the H–Br bond to Br (heterolytic fission, producing Br⁻). Step 2: A curly arrow goes from a lone pair on Br⁻ to the positively charged carbon (the carbocation), forming the C–Br bond. (c) The intermediate after step 1 is a carbocation (CH₃CH₂⁺), where one carbon bears a positive charge because it lost the π electrons to form the new C–H bond. This carbocation is then attacked by Br⁻ in step 2 to give CH₃CH₂Br.

Question 5: CRITICAL THINKING

A student claims: 'Since water (H₂O) has lone pairs on oxygen, it can only act as a nucleophile in electron-pair sharing reactions.' Evaluate this claim. Provide at least two examples to support your argument.

Show answer

The student's claim is incorrect. Water is amphoteric in the Lewis sense — it can act as both a nucleophile and an electrophile depending on the reaction partner. As a nucleophile (Lewis base), water donates a lone pair to electrophiles such as H⁺, forming H₃O⁺ (hydronium ion). For example, in the hydration of carbocations during SN1 reactions, water attacks the Cδ⁺ using its lone pair. However, water can also act as an electrophile (Lewis acid) when it accepts an electron pair from a very strong base. For instance, when metallic sodium reacts with water, the Na donates electrons to the partially positive hydrogen in water, and the O–H bond breaks heterolytically. Furthermore, in reactions with extremely strong nucleophiles like organolithium reagents (e.g., CH₃Li), the carbon in CH₃Li is so electron-rich that it attacks the hydrogen of water. This demonstrates that nucleophilic or electrophilic behaviour is relative — it depends on the reaction partner, not solely on the presence of lone pairs.

SUMMARY

Lesson Summary

Summary

Electron-pair sharing reactions are the foundation of organic reaction mechanisms. Every such reaction involves a nucleophile (electron-pair donor) and an electrophile (electron-pair acceptor). The movement of electron pairs is tracked using curly arrows, which always start at the electron source (tail) and end at the electron sink (head). When a bond breaks and both electrons go to one atom, this is heterolytic fission. The new bond formed when both electrons come from one atom is a coordinate (dative) covalent bond.

The three major IB reaction types are nucleophilic substitution (SN2) (a nucleophile replaces a leaving group on an sp³ carbon in a concerted single step), electrophilic addition (an electrophile attacks a π bond in an alkene, proceeding via a carbocation intermediate), and nucleophilic addition (a nucleophile attacks a polar C=O bond, giving a tetrahedral product). These three reaction types — and their curly-arrow mechanisms — are the core of IB Reactivity 3.4. Frontier molecular orbital theory (HOMO–LUMO) and the SN1 mechanism go beyond the IB specification and are extension material only. Comparing heterolytic vs. homolytic fission clarifies why electron-pair sharing reactions differ fundamentally from free-radical processes. Mastering curly-arrow notation is essential for success in IB Chemistry and beyond.