AQA A-Level Organic Chemistry 3.3.1 Introduction to Organic Chemistry Organic Reaction Types and Bond Fission
3.3.1

Organic Reaction Types and Bond Fission

Covalent bond splitting, curly arrow notation, reagent classifications, and reaction types.

Chemical reactions occur when chemical bonds break and new bonds form. To understand organic reaction mechanisms, we need to know how covalent bonds can break (bond fission), how to trace the movement of electrons using curly arrows, and how to classify the organic reagents and reaction pathways.

🔑 Key Principle

Curly arrows always show the movement of electrons (never atoms). A full-headed arrow represents a pair of electrons moving, while a single-headed arrow (fishhook) represents a single electron moving.

Bond Fission

A covalent bond consists of a shared pair of electrons. When this bond is broken, it can split in two ways:

1. Homolytic Fission

During homolytic fission, the shared pair of electrons in the covalent bond breaks evenly, with each bonded atom taking one electron from the shared pair. This process forms highly reactive neutral species containing an unpaired electron, called free radicals. We represent homolytic fission using single-headed curly arrows (fishhook arrows).

\( \text{X-Y} \rightarrow \text{X}^\bullet + \text{Y}^\bullet \)

2. Heterolytic Fission

During heterolytic fission, the shared pair of electrons in the covalent bond breaks unevenly, with one of the bonded atoms taking both of the shared electrons. This results in the formation of charged species: a positive ion (cation) and a negative ion (anion). We represent heterolytic fission using a standard full-headed curly arrow pointing to the more electronegative atom.

\( \text{X-Y} \rightarrow \text{X}^+ + \text{:Y}^- \)

Homolytic fission

The breaking of a covalent bond so that each bonded atom receives one electron from the bonded pair, forming two radicals.

Heterolytic fission

The breaking of a covalent bond so that both bonded electrons go to one atom, forming a cation and an anion.

Free radical

A highly reactive chemical species containing an unpaired valence electron.

Homolytic vs Heterolytic Bond Fission Homolytic Fission Cl Cl 2 Cl• Two neutral radicals formed Heterolytic Fission H Cl H⁺ + Cl⁻ Cation and anion formed

Curly Arrow Notation

Curly arrows show the path of valence electrons during chemical processes. When drawing reaction mechanisms, ensure your arrows adhere strictly to these conventions:

📝 AQA Examiner Tip

Never start a curly arrow from a positive charge (like \( \text{H}^+ \)) or from an element symbol itself unless it has a drawn lone pair. This is a common error that instantly loses marks in mechanism questions. Always draw the lone pair of electrons (e.g. on \( \text{:OH}^- \) or \( \text{:NH}_3 \)) to show exactly where the arrow begins.

Classifying Reagents

In organic mechanisms, species attacking organic molecules are classified as:

Nucleophiles (Nucleus-loving)

An electron pair donor. They are electron-rich species containing a lone pair of electrons. They attack delta positive (\( \delta^+ \)) carbons.

Examples: \( \text{:OH}^- \), \( \text{:NH}_3 \), \( \text{H}_2\text{O:} \), \( \text{:CN}^- \)

Electrophiles (Electron-loving)

An electron pair acceptor. They are electron-deficient species (often positive ions or molecules with polar bonds) that attack electron-rich sites.

Examples: \( \text{NO}_2^+ \), \( \text{H}^+ \), polar molecules like \( \text{H-Br} \)

Nucleophile

An electron pair donor that attacks a positively charged or electron-deficient carbon atom.

Electrophile

An electron pair acceptor that attacks an electron-rich site (such as a double bond or aromatic ring).

Reaction Types Overview

You must be able to classify organic reactions based on what happens to the reactants and products:

Reaction Type Description General Form Example
Addition Two reactant molecules combine to form a single product. Typical of alkenes. \( \text{A} + \text{B} \rightarrow \text{C} \) \( \text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br} \)
Substitution An atom or group of atoms in a molecule is replaced by another atom or group. \( \text{A-B} + \text{C} \rightarrow \text{A-C} + \text{B} \) \( \text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^- \)
Elimination A small molecule (e.g. \( \text{H}_2\text{O} \), \( \text{HCl} \)) is removed from a single reactant, forming a double bond. \( \text{A} \rightarrow \text{B} + \text{C} \) \( \text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} + \text{Br}^- \)
Oxidation The addition of oxygen, removal of hydrogen, or loss of electrons. Represented as \( \text{[O]} \). \( \text{Reactant} + \text{[O]} \) \( \text{CH}_3\text{CH}_2\text{OH} + \text{[O]} \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O} \)
Hydrolysis The breaking of a covalent bond using water or hydroxide ions. \( \text{R-X} + \text{H}_2\text{O} \rightarrow \text{R-OH} + \text{HX} \) \( \text{CH}_3\text{COOCH}_3 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{COOH} + \text{CH}_3\text{OH} \)

Worked Examples

✏️ Worked Example 1
Write equation equations to show:
  1. The homolytic fission of a chlorine molecule (\( \text{Cl}_2 \)) under UV light.
  2. The heterolytic fission of the C-Br bond in bromoethane (\( \text{CH}_3\text{CH}_2\text{Br} \)).

Solution:

  1. Homolytic fission of chlorine splits the bond evenly to form chlorine radicals: \[ \text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}^\bullet \]
  2. Heterolytic fission of bromoethane splits the polar bond unevenly, with the bromine atom taking both electrons because it is more electronegative: \[ \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2^+ + \text{Br}^- \]
✏️ Worked Example 2
Classify the reaction types and identify the attacking reagent as a nucleophile, electrophile, or free radical:
  1. \( \text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br} \)
  2. \( \text{CH}_4 + \text{Cl}^\bullet \rightarrow \text{CH}_3^\bullet + \text{HCl} \)

Solution:

  1. Two reactant molecules combine into one. This is an addition reaction. The alkene has an electron-rich double bond which is attacked by the electron-deficient \( \text{H}^{\delta+} \) of \( \text{H-Br} \), so the attacking reagent is an electrophile (specifically, this is electrophilic addition).
  2. An atom is substituted. This is a substitution reaction (propagation step). The attacking chlorine species Cl• has an unpaired electron, so it is a free radical (free radical substitution).
Study this topic on the go

Get flashcards and quizzes in ChemEasy, or plan your revision with ChemPlan IB.

See our apps →
← Isomerism Back to 3.3.1 Index →