Chemistry | Introductory Organic Chemistry and Alkanes

Hello Future Organic Chemists!

Welcome to the fascinating world of Organic Chemistry! This chapter is your essential starting point, dealing with the chemistry of carbon and the simplest family of organic molecules: the Alkanes.

Don't worry if this seems like a whole new language at first. We will break down every definition, rule, and concept step-by-step. By the end of this chapter, you will be able to draw, name, and predict the basic reactions of some common organic compounds. Let’s get started!

1. The Foundations of Organic Chemistry

1.1 What Makes Carbon Special?

Organic chemistry is defined as the chemistry of carbon compounds, primarily those containing carbon and hydrogen. Why is carbon so important?

• Tetravalency: Carbon is in Group 4, meaning it always forms four covalent bonds. This allows it to connect to many other atoms.
• Catenation: This is carbon’s unique ability to bond strongly to other carbon atoms, forming long chains, branched structures, and rings. This is why millions of organic compounds exist!

1.2 Defining Key Terms

a) Hydrocarbons and Functional Groups

A hydrocarbon is the simplest type of organic molecule, containing only carbon and hydrogen atoms.

A functional group is a specific atom or group of atoms within a molecule that determines its characteristic chemical properties. Think of it as the "action centre" of the molecule.

Example: If the molecule contains a C=C double bond, it will behave like an alkene, regardless of how long the carbon chain is.

b) Homologous Series

A homologous series is a family of organic compounds that:

• Share the same general formula.
• Differ from the next member by a \(\text{CH}_2\) group.
• Show a gradual trend in physical properties (e.g., boiling point).
• Have similar chemical properties (because they share the same functional group).

c) Saturated vs. Unsaturated

These terms describe the nature of the bonds within a hydrocarbon:

• Saturated: Contains only single C–C bonds. Alkanes are saturated hydrocarbons.
• Unsaturated: Contains at least one C=C double bond or \(\text{C}\equiv\text{C}\) triple bond.

2. Representing Organic Molecules

There are several ways to write down the structure of an organic molecule. We move from the most basic information to the most detailed structure.

Let’s use the example of Propane (3 carbons):

• Empirical Formula: The simplest whole-number ratio of atoms (Rarely used for complex organic molecules).
• Molecular Formula: Shows the actual number of atoms of each element in the molecule. Example: \(\text{C}_3\text{H}_8\)
• General Formula: A formula that represents all members of a homologous series. Example for alkanes: \(\text{C}_n\text{H}_{2n+2}\)
• Structural Formula: Shows how the atoms are arranged, usually by grouping atoms together. Example: \(\text{CH}_3\text{CH}_2\text{CH}_3\)
• Displayed Formula: Shows all the atoms and all the bonds explicitly drawn out (like a detailed blueprint).

2.1 Introducing Skeletal Formulae (A-Level Essential)

As molecules get bigger, drawing every C and H becomes tedious. Skeletal formulae are used to simplify diagrams:

• Carbon atoms are represented by the end of a line or a bend in the line.
• Hydrogen atoms attached to carbon are not shown (we assume carbon has four bonds, so we fill in the hydrogens automatically).
• Atoms other than C and H (heteroatoms) are shown explicitly.

Analogy: A molecular formula is like the headcount of people in a room; a displayed formula is a detailed floor plan showing where everyone is standing and holding hands.

3. Naming Simple Organic Compounds (Nomenclature)

We use the IUPAC (International Union of Pure and Applied Chemistry) system to ensure every molecule has one unique name.

3.1 Core Prefixes (The Root Chain)

The first step is always identifying the number of carbon atoms in the longest continuous chain. You must memorise these prefixes:

• 1 Carbon: Meth-
• 2 Carbons: Eth-
• 3 Carbons: Prop-
• 4 Carbons: But-
• 5 Carbons: Pent-
• 6 Carbons: Hex-

Memory Aid: Monkeys Eat Peeled Bananas (Meth, Eth, Prop, But).

3.2 Naming Branched Alkanes (IUPAC Rules)

The name consists of three parts: (1) Substituent/Branch, (2) Root Chain, (3) Suffix.

For alkanes, the suffix is always -ane.

Step-by-Step Naming Process:

1. Find the Longest Chain (The Root): Identify the longest continuous chain of carbon atoms. This determines the root name (e.g., propane, butane).
2. Identify Substituents (Branches): Any carbon groups hanging off the main chain are called alkyl groups (e.g., \(\text{CH}_3\) is a methyl group).
3. Number the Chain: Start numbering the main chain from the end that gives the substituents the lowest possible numbers.
4. Assemble the Name:
   • List the substituents alphabetically (e.g., ethyl before methyl).
   • Use numbers to indicate where the substituents are attached.
   • Use prefixes (di-, tri-, tetra-) if the same group appears multiple times.

Important Formatting Rule: Use hyphens (-) between numbers and letters, and commas (,) between numbers.

4. Isomerism

Isomers are molecules that have the same molecular formula but different arrangement of atoms (different structural formulae).

Isomers have different physical properties (like boiling points) and sometimes different chemical properties.

4.1 Structural Isomerism

Structural isomerism (also called constitutional isomerism) is when the atoms are connected in completely different ways.

There are three main types relevant at AS level:

a) Chain Isomerism

The carbon skeleton (the chain) is arranged differently. This involves having branched vs. unbranched chains.

Example: Pentane (\(\text{C}_5\text{H}_{12}\)) can exist as linear pentane, methylbutane (one branch), or dimethylpropane (two branches).

b) Position Isomerism

The functional group (or substituent) is attached to a different carbon atom on the main chain.

Example: 1-chloropropane vs. 2-chloropropane (The chlorine atom is in a different position).

c) Functional Group Isomerism

The molecules have the same molecular formula but different functional groups.

Example: Ethanol (an alcohol, \(\text{C}_2\text{H}_6\text{O}\)) and Dimethyl ether (an ether, \(\text{C}_2\text{H}_6\text{O}\)).

5. Alkanes: Structure, Bonding, and Properties

Alkanes are the simplest homologous series. They are saturated hydrocarbons with the general formula \(\text{C}_n\text{H}_{2n+2}\).

5.1 Structure and Bonding in Alkanes

Every bond in an alkane is a single covalent bond. These are specifically called sigma (\(\sigma\)) bonds, formed by the direct overlap of orbitals.

• Geometry: Around every carbon atom in an alkane, there are four single bonds. These four electron pairs repel each other equally.
• Shape: This repulsion results in a tetrahedral shape around each carbon atom.
• Bond Angle: The bond angle is approximately \(109.5^\circ\).

5.2 Physical Properties and Trends

Alkanes are generally non-polar molecules because the C–C bonds are non-polar, and the C–H bonds have very low polarity. Furthermore, the symmetrical structure means any minor polarity cancels out.

Explaining Boiling Point Trends

Since alkanes are non-polar, the only intermolecular forces acting between the molecules are the weak van der Waals forces (specifically, London/dispersion forces).

Trend: As the chain length increases (e.g., going from methane to octane), the boiling point increases.

Explanation:

1. A longer chain means a larger molecule with more electrons.
2. This results in a larger surface area and stronger van der Waals forces.
3. More energy is required to overcome these stronger forces, hence a higher boiling point.

Effect of Branching on Boiling Point

For isomers (molecules with the same molecular formula):

• Straight-chain isomers have higher boiling points.
• Branched isomers have lower boiling points.

Explanation: Branching makes the molecule more spherical and compact, reducing the surface area available for close contact between molecules. This weakens the van der Waals forces.

6. Reactions of Alkanes

Alkanes are known for being chemically inert (unreactive) due to the strength of the non-polar C–C and C–H sigma bonds. However, they undergo two main types of reaction:

6.1 Complete and Incomplete Combustion

Alkanes are excellent fuels. They burn (combust) in the presence of oxygen.

a) Complete Combustion (Excess Oxygen)

The alkane burns completely to form only carbon dioxide and water, releasing large amounts of heat (exothermic).

$$ \text{Alkane} + \text{O}_2 \longrightarrow \text{CO}_2 + \text{H}_2\text{O} $$ Example: \(\text{CH}_4 (g) + 2\text{O}_2 (g) \longrightarrow \text{CO}_2 (g) + 2\text{H}_2\text{O} (l)\)

b) Incomplete Combustion (Limited Oxygen)

If there is insufficient oxygen, combustion produces dangerous products like carbon monoxide (\(\text{CO}\)) and/or carbon (soot, C).

$$ \text{Alkane} + \text{O}_2 \longrightarrow \text{CO} / \text{C} + \text{H}_2\text{O} $$

Danger! Carbon monoxide is toxic because it binds irreversibly to haemoglobin in the blood.

6.2 Free-Radical Substitution of Alkanes

Alkanes can react with halogens (like chlorine, \(\text{Cl}_2\), or bromine, \(\text{Br}_2\)) under specific conditions. This is a substitution reaction because a hydrogen atom is replaced by a halogen atom.

Condition Required: High temperature OR ultraviolet (UV) light.

$$ \text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_3\text{Cl} + \text{HCl} $$

The Mechanism (Step-by-Step)

This reaction proceeds via a free-radical mechanism. A free radical is a species with an unpaired electron, making it highly reactive (shown by a dot, e.g., \(\text{Cl}\cdot\)).

Step 1: Initiation

The UV light provides energy to break the halogen molecule (\(\text{Cl}_2\)) bond. The bond breaks homolytically (equally), forming two free radicals.

$$ \text{Cl}_2 \xrightarrow{\text{UV}} 2\text{Cl}\cdot $$ Analogy: This is like "starting the engine" of the reaction.

Step 2: Propagation (The Main Reaction Cycle)

This is where the substitution happens. Free radicals are consumed and created in equal numbers, keeping the cycle going.

a) A chlorine radical attacks the alkane, pulling off a hydrogen atom to form \(\text{HCl}\) and creating an alkyl radical (\(\text{R}\cdot\)).

$$ \text{Cl}\cdot + \text{CH}_4 \longrightarrow \text{HCl} + \text{CH}_3\cdot $$

b) The new methyl radical attacks another halogen molecule, forming the product and generating a new chlorine radical.

$$ \text{CH}_3\cdot + \text{Cl}_2 \longrightarrow \text{CH}_3\text{Cl} + \text{Cl}\cdot $$

Step 3: Termination

The reaction stops when two free radicals meet and bond together, removing the reactive species from the system.

$$ \text{Cl}\cdot + \text{Cl}\cdot \longrightarrow \text{Cl}_2 $$ $$ \text{CH}_3\cdot + \text{Cl}\cdot \longrightarrow \text{CH}_3\text{Cl} $$ $$ \text{CH}_3\cdot + \text{CH}_3\cdot \longrightarrow \text{C}_2\text{H}_6 \text{ (Formation of trace impurities)} $$