Chemistry | Alkenes

Welcome to the Exciting World of Alkenes!

Hello future organic chemists! Don's worry if organic chemistry seems overwhelming; we will break down the fundamental concepts step-by-step. This chapter introduces Alkenes, a family of molecules that are incredibly important for everything from making plastics to ripening fruit.

In this section, we will explore the unique structure of alkenes, how to name them correctly, and why their defining feature—the double bond—makes them much more reactive than their alkane cousins.

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1. Structure and Bonding in Alkenes

What are Alkenes?

Alkenes are a class of unsaturated hydrocarbons. This means they are organic compounds containing only carbon and hydrogen atoms, but they contain at least one carbon-carbon double bond (\(\text{C}=\text{C}\)).

• Hydrocarbon: Contains only H and C.
• Unsaturated: Contains at least one double or triple bond (meaning it is not 'saturated' with hydrogen atoms).

The General Formula

The general formula for non-cyclic alkenes (those with only one double bond) is: $$ \text{C}_n\text{H}_{2n} $$ Example: Ethene has 2 carbons, so it is \(\text{C}_2\text{H}_{4}\).

The Anatomy of the Double Bond (\(\text{C}=\text{C}\))

The double bond is not just two single bonds side-by-side; it consists of two distinct types of covalent bonds:

a) The Sigma (\(\sigma\)) Bond

The first bond formed between the two carbon atoms is a sigma bond. This is formed by the head-on overlap of orbitals. It is a very strong bond.

Analogy: Think of the sigma bond as a firm handshake—it holds the atoms tightly together along the axis.

b) The Pi (\(\pi\)) Bond

The second bond is a pi bond. This is formed by the sideways overlap of the p-orbitals, resulting in electron density concentrated above and below the plane of the \(\sigma\)-bond.

The \(\pi\)-bond is:

• Weaker than the \(\sigma\)-bond.
• More exposed to attack by other molecules.
• Responsible for the high reactivity of alkenes.

Key Takeaway: The weak, exposed pi bond is the Achilles' heel of the alkene molecule, making it very eager to react and turn the double bond into a single bond.

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2. Nomenclature: Naming Alkenes

Naming alkenes follows the standard IUPAC rules for organic chemistry, with a few important adjustments.

Step-by-Step Naming Rules

1. Identify the Longest Chain: Find the longest continuous carbon chain that must include both carbon atoms of the double bond.
2. Base Name: Replace the '-ane' ending of the corresponding alkane with '-ene'. (e.g., Propane becomes Propene).
3. Number the Chain: Number the carbon chain starting from the end that gives the double bond the lowest possible number.
4. Locate the Double Bond: Place the number of the first carbon atom in the double bond immediately before the '-ene' suffix or directly before the base name.
5. Add Substituents: Name and number any side chains (substituents) as usual.

Example 1: A four-carbon chain with a double bond between C1 and C2 is But-1-ene.
Example 2: A four-carbon chain with a double bond between C2 and C3 is But-2-ene.

Common Mistake to Avoid: You must include the double bond in the main numbered chain, even if a slightly longer chain exists elsewhere in the molecule!

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3. Stereoisomerism (E/Z Isomerism)

Don't worry if this sounds complicated—we'll tackle stereoisomerism using a simple checklist!

What is Stereoisomerism?

Stereoisomers are molecules that have the same molecular formula and the same structural formula (the atoms are connected in the same order) but have a different arrangement of atoms in space.

E/Z isomerism (sometimes called Geometric Isomerism) happens only in alkenes because the pi bond prevents free rotation around the \(\text{C}=\text{C}\) axis. The molecule is effectively "locked" in place.

Conditions for E/Z Isomerism

E/Z isomers can exist only if:

Each carbon atom in the double bond is bonded to two different groups.

If one carbon has two identical groups attached (e.g., two hydrogens), E/Z isomerism is not possible (e.g., propene cannot have E/Z isomers).

Applying the E/Z System (Cahn-Ingold-Prelog Rules)

To determine if an isomer is E or Z, we use priority rules based on the atomic number of the atoms directly attached to the double-bonded carbons.

1. Assign Priority: For each carbon in the double bond, look at the two groups attached. The group with the atom of higher atomic number gets the higher priority.
2. Compare Positions: Look at the positions of the two high-priority groups relative to each other.

Z Isomer: Zusammen (Same Side)

If the two high-priority groups are on the Same side of the double bond (both above or both below the plane), it is the Z isomer.

Memory Trick: Z = Zame side.

E Isomer: Entgegen (Opposite Side)

If the two high-priority groups are on Opposite sides (one above, one below), it is the E isomer.

Example: In 1-bromo-2-chloroethene, Bromine (Br) has a higher atomic number than Hydrogen (H), and Chlorine (Cl) has a higher atomic number than Hydrogen (H). If Br and Cl are on the same side, it is the Z isomer.

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4. Reactivity of Alkenes: Electrophilic Addition

Alkenes are highly reactive because of that exposed \(\pi\)-bond. They undergo Addition Reactions, where the double bond breaks, and new atoms are added across the carbons, turning the alkene into a saturated molecule (like an alkane).

The Mechanism: Electrophilic Addition

Since the \(\pi\)-bond is rich in electrons, it attracts species that are deficient in electrons. These electron-loving species are called electrophiles.

Key Definitions:

• Electrophile: An electron-pair acceptor. They are positively charged ions (\(\text{H}^+\)) or molecules with a partially positive region ($\delta+$).
• Nucleophile: An electron-pair donor (likes positive centres).

Step 1: Attack by the Electrophile

The electron pair in the \(\pi\)-bond attacks the electrophile. This breaks the \(\pi\)-bond and simultaneously breaks the bond within the attacking electrophile molecule.

Step 2: Formation of the Carbocation Intermediate

One carbon atom forms a bond with the electrophile, leaving the other carbon atom positively charged. This positively charged ion is called a carbocation.

Step 3: Attack by the Nucleophile

The remaining negatively charged ion (the nucleophile, often a halide ion like \(\text{Br}^-\) or \(\text{Cl}^-\)) attacks the positive carbocation, forming the final product.

Don't worry if this seems tricky at first! The movement of electrons is always shown using curly arrows, starting from an electron source (like the \(\pi\)-bond) and pointing to the electron sink (the positive centre/electrophile).

Key Takeaway: Alkenes react by electrophilic addition, meaning the double bond uses its electrons to attack an electron-deficient species first.

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5. Key Addition Reactions of Alkenes

All of the following reactions involve the breaking of the \(\pi\)-bond and the formation of a saturated product.

5.1. Addition of Hydrogen (Hydrogenation)

• Reagents: \(\text{H}_2\) gas.
• Conditions: Nickel (Ni) catalyst, 150°C (or Platinum/Palladium at room temperature).
• Product: An Alkane.
• Use: This process is used industrially to convert unsaturated vegetable oils (liquids) into saturated fats (solids), like margarine.

$$ \text{Alkene} + \text{H}_2 \xrightarrow{\text{Ni Catalyst}} \text{Alkane} $$

5.2. Addition of Halogens (Halogenation)

• Reagents: Halogen molecules like \(\text{Br}_2\), \(\text{Cl}_2\).
• Conditions: Room temperature, no catalyst or UV light needed. Often carried out using the halogen dissolved in an organic solvent (e.g., non-polar \(\text{CCl}_4\) or aqueous solution).
• Product: Dihaloalkane (e.g., 1,2-dibromoethane).

The Test for Unsaturation

This reaction forms the basis of the chemical test to distinguish between alkanes and alkenes:

If you bubble gas or shake liquid with Bromine water (which is orange/brown):

1. Alkane: No reaction (remains orange/brown).
2. Alkene: The bromine adds across the double bond, and the orange colour immediately disappears (turns colourless).

5.3. Addition of Hydrogen Halides (HX)

• Reagents: Hydrogen bromide (\(\text{HBr}\)), hydrogen chloride (\(\text{HCl}\)).
• Conditions: Room temperature.
• Product: Haloalkane.

The "Rich Get Richer": Markovnikov's Rule

When an asymmetric alkene (like propene) reacts with an asymmetric reagent (like \(\text{HBr}\)), two different products are possible. One product is usually formed in a much greater quantity (the major product).

Markovnikov's Rule predicts the major product:

When an electrophilic addition reaction occurs to an unsymmetrical alkene, the hydrogen atom of the attacking species adds to the carbon atom of the double bond that already has the most hydrogen atoms.

This happens because adding the hydrogen to the "richer" carbon forms a more stable carbocation intermediate (tertiary > secondary > primary).

Example: Propene + \(\text{HBr}\) primarily yields 2-bromopropane (major product), not 1-bromopropane (minor product).

5.4. Addition of Steam (Hydration)

• Reagents: Steam (\(\text{H}_2\text{O}\)).
• Conditions: High temperature (300°C), high pressure (60 atm), and an acidic catalyst (usually phosphoric(V) acid, \(\text{H}_3\text{PO}_4\)).
• Product: Alcohol.
• Use: This is the industrial method for producing ethanol from ethene.

$$ \text{Alkene} + \text{H}_2\text{O} \xrightarrow{\text{H}_3\text{PO}_4, \text{heat}} \text{Alcohol} $$

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6. Polymerisation of Alkenes

One of the most important uses of alkenes is their ability to form addition polymers.

What is Polymerisation?

Polymerisation is the process of joining many small, reactive molecules (monomers) together to form a very long chain molecule (a polymer).

Alkenes, like ethene, are ideal monomers because the double bond can easily break open to form two new single bonds, linking it to the next molecule.

The Process

In addition polymerisation:

1. High temperature, high pressure, and an initiator (often a catalyst that forms free radicals) are required.
2. The \(\pi\)-bond in the alkene monomer breaks.
3. The broken bonds link up with adjacent monomers, forming a huge chain.

Example: Ethene monomers polymerise to form poly(ethene), commonly known as polythene or polyethylene (used in plastic bags and films).

The general equation shows 'n' number of monomers linking up:

$$ n \cdot \text{ethene} \rightarrow \text{poly(ethene)} $$