Chemistry (9701) | Alkenes

📚 Chemistry 9701 Study Notes: Alkenes (Unsaturated Hydrocarbons)

Welcome to the exciting world of Alkenes! This chapter is crucial because it introduces you to the chemistry of the carbon-carbon double bond ($\text{C}=\text{C}$). Unlike the very stable alkanes, alkenes are highly reactive, and understanding *why* they are reactive is key to mastering organic chemistry.
We will explore their unique structure, the type of reactions they undergo (Electrophilic Addition), and how they are used to make huge polymer molecules!

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1. Structure and Bonding in Alkenes

1.1 The Carbon-Carbon Double Bond ($\text{C}=\text{C}$)

Alkenes are defined as unsaturated hydrocarbons, meaning they contain at least one $\text{C}=\text{C}$ double bond. This double bond is the source of all their unique chemical properties.

Key Bonding Concepts: $\sigma$ and $\pi$ bonds

The double bond is not just one big bond; it's actually made of two different types of bonds:

• Sigma ($\sigma$) Bond: This is formed by the direct, head-on overlap of two $sp^2$ hybrid orbitals (one from each carbon atom). This bond is very strong and lies directly between the two carbon nuclei.
• Pi ($\pi$) Bond: This is formed by the sideways overlap of the unhybridized $p$-orbitals above and below the plane of the $\sigma$ bond.

Analogy: Think of the $\sigma$ bond as a strong central pin holding two atoms together firmly. The $\pi$ bond is like a loose, electron cloud blanket sitting above and below the pin. It holds the atoms together, but it is much weaker and more exposed.

Hybridisation and Shape (13.3)

In an alkene (like ethene, $\text{C}_2\text{H}_4$), each carbon atom in the double bond is $sp^2$ hybridised:

• Three $sp^2$ orbitals form three $\sigma$ bonds (two to hydrogen atoms, one to the other carbon).
• One unhybridized $p$ orbital forms the one $\pi$ bond.

This $sp^2$ hybridisation leads to a specific geometry:

• Shape: Trigonal planar around each carbon atom in the double bond.
• Bond Angle: Approximately $\text{120}^\circ$ (ideal for trigonal planar geometry).
• Planarity: The double bond and the four atoms attached to it all lie in the same plane (they are planar).

Key Takeaway: The $\pi$ bond is the Achilles' heel of the alkene. It is weaker than the $\sigma$ bond and contains a region of high electron density, making it extremely attractive to electron-seeking species (electrophiles).

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2. Nomenclature and Stereoisomerism (13.4)

2.1 Naming Alkenes

Alkenes follow the same rules as general organic naming (systematic nomenclature, up to six carbons):

• The parent chain must contain the $\text{C}=\text{C}$ bond.
• The ending changes from -ane (for alkane) to -ene.
• You must number the chain so the double bond gets the lowest possible number.
• Example: $\text{CH}_3\text{CH}=\text{CHCH}_3$ is But-2-ene.

2.2 Geometrical (Cis/Trans) Isomerism

Geometrical isomerism (also known as cis-trans isomerism, or E/Z isomerism, although E/Z nomenclature is acceptable but not required) is a type of stereoisomerism. It occurs because the $\pi$ bond prevents rotation around the double bond.

Origin: Restricted Rotation (13.4.3)

The presence of the rigid $\pi$ bond prevents free rotation of the groups attached to the double-bonded carbons. If you try to rotate one side, you would have to break the $\pi$ bond, which requires significant energy. This locks the molecule into a fixed structure.

Conditions for Geometrical Isomerism:

For cis/trans isomerism to exist, each carbon atom in the $\text{C}=\text{C}$ bond must be attached to two different groups. If either carbon atom has two identical groups, geometrical isomerism is impossible.

Identifying Isomers:

• Cis Isomer: The two identical (or priority) groups are on the same side of the double bond.
• Trans Isomer: The two identical (or priority) groups are on opposite sides of the double bond.

Example: But-2-ene ($\text{CH}_3\text{CH}=\text{CHCH}_3$)

If the two $\text{CH}_3$ groups are on the same side $\rightarrow$ Cis-but-2-ene
If the two $\text{CH}_3$ groups are on opposite sides $\rightarrow$ Trans-but-2-ene

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3. Preparation of Alkenes (14.2.1)

Alkenes are typically made via elimination reactions, where a small molecule is removed from an adjacent carbon structure.

3.1 Elimination of Hydrogen Halide ($\text{HX}$) from a Halogenoalkane

This is often called dehydrohalogenation (removal of $\text{H}$ and $\text{X}$).

• Starting Material: Halogenoalkane ($\text{R}-\text{X}$).
• Reagent: Hot, concentrated ethanolic sodium hydroxide ($\text{NaOH}$ in ethanol, heat).
• Function: The $\text{OH}^-$ ion acts as a strong base, removing a hydrogen atom from a carbon adjacent to the $\text{C}-\text{X}$ bond, forming the $\text{C}=\text{C}$ bond.
• Equation Example: $\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} (\text{ethanolic, heat}) \rightarrow \text{CH}_2=\text{CH}_2 + \text{NaBr} + \text{H}_2\text{O}$

3.2 Dehydration of Alcohols

This is the removal of a water molecule ($\text{H}_2\text{O}$).

• Starting Material: Alcohol ($\text{R}-\text{OH}$).
• Reagent/Conditions: Heated catalyst, such as concentrated sulfuric acid ($\text{conc. H}_2\text{SO}_4$) or aluminium oxide ($\text{Al}_2\text{O}_3$).
• Equation Example (Ethanol): $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4, \text{heat}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$

3.3 Cracking of Longer Chain Alkanes (14.2.1c)

Cracking breaks large, less useful alkanes (from crude oil) into smaller, more valuable molecules, often including alkenes.

• Conditions: High heat, often with an $\text{Al}_2\text{O}_3$ catalyst.
• Product: Produces a mixture of smaller alkanes and alkenes.

Key Takeaway: Alkenes are typically made by removing two groups from adjacent carbons, often needing harsh conditions (heat, concentrated acid, or ethanolic alkali).

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4. Characteristic Reactions: Electrophilic Addition

Since the $\pi$ bond is an area of high electron density, alkenes are susceptible to attack by electrophiles (species that are attracted to electrons).

4.1 What is an Electrophile? (13.2.1e)

A $\text{Nucleophile}$ is an electron-rich species (nucleus-loving).
A $\text{Electrophile}$ is an electron-deficient species (electron-loving).
Electrophiles are electron pair acceptors. They are either positively charged ions ($\text{H}^+$, $\text{Br}^+$) or highly polarised molecules ($\text{Br}_2$).

4.2 The Mechanism of Electrophilic Addition (14.2.4)

The mechanism is a two-step process:

1. Step 1: Attack and Carbocation Formation
   The electron-rich $\pi$ bond attacks the electrophile ($\text{E}^+)$. The $\pi$ bond breaks (heterolytic fission), forming a bond between one carbon atom and the electrophile (E), leaving the other carbon atom electron-deficient. This electron-deficient carbon atom forms a transient intermediate called a carbocation (or carbonium ion).
   Example: Addition of $\text{HBr}$ to Propene. The $\pi$ bond breaks and forms a bond with the $\text{H}$ atom of $\text{HBr}$. The pair of electrons in the $\text{H}-\text{Br}$ bond moves to the $\text{Br}$ atom, forming a $\text{Br}^-$ ion.
2. Step 2: Nucleophilic Attack
   The positively charged carbocation is instantly attacked by the nucleophile ($\text{Nu}^-$ or $\text{Br}^-$ in the example above), forming the final product.

Encouragement: Don't worry if drawing the curly arrows seems tricky at first! Remember the fundamental rule: **Curly arrows always start at a bond or a lone pair of electrons and point to an atom or a positive centre.**

4.3 Explaining Markovnikov Addition (14.2.5)

When you add an unsymmetrical reagent ($\text{HX}$, like $\text{HBr}$) to an unsymmetrical alkene (like propene, $\text{CH}_3\text{CH}=\text{CH}_2$), two possible products can form. Markovnikov's Rule helps predict the major product.

Markovnikov's Rule (Simplified):
When a hydrogen halide ($\text{HX}$) adds to an unsymmetrical alkene, the hydrogen atom adds to the carbon atom of the double bond that already has the most hydrogen atoms attached.

Example: $\text{HBr}$ addition to Propene

• $\text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow$ Major Product: 2-bromopropane ($\text{CH}_3\text{CH}(\text{Br})\text{CH}_3$)
• The $\text{H}$ goes to the end carbon ($\text{CH}_2$) because it had two $\text{H}$s (more than the middle carbon's one $\text{H}$).

The Chemical Explanation: Carbocation Stability (14.2.5)

Markovnikov's rule is explained by the stability of the intermediate carbocation formed in Step 1.

• Tertiary Carbocation ($\text{3}^\circ$): Carbon attached to three alkyl groups. Most Stable.
• Secondary Carbocation ($\text{2}^\circ$): Carbon attached to two alkyl groups. Intermediate Stability.
• Primary Carbocation ($\text{1}^\circ$): Carbon attached to one alkyl group. Least Stable.

Reason for Stability (Inductive Effect): Alkyl groups ($\text{R}$) are slightly electron-donating (they have a positive inductive effect). When a carbocation forms, these groups push electron density towards the positively charged carbon, dispersing the positive charge and making the ion more stable. The more alkyl groups present, the more stable the carbocation.

Since the formation of the more stable carbocation has a lower activation energy, it forms faster. Therefore, the pathway through the most stable carbocation intermediate determines the major product.

Key Takeaway: Stability order: $3^\circ > 2^\circ > 1^\circ$. Alkenes prefer to form the most stable carbocation possible.

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5. Specific Addition Reactions of Alkenes (14.2.2a)

5.1 Addition of Hydrogen ($\text{H}_2$) - Hydrogenation

• Product: Alkane (Saturated hydrocarbon).
• Reagents/Conditions: $\text{H}_2$ gas, Platinum ($\text{Pt}$) or Nickel ($\text{Ni}$) catalyst, and heat.
• Mechanism Type: Addition.
• Real-world Use: Used industrially to convert unsaturated vegetable oils (liquids) into saturated fats (solids, like margarine).

5.2 Addition of Halogens ($\text{X}_2$) - Halogenation

• Product: Dihaloalkane.
• Reagents/Conditions: Halogen ($\text{Br}_2$ or $\text{Cl}_2$), usually at room temperature. No catalyst needed.
• Mechanism Type: Electrophilic addition (The non-polar $\text{Br}_2$ molecule becomes temporarily polarised near the electron-rich $\pi$ bond, creating an electrophilic end).

The Test for Unsaturation (14.2.3)

This is a vital identification reaction:

• Reagent: Aqueous Bromine ($\text{Br}_2 (\text{aq})$), usually in an organic solvent like an inert hydrocarbon or water.
• Observation: The orange/brown colour of the bromine water quickly disappears (is decolourised) as the bromine adds across the $\text{C}=\text{C}$ bond.
• Result: This rapid decolourisation shows the presence of the $\text{C}=\text{C}$ bond (unsaturation). Alkanes require $\text{UV}$ light to react with bromine via free radical substitution, so they do not decolourise bromine water in the dark.

5.3 Addition of Steam ($\text{H}_2\text{O(g)}$) - Hydration

• Product: Alcohol.
• Reagents/Conditions: Steam ($\text{H}_2\text{O}(\text{g})$) and a phosphoric acid ($\text{H}_3\text{PO}_4$) catalyst, high temperature and pressure.
• Real-world Use: Industrial manufacture of ethanol from ethene.

Did you know? Since this uses an unsymmetrical reagent ($\text{H}-\text{OH}$) on an alkene, it also follows Markovnikov's rule. The $\text{H}$ adds to the carbon with more $\text{H}$s, and the $\text{OH}$ group adds to the other carbon.

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6. Oxidation Reactions and Identification

Oxidation of alkenes involves adding oxygen atoms to the molecule, breaking the $\pi$ bond. We study two main types using acidified potassium manganate(VII) ($\text{KMnO}_4$), which is a strong oxidising agent.

6.1 Mild Oxidation to form a Diol (14.2.2b)

• Reagents: Cold, dilute, acidified $\text{KMnO}_4$.
• Product: A diol (a molecule with two $\text{OH}$ groups).
• Mechanism: The double bond breaks and an $\text{OH}$ group is added to each carbon atom.
• Observation/Test: The purple colour of the $\text{KMnO}_4$ solution disappears, often leaving a brown precipitate of $\text{MnO}_2$. This reaction is another confirmatory test for unsaturation.

$$ \text{RCH}=\text{CHR}' + [\text{O}] + \text{H}_2\text{O} \rightarrow \text{RCH}(\text{OH})\text{CH}(\text{OH})\text{R}' $$

6.2 Strong Oxidation (Cleavage) for Linkage Determination (14.2.2c)

When the conditions are hot and concentrated, the $\text{C}=\text{C}$ bond completely ruptures (breaks apart) into two separate molecules. This is used to determine the exact position of the double bond in a larger molecule.

• Reagents: Hot, concentrated, acidified $\text{KMnO}_4$.
• Result: The products depend on how many alkyl groups are attached to the $\text{C}=\text{C}$ bond carbons:
• If a $\text{C}$ atom has $\text{H}$ atoms: it is oxidised further.
  • $\text{C}$ with two $\text{H}$s ($\text{CH}_2$) $\rightarrow$ becomes $\text{CO}_2$ and $\text{H}_2\text{O}$.
  • $\text{C}$ with one $\text{H}$ ($\text{RCH}$) $\rightarrow$ becomes a carboxylic acid ($\text{RCOOH}$).
• If a $\text{C}$ atom has NO $\text{H}$ atoms ($\text{R}_2\text{C}$) $\rightarrow$ becomes a ketone ($\text{R}_2\text{CO}$).

Key Takeaway: Mild oxidation makes diols; strong oxidation breaks the chain entirely, producing ketones, carboxylic acids, or $\text{CO}_2$.

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7. Addition Polymerisation (14.2.2d)

Alkenes are the building blocks (monomers) for a critically important class of materials: addition polymers (plastics).

7.1 The Process

Addition Polymerisation is the process where many small unsaturated monomer molecules join together to form one very large saturated polymer molecule. The only functional group involved in this process is the $\text{C}=\text{C}$ bond.

• The double bond of the monomer opens up.
• The electrons from the broken $\pi$ bond are used to form new $\sigma$ bonds connecting the monomers in a long chain.

Example 1: Ethene Polymerisation
Monomer: Ethene ($\text{CH}_2=\text{CH}_2$) $\rightarrow$ Polymer: Poly(ethene) (often called polythene, used for plastic bags and bottles).

Example 2: Propene Polymerisation
Monomer: Propene ($\text{CH}_3\text{CH}=\text{CH}_2$) $\rightarrow$ Polymer: Poly(propene) (used for ropes and plastic chairs).

7.2 Identifying Monomers and Repeat Units (20.1)

• The repeat unit is the smallest section of the polymer that repeats, and it always contains two carbon atoms from the original double bond.
• To find the monomer from a polymer chain, find the repeat unit and then insert a double bond between the two carbon atoms in the chain.

7.3 Environmental Consequences (20.1.4)

Most poly(alkenes) (plastics) are large, non-polar, saturated molecules (once polymerised). This structure makes them chemically inert and non-biodegradable.

• Disposal Issue 1: Non-biodegradability: They do not rot naturally, leading to massive landfill problems.
• Disposal Issue 2: Harmful Combustion Products: When burnt, they can release harmful combustion products. For example, poly(chloroethene) ($\text{PVC}$) releases toxic $\text{HCl}$ gas when incinerated.

Key Takeaway: Alkenes readily undergo addition polymerisation to form long chains, which are environmentally problematic due to their stability.