Hydrocarbons: Alkanes, Alkenes and Arenes

Introduction

Assalam-o-Alaikum, students. I am Dr. Fatima Malik. As you prepare for your A Level examinations, understanding hydrocarbons is absolutely fundamental. These compounds, consisting solely of carbon and hydrogen, form the backbone of organic chemistry and are the lifeblood of Pakistan's economy, from the natural gas extracted at Sui to the fuels sold by PSO and the raw materials used by companies like Engro for producing polymers and fertilisers.

This chapter explores three key families: Alkanes, the saturated and relatively unreactive workhorses; Alkenes, with their reactive C=C double bond ripe for addition; and Arenes, the special case of benzene and its derivatives, which possess unique aromatic stability. Mastering their structures and characteristic reactions is not just about passing an exam; it is about understanding the chemistry that shapes our modern world. We will focus on the *why* behind the reactions by dissecting their mechanisms step-by-step.

Core Theory

Alkanes: Free-Radical Substitution

Alkanes are saturated hydrocarbons (C-C single bonds) and are generally unreactive. Their main reaction is combustion, but they also undergo free-radical substitution in the presence of UV light. Let's examine the chlorination of methane (CH4).

The mechanism occurs in three stages:

1. Initiation: UV light provides the energy for homolytic fission of a chlorine molecule, creating two highly reactive chlorine free radicals.

Cl-Cl --(UV light)--> 2Cl•

1. Propagation: These radicals attack methane molecules in a chain reaction. This stage consists of two steps that repeat.

Step 1: Cl• + CH4 -> HCl + •CH3 (methyl radical)

Step 2: •CH3 + Cl2 -> CH3Cl + Cl•

The chlorine radical is regenerated in Step 2, allowing the chain to continue.

1. Termination: The reaction stops when two free radicals collide and combine. There are three possible termination steps:

Cl• + Cl• -> Cl2

•CH3 + •CH3 -> C2H6 (ethane)

Cl• + •CH3 -> CH3Cl (chloromethane)

A key limitation is the lack of control. The reaction produces a mixture of products (CH3Cl, CH2Cl2, CHCl3, CCl4) because the product, chloromethane, can itself be attacked by a chlorine radical.

Alkenes: Electrophilic Addition

Alkenes are unsaturated due to the C=C double bond. This bond is an area of high electron density (a π-bond), making it susceptible to attack by electrophiles (electron-pair acceptors).

Mechanism: Addition of HBr to Propene

The double bond attacks the partially positive hydrogen in H-Br.

CH3-CH=CH2 + H-Br -> [Intermediate Carbocation] + Br-

The H can add to C1 or C2.

* Path A (H adds to C1): Forms a secondary carbocation (CH3-C+H-CH3). This is more stable.

* Path B (H adds to C2): Forms a primary carbocation (CH3-CH2-C+H2). This is less stable.

The stability of carbocations is: tertiary > secondary > primary. This is due to the positive inductive effect of alkyl groups, which helps to spread out and stabilise the positive charge.

According to Markovnikov's Rule, the hydrogen atom adds to the carbon atom of the double bond that already has more hydrogen atoms. Therefore, Path A is favoured.

The bromide ion (Br-) then acts as a nucleophile and attacks the stable secondary carbocation.

CH3-C+H-CH3 + Br- -> CH3-CH(Br)-CH3 (2-bromopropane - major product)

Other Electrophilic Addition Reactions:

* With Br2(aq): The orange-brown colour of bromine water is decolourised. This is the standard test for unsaturation.

* With H2O(g)/H3PO4 catalyst: Industrial production of ethanol from ethene.

* With cold, dilute, alkaline KMnO4: The purple solution turns colourless (or a brown precipitate of MnO2 forms). A diol is formed. This is Baeyer's test.

CH2=CH2 + [O] + H2O -> HO-CH2-CH2-OH (ethane-1,2-diol)

Arenes: Benzene and Electrophilic Substitution

Benzene (C6H6) has a unique structure. Evidence (equal C-C bond lengths, lower-than-expected enthalpy of hydrogenation) disproved the simple alternating double bond Kekulé structure. The modern model describes a planar ring of 6 carbon atoms, each bonded to one hydrogen. The remaining p-orbital on each carbon overlaps sideways to form a delocalised π-system of electrons above and below the plane of the ring.

This delocalisation gives benzene exceptional stability. It does not undergo electrophilic addition because this would destroy the stable delocalised ring. Instead, it undergoes electrophilic substitution.

General Mechanism:

1. An electrophile (E+) is generated.
2. The π-system of benzene attacks the electrophile, forming an unstable intermediate carbocation (Wheland intermediate) where the delocalisation is temporarily broken.
3. A C-H bond breaks, and the pair of electrons returns to the π-system, restoring aromatic stability. H+ is released.

(ASCII Diagram of Benzene Electrophilic Substitution)

H

|

/= C /= | | C | | + E+ --->

=/ C =/

|

H

H E

| /

/= C+--

| | C

=/ C ... (delocalised positive charge)

E

|

/= C /= | | C | | + H+

=/ C =/

Key Substitution Reactions:

* Nitration: Reagents: conc. HNO3 and conc. H2SO4 (catalyst). Conditions: 55°C. Electrophile: NO2+ (nitronium ion). Product: Nitrobenzene.

* Halogenation: Reagents: Halogen (e.g., Br2) and a halogen carrier catalyst (e.g., AlCl3 or FeBr3). Conditions: RTP. Electrophile: Br+.

* Friedel-Crafts Alkylation: Reagents: Haloalkane (e.g., CH3Cl) and AlCl3 catalyst. Electrophile: CH3+.

Reactivity Comparison: Benzene vs. Cyclohexene

* Cyclohexene has a localised π-bond. It readily decolourises bromine water at RTP via electrophilic addition.

* Benzene has a delocalised π-system. The electron density is lower than in cyclohexene's double bond. It does not react with bromine water. It requires a halogen carrier catalyst to polarise the Br2 molecule enough to create a strong Br+ electrophile for substitution.

Key Definitions

* Hydrocarbon: An organic compound consisting entirely of hydrogen and carbon atoms.

* Saturated: A hydrocarbon containing only single C-C bonds (e.g., alkanes).

* Unsaturated: A hydrocarbon containing one or more C=C or C≡C multiple bonds (e.g., alkenes, alkynes).

* Free Radical: A highly reactive species with an unpaired electron, formed by homolytic fission.

* Electrophile: An electron-deficient species (e.g., H+, NO2+, Br+) that is attracted to areas of high electron density and accepts a pair of electrons.

* Carbocation: An ion with a positively charged carbon atom. Stability order: tertiary > secondary > primary.

* Markovnikov's Rule: In the addition of a protic acid (HX) to an unsymmetrical alkene, the hydrogen atom attaches to the carbon with the greater number of hydrogen substituents.

* Delocalisation: The spreading of π-electrons over several atoms in a molecule, rather than being localised between two atoms. This increases stability.

* Electrophilic Substitution: A reaction in which an electrophile attacks an aromatic ring and substitutes one of the hydrogen atoms.

* Electrophilic Addition: A reaction in which an electrophile attacks a C=C double bond, which breaks to form a new single bond, and a second species adds to the other carbon.

Worked Examples (Pakistani Context)

Example 1: Combustion Calculation at a CNG Station

A typical CNG cylinder for a car in Karachi contains natural gas which is approximately 95% methane (CH4) by volume. If a cylinder holds 12 kg of CNG, calculate the volume of carbon dioxide gas produced, in m³, when this fuel is completely combusted at standard temperature and pressure (STP: 273 K, 101 kPa). (Molar mass of CH4 = 16.0 g mol⁻¹; Molar volume of a gas at STP = 22.4 dm³ mol⁻¹).

Solution:

1. Calculate mass of methane:

Mass of CH4 = 12 kg * 0.95 = 11.4 kg = 11400 g

1. Calculate moles of methane:

Moles of CH4 = Mass / Molar Mass = 11400 g / 16.0 g mol⁻¹ = 712.5 mol

1. Write the balanced combustion equation:

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)

1. Use stoichiometry to find moles of CO2:

From the equation, the mole ratio of CH4 : CO2 is 1:1.

Therefore, moles of CO2 produced = 712.5 mol

1. Calculate the volume of CO2 at STP:

Volume of CO2 = Moles * Molar Volume

Volume = 712.5 mol * 22.4 dm³ mol⁻¹ = 15960 dm³

1. Convert dm³ to m³:

Since 1 m³ = 1000 dm³,

Volume in m³ = 15960 / 1000 = 15.96 m³

Answer: 15.96 m³

Example 2: Industrial Synthesis at Engro Polymer & Chemicals

Engro Polymer & Chemicals Ltd. in Port Qasim, Karachi, is a major producer of PVC (poly(chloroethene)). A potential starting material is ethene, sourced from the cracking of naphtha from a local refinery. Outline the two-step synthesis of chloroethene from ethene, showing the mechanism for the first step.

Solution:

Overall Synthesis:

Step 1: Ethene -> 1,2-dichloroethane

Step 2: 1,2-dichloroethane -> Chloroethene

Step 1: Electrophilic Addition of Chlorine to Ethene

* Reaction: CH2=CH2 + Cl2 -> Cl-CH2-CH2-Cl

* Conditions: Room temperature and pressure.

* Mechanism:

1. As a Cl2 molecule approaches the electron-rich C=C bond of ethene, the π-electrons repel the electrons in the Cl-Cl bond, inducing a dipole: Clδ+—Clδ-.
2. The π-bond acts as a nucleophile, attacking the Clδ+ atom. The Cl-Cl bond breaks heterolytically. This forms a carbocation intermediate and a chloride ion.

(ASCII Diagram)

H2C=CH2 --> +CH2-CH2-Cl + Cl-

^

|

Clδ+-Clδ-

1. The highly reactive chloride ion (Cl-) then quickly attacks the positive carbon atom of the carbocation, forming the final product, 1,2-dichloroethane.

+CH2-CH2-Cl + Cl- --> Cl-CH2-CH2-Cl

Step 2: Elimination Reaction

* Reaction: Cl-CH2-CH2-Cl -> CH2=CHCl + HCl

* Conditions: High temperature (thermal cracking, ~500°C). This is an elimination reaction, which you will study in more detail later.

Exam Technique

* Mechanisms are Crucial: For Paper 2 and 4, you MUST be able to draw mechanisms perfectly. Use curly arrows originating from a bond or a lone pair of electrons and pointing to the atom/delta-positive charge being attacked. Show all intermediate charges (e.g., C+) and relevant lone pairs (e.g., on Br-).

* State Conditions: Marks are always awarded for stating correct reagents and conditions. For nitration, "conc. H2SO4 / conc. HNO3 at 55°C" is a full-mark answer. Just writing "H2SO4/HNO3" may lose you marks.

* Distinguish Reaction Types: A common mistake is confusing addition and substitution. Alkenes add, breaking the π-bond. Benzene substitutes, preserving the stable ring. Be explicit in your answers, e.g., "Benzene undergoes electrophilic substitution whereas cyclohexene undergoes electrophilic addition."

* Markovnikov's Rule: When asked to predict the major product of addition to an unsymmetrical alkene, always explain your choice in terms of the stability of the carbocation intermediate (e.g., "The major product is formed via the more stable secondary carbocation intermediate.").

* Paper 4 Synthesis: Expect multi-step synthesis questions. You might be asked to convert an alkane to an alcohol. This would involve: Alkane -> (Free-radical substitution) -> Haloalkane -> (Elimination) -> Alkene -> (Electrophilic addition of H2O/steam) -> Alcohol. Practice linking reactions from different chapters.