Lesson 10.1: Addition and Elimination Reactions

Definition of Addition Reactions: These reactions occur when atoms are added across a double or triple bond in an unsaturated hydrocarbon. This process involves the breaking of a multiple bond and its replacement by two substituents derived from the chemical reagent used.
Unsaturated vs. Saturated Hydrocarbons: * Unsaturated: Contain double or triple bonds between carbon atoms. * Saturated: Contain only single carbon-carbon bonds (referencing Lesson 9.1).
Classification: Addition reactions include hydrogenation, halogenation, hydrohalogenation, hydration, and addition polymerisation.

Process: The addition of hydrogen ( $H_2$ ) across a carbon-carbon double or triple bond, adding one hydrogen atom to each carbon in the multiple bond. This is also classified as a reduction reaction (referencing Lesson 10.3).
Required Conditions: * Temperature: High temperatures between $150-300\,^\circ\text{C}$ . * Catalyst: Requires a metal catalyst such as finely divided platinum ( $Pt$ ), palladium ( $Pd$ ), or nickel ( $Ni$ ).
Chemical Equation Formatting: When writing equations for hydrogenation, the specific catalyst must be written above the arrow, and the word "heat" must be written below the arrow.
Example (Ethene to Ethane): * ${Ethene} + {H_2} \xrightarrow[Heat]{Pt} {Ethane}$ * $CH_2=CH_2 + H_2 \xrightarrow[Heat]{Pt} CH_3-CH_3$

Halogenation Definition: The addition of a halogen such as Bromine ( $Br_2$ ) or Chlorine ( $Cl_2$ ) across a multiple bond.
Conditions: Unlike other addition reactions, halogenation does not require heat or specific catalysts.
Reactivity Tip: Halogens are highly electronegative and react to accept electrons, allowing them to react easily with the electrons in double and triple bonds without external energy.
Example (Chlorination of Ethene): * $CH_2=CH_2 + Cl_2 \rightarrow CH_2Cl-CH_2Cl$ ( $1,2\text{-dichloroethane}$ ).
Chemical Test for Saturation (Bromine Test): * Reagent: Bromine dissolved in water (bromine water) or as a $1\%$ solution in carbon tetrachloride ( $CCl_4$ ). * Observations: The bromine reagent is initially a bright orange color. * Unsaturated Results: The color disappears rapidly as the bromine adds across multiple bonds. * Saturated Results: If the orange color remains, the substance is likely an alkane (saturated) or an aromatic compound. Aromatic compounds undergo addition reactions less easily and would require heat and a catalyst to react.

Definition: The addition of a hydrohalide like hydrogen bromide ( $HBr$ ) or hydrogen chloride ( $HCl$ ) across a double bond. This reaction requires no heat or catalyst.
Vladimir Markovnikov: A Russian chemist who identified trends in the addition of asymmetrical reagents to asymmetrical hydrocarbons.
Markovnikov's Rule: In the addition of an asymmetrical reagent, the hydrogen atom will bond to the carbon atom that already has the most hydrogen atoms. * Analogy: "The rich get richer." * Mechanism: The hydrogen typically bonds to the carbon at the end of the chain, while the halogen bonds to the carbon with fewer hydrogens (typically the second carbon in the chain).
Product Distribution: * Usually, reactions produce approximately $90\%$ of the Markovnikov product (main product). * Approximately $10\%$ is the anti-Markovnikov product. * Exact percentages depend on specific reactants and conditions; separation and purification are required to isolate the main product.
Example (Propene + HCl): * $CH_3-CH=CH_2 + HCl \rightarrow CH_3-CHCl-CH_3$ ( $2\text{-chloropropane}$ , major product). * Minor product: $CH_3-CH_2-CH_2Cl$ ( $1\text{-chloropropane}$ ).

Definition: The addition of water ( $H_2O$ ) across a double bond to produce an alcohol. This involves adding one hydrogen atom and one hydroxyl ( $OH$ ) group.
Regioselectivity: Hydration of asymmetrical alkenes follows Markovnikov's rule; the hydroxyl group bonds to the carbon with fewer hydrogen atoms.
Conditions for Alcohol Production (Summary of Table 1): * Primary Alcohol (e.g., Ethanol from Ethene): Requires $300\,^\circ\text{C}$ , $6000-7000\,\text{kPa}$ pressure, and a trace amount of Phosphoric acid ( $H_3PO_4$ ) catalyst. * Secondary Alcohol (e.g., Propan-2-ol from Propene): Requires $100-250\,^\circ\text{C}$ , $500\,\text{kPa}$ pressure, and a Sulfuric acid ( $H_2SO_4$ ) or Phosphoric acid ( $H_3PO_4$ ) catalyst. * Tertiary Alcohol (e.g., 2-methylpropan-2-ol from 2-methylpropene): Requires $25-100\,^\circ\text{C}$ , $100\,\text{kPa}$ pressure, and a Sulfuric acid ( $H_2SO_4$ ) catalyst.
Example Equation: * $CH_2=CH_2 + H-OH \xrightarrow[300\,^\circ\text{C}, 6000\,\text{kPa}]{H_3PO_4} CH_3-CH_2-OH$

Mechanism: Successive addition reactions of alkene monomers. It is a "self-addition" reaction; the presence of other reagents stops the process.
Key Terms: * Monomer: A single unit within a polymer. * Polymer: A large molecule made of many monomers.
Conditions: Requires heat, high pressure, and a catalyst that generates unpaired non-bonding electrons to break double bonds.
Occurrence: * Synthetic: Plastics, rubbers, and fibres. * Natural: Carbohydrates, proteins, and nucleic acids.
Representation: Brackets and the subscript $n$ (representing number of repeating units) are used. The reaction should show $n$ moles of alkene forming a chain.
Examples: * $n(Ethene) \xrightarrow[High\,Pressure]{Heat} Polyethene$ * $n(Chloroethene) \xrightarrow[High\,Pressure]{Heat} Poly(chloroethene)$ * $n(Propene) \xrightarrow[High\,Pressure]{Heat} Polypropene$

Definition: The removal of substituents from two adjacent carbon atoms in a haloalkane to form a multiple bond. It is conceptually the opposite of an addition reaction.
Key Conditions: * Reagents: Concentrated solution of strong base ( $NaOH$ or $KOH$ ). * Solvent: Ethanol. * Reflux: A technique where the reaction mixture is kept boiling, and the vapors are condensed and collected back into the mixture.
Haloalkane Reactivity: * Tertiary Haloalkanes: Undergo elimination reactions almost exclusively. * Secondary Haloalkanes: Undergo a mixture of elimination and substitution reactions (referencing Lesson 10.6).
Factors Favoring Elimination: Using ethanol as a solvent, higher temperatures, and higher base concentrations.
Example (2-bromopropane): * $CH_3CH(Br)CH_3 + NaOH \xrightarrow[Ethanol]{Heat} CH_2=CHCH_3 + NaBr + H_2O$ ( $Propene$ is the product).