Organic Reactions: Your Guide to Making and Breaking Molecules!
Hey everyone! Welcome to one of the most fascinating topics in Chemistry: organic reactions. Think of yourself as a molecular architect or a chef. You'll learn how to take simple carbon compounds and transform them into new, exciting substances by swapping atoms, adding new ones, reducing, oxidising, or even breaking molecules apart. This is the heart of how we make medicines, plastics, and even the flavourings in your favourite snacks!
Don't worry if it seems like a lot at first. We'll break down each type of reaction step-by-step with simple explanations and real-world examples. Let's get started!
Substitution Reactions: The Great Swap!
Imagine you're at a dance, and two partners decide to swap. That's exactly what a substitution reaction is! One atom or group of atoms on a molecule is replaced by another atom or group.
1. Substitution in Alkanes (Free-Radical Substitution)
Alkanes are known as saturated hydrocarbons because all their carbon atoms are bonded with the maximum number of hydrogen atoms—they're "full"! This means you can't just add more atoms to them. Instead, you have to swap one out.
• What you need: An alkane (like methane, CH₄) and a halogen (like chlorine, Cl₂).
• The special condition: This reaction is picky! It won't happen in the dark. It needs ultraviolet (UV) light (or sunlight/high temperature) to kick things off.
• What happens: A chlorine atom swaps with a hydrogen atom on the methane molecule.
The Equation:
Methane + Chlorine → Chloromethane + Hydrogen Chloride
\(CH_4(g) + Cl_2(g) \xrightarrow{UV\ light} CH_3Cl(g) + HCl(g)\)Did you know? This reaction can continue, swapping more hydrogens for chlorines to form dichloromethane (CH₂Cl₂), trichloromethane (CHCl₃), and eventually tetrachloromethane (CCl₄).
2. Substitution in Haloalkanes
A haloalkane is an alkane with a halogen atom (like Cl, Br, I) attached. We can swap that halogen for a hydroxyl group (-OH) to make an alcohol.
• What you need: A haloalkane (like chloroethane, CH₃CH₂Cl) and an aqueous alkali (like sodium hydroxide solution, NaOH(aq)).
• The special condition: Heat under reflux.
• What happens: The hydroxide ion (OH⁻) swaps with the halogen atom.
The Equation:
Chloroethane + Sodium Hydroxide → Ethanol + Sodium Chloride
\(CH_3CH_2Cl(l) + NaOH(aq) \xrightarrow{reflux} CH_3CH_2OH(aq) + NaCl(aq)\)Key Takeaway: Substitution
• In Alkanes: Swap H for a Halogen. Condition: UV light.
• In Haloalkanes: Swap Halogen for an -OH group. Condition: Aqueous alkali (e.g. NaOH(aq)) and heat under reflux.
Addition Reactions: Opening the Gates!
Unlike alkanes, alkenes have a carbon-carbon double bond (C=C). Alkenes are unsaturated because they can hold more atoms.
In an addition reaction, the double bond breaks open, and atoms are added across the two carbon atoms.
1. Adding Hydrogen (Hydrogenation)
• What you need: An alkene (like ethene, CH₂=CH₂) and hydrogen gas (H₂).
• Condition: A nickel (Ni) catalyst and heat (around 150°C), or platinum (Pt) catalyst at room temperature.
\(CH_2=CH_2(g) + H_2(g) \xrightarrow{Ni,\ heat} CH_3CH_3(g)\)2. Adding a Halogen (Halogenation)
• What you need: An alkene and bromine (Br₂) dissolved in an organic solvent (or aqueous bromine water).
• Condition: Room temperature, no light required.
\(CH_2=CH_2(g) + Br_2(\text{in organic solvent}) \rightarrow CH_2BrCH_2Br(l)\)The Observation: When an alkene is mixed with bromine, the reddish-brown/orange colour quickly decolourises. Alkanes do not react under room lighting without UV light, making this the classic test for unsaturation.
3. Adding Hydrogen Halides and Markovnikov's Rule
When adding an unsymmetrical reagent (like HBr or H₂O) to an asymmetric alkene (like propene, CH₃CH=CH₂), two different products can form.
• Markovnikov's Rule: The hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms ("the rich get richer").
• Example: Reaction of propene with HBr:
\(CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3\text{ (major product: 2-bromopropane)}\) \(CH_3CH=CH_2 + HBr \rightarrow CH_3CH_2CH_2Br\text{ (minor product: 1-bromopropane)}\)4. Reaction with Acidified Potassium Permanganate (KMnO₄/H⁺)
Cold, dilute acidified KMnO₄ oxidises alkenes to diols, decolourising the solution from purple to colourless. This is another key distinguishing test for C=C double bonds!
Key Takeaway: Addition
• What reacts? Alkenes (C=C double bonds).
• What happens? The double bond breaks open and new atoms add across it.
• Asymmetric alkenes: Use Markovnikov's rule to identify the major product.
• Distinguishing tests: Decolourises Br₂ (reddish-brown to colourless) and cold acidified KMnO₄ (purple to colourless).
Oxidation and Reduction Reactions
1. Oxidation of Alcohols
Common oxidising agents in HKDSE: acidified potassium dichromate(VI) (K₂Cr₂O₇/H⁺) and acidified potassium permanganate (KMnO₄/H⁺).
• Observation for K₂Cr₂O₇/H⁺: Orange (Cr₂O₇²⁻) turns Green (Cr³⁺).
• Observation for KMnO₄/H⁺: Purple (MnO₄⁻) turns Colourless (Mn²⁺).
• Primary (1°) Alcohols:
- Gentle oxidation (distil off product immediately): forms an aldehyde (e.g. ethanol → ethanal).
- Prolonged heating under reflux with excess oxidising agent: forms a carboxylic acid (e.g. ethanol → ethanoic acid).
\(CH_3CH_2OH + 2[O] \xrightarrow{K_2Cr_2O_7/H^+,\ reflux} CH_3COOH + H_2O\)• Secondary (2°) Alcohols:
- Oxidise to form a ketone (e.g. propan-2-ol → propanone).
\(CH_3CH(OH)CH_3 + [O] \xrightarrow{K_2Cr_2O_7/H^+,\ reflux} CH_3COCH_3 + H_2O\)• Tertiary (3°) Alcohols:
- Cannot be oxidised under these conditions because the carbon bonded to -OH has no hydrogen atoms attached.
2. Reduction Reactions
Reduction in organic chemistry corresponds to gaining hydrogen or losing oxygen.
• Reducing agents: Sodium borohydride (NaBH₄) in aqueous/alcoholic solution, or lithium aluminium hydride (LiAlH₄) in dry ether.
• Aldehydes: Reduced to primary alcohols (e.g. ethanal → ethanol).
• Ketones: Reduced to secondary alcohols (e.g. propanone → propan-2-ol).
• Carboxylic acids: Reduced to primary alcohols using the stronger reducing agent LiAlH₄ (e.g. ethanoic acid → ethanol).
Esterification, Amide Formation, and Hydrolysis
1. Esterification: Making Fruity Esters
An esterification reaction is when a carboxylic acid reacts with an alcohol in the presence of concentrated H₂SO₄ catalyst and heat.
\(CH_3COOH(l) + CH_3CH_2OH(l) \xrightleftharpoons{conc.\ H_2SO_4,\ heat} CH_3COOCH_2CH_3(l) + H_2O(l)\)Ethanoic Acid + Ethanol ⇌ Ethyl Ethanoate + Water
2. Hydrolysis of Esters
a) Acid Hydrolysis: Heating with dilute acid (e.g., dilute H₂SO₄). It is a reversible reaction forming the carboxylic acid and alcohol.
\(CH_3COOCH_2CH_3(l) + H_2O(l) \xrightleftharpoons{H^+,\ heat} CH_3COOH(aq) + CH_3CH_2OH(aq)\)b) Alkaline Hydrolysis (Saponification): Heating with dilute alkali (e.g., NaOH(aq)). It is irreversible (goes to completion) and yields a carboxylate salt and alcohol.
\(CH_3COOCH_2CH_3(l) + NaOH(aq) \xrightarrow{heat} CH_3COONa(aq) + CH_3CH_2OH(aq)\)3. Amide Formation and Hydrolysis
• Amide Formation: Heating a carboxylic acid or ester with ammonia (NH₃) or an amine forms an amide.
\(CH_3COOH + NH_3 \rightarrow CH_3CONH_2 + H_2O\)• Hydrolysis of Amides: Heating an amide under reflux with dilute acid gives a carboxylic acid and ammonium salt; heating with dilute alkali gives a carboxylate salt and releases ammonia / amine gas.
The Big Picture: Inter-conversions & Synthesis
All these reaction types link together to form pathways for converting one carbon compound into another!
Synthesis Route Challenge: Prepare ethyl ethanoate from ethene.
Plan:
1. Hydration of alkene: React ethene with steam to form ethanol.
\(CH_2=CH_2(g) + H_2O(g) \xrightarrow{H_3PO_4,\ 300^\circ C,\ 60\text{ atm}} CH_3CH_2OH(g)\)2. Oxidation of alcohol: Oxidise part of the ethanol to ethanoic acid by heating under reflux with acidified K₂Cr₂O₇.
\(CH_3CH_2OH + 2[O] \xrightarrow{K_2Cr_2O_7/H^+,\ reflux} CH_3COOH + H_2O\)3. Esterification: React the ethanol and ethanoic acid together in the presence of concentrated H₂SO₄.
\(CH_3COOH(l) + CH_3CH_2OH(l) \xrightleftharpoons{conc.\ H_2SO_4,\ heat} CH_3COOCH_2CH_3(l) + H_2O(l)\)