characteristic reactions of organic families including substitution (haloalkanes from alkanes and from alcohols), addition (alkenes), oxidation (alcohols to aldehydes/ketones/carboxylic acids), condensation (esterification) and hydrolysis (of esters and amides), and the design of multi-step reaction pathways linking functional-group families

What this dot point is asking

VCAA wants you to know the characteristic reactions of each main organic family (substitution, addition, oxidation, condensation/esterification, hydrolysis) including reagents and conditions, and to design multi-step pathways linking functional groups (alkane to alkene to alcohol to carboxylic acid to ester is the classic backbone).

The answer

The five reaction types you need

1. Substitution (R-X to R-Y). One atom or group on a saturated carbon is replaced by another. The skeleton stays the same.

• Alkane to haloalkane: CH4 + Cl2 -> CH3Cl + HCl (UV light required; radical mechanism, gives a mixture of products in practice).
• Alcohol to haloalkane: CH3CH2OH + HBr -> CH3CH2Br + H2O (reflux; concentrated acid). Also possible with PBr3 or SOCl2 (which give cleaner conversions).

2. Addition (across a multiple bond). Two atoms or groups add across C=C or C triple C, converting the molecule from unsaturated to (more) saturated.

• Hydrogenation of an alkene: CH2=CH2 + H2 -> CH3CH3 (Ni or Pt catalyst, ~150 deg C, high pressure).
• Halogenation: CH2=CH2 + Br2 -> CH2BrCH2Br (room temperature; the orange Br2 decolourises, used as a positive test for unsaturation).
• Hydrohalogenation: CH2=CH2 + HBr -> CH3CH2Br. Markovnikov's rule applies to unsymmetric alkenes.
• Hydration: CH2=CH2 + H2O -> CH3CH2OH (steam, H2SO4 catalyst, ~300 deg C; or dilute acid for lab scale).

3. Oxidation (mainly of alcohols). Acidified K2Cr2O7 / H2SO4 (or KMnO4 / H2SO4) under reflux.

• Primary alcohol to aldehyde (limited oxidant, distillation removes aldehyde quickly) to carboxylic acid (excess oxidant, reflux):
  CH3CH2OH -> CH3CHO -> CH3COOH
• Secondary alcohol to ketone (does not oxidise further):
  CH3CH(OH)CH3 -> CH3COCH3
• Tertiary alcohols do not oxidise under these conditions (no H on the OH carbon to remove).

Colour change for K2Cr2O7: orange to green (Cr(VI) -> Cr(III)). Colour change for KMnO4: purple to colourless. Both are classic alcohol tests.

4. Condensation (esterification). Two small molecules combine into a larger molecule with loss of a small molecule (usually water). The key example is esterification:

CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O

Conditions: heat under reflux with concentrated H2SO4 as acid catalyst. It is reversible, so use excess of one reagent or remove the water to maximise yield. Esters are sweet-smelling and used as flavourings and solvents.

Amide formation (a condensation between a carboxylic acid and an amine) follows the same pattern:
CH3COOH + CH3NH2 -> CH3CONHCH3 + H2O

5. Hydrolysis (the reverse of condensation). Adding water (often with acid or base catalyst) cleaves an ester or amide bond back to the carboxylic acid + alcohol (or carboxylic acid + amine).

CH3COOCH2CH3 + H2O -> CH3COOH + CH3CH2OH (acid catalysis: reflux with dilute H2SO4)

Base-catalysed hydrolysis (saponification, used for making soaps from fats):
CH3COOCH2CH3 + NaOH -> CH3COONa + CH3CH2OH

Saponification is irreversible because the salt does not equilibrate back.

Markovnikov's rule

For an unsymmetric alkene reacting with an unsymmetric reagent like HBr or H2O, the H goes to the carbon with more hydrogens (the less-substituted end), and the other group (Br, OH) goes to the carbon with fewer hydrogens (the more-substituted end). The major product is the one with the more-substituted carbocation intermediate.

Multi-step pathway design

The classic alkane-to-ester pathway is:

Alkane to (1) Haloalkane to (2) Alcohol to (3) Carboxylic acid to (4) Ester

with the reactions and reagents:

1. Substitution with Cl2 or Br2 in UV light.
2. Substitution with NaOH (aq) or dilute KOH (aq): R-Br + NaOH (aq) -> R-OH + NaBr.
3. Oxidation with acidified K2Cr2O7 / H2SO4 under reflux.
4. Esterification with an alcohol and concentrated H2SO4.

There is also a parallel path through alkenes:

Alkane to Haloalkane (substitution), Haloalkane to Alkene (elimination), Alkene to Alcohol (addition), Alcohol to Acid (oxidation), Acid to Ester (esterification)

When designing a pathway:

• Always state reagent, conditions (temperature, catalyst, solvent), and reaction type.
• Note whether the reaction is reversible (esterification, hydrolysis) and how yield can be improved (excess reagent, remove product, use Dean-Stark trap to remove water, etc.).
• Be explicit about the type of alcohol being oxidised (primary to acid; secondary to ketone; tertiary not oxidised).

Reagent-condition summary table

Examples in context

Example 1. Biodiesel transesterification at Macquarie Oil Geelong. Macquarie Oil's regional refinery converts canola oil and waste cooking oil to biodiesel by base-catalysed transesterification with methanol: $\text{triglyceride} + 3 \text{CH}_3 \text{OH} \to 3 \text{methyl ester} + \text{glycerol}$, with KOH catalyst at $60^{\circ}\text{C}$. This is a condensation-style reaction where each ester linkage of the triglyceride is exchanged from a $\text{C}_3$ glycerol back-bone to three separate methyl groups, producing fatty acid methyl esters (FAME) used as fuel. Per tonne of canola oil ($\sim 0.95 \, \text{kg/L}$ density), the reaction yields $\sim 1.0 \, \text{tonne}$ of biodiesel and $0.11 \, \text{tonne}$ of glycerol. Glycerol is sold as a by-product to the cosmetic and pharmaceutical industries. The FAME meets Australian standard AS 3570.

Example 2. Aspirin synthesis at WEHI clinical-trials chemistry. WEHI medicinal chemists prepare a $50 \, \text{g}$ batch of pharmaceutical-grade acetylsalicylic acid (aspirin) for a cardiovascular trial. The reaction is an esterification: salicylic acid (the alcohol-bearing reagent at the phenol -OH) reacts with acetic anhydride (acetyl donor) with concentrated $\text{H}_2 \text{SO}_4$ catalyst at $90^{\circ}\text{C}$ for $30 \, \text{minutes}$. Stoichiometry: $\text{C}_7 \text{H}_6 \text{O}_3 + (\text{CH}_3 \text{CO})_2 \text{O} \to \text{C}_9 \text{H}_8 \text{O}_4 + \text{CH}_3 \text{COOH}$. From $38.2 \, \text{g}$ of salicylic acid ($0.277$ mol) and excess anhydride, theoretical yield is $0.277 \times 180 = 49.8 \, \text{g}$ of aspirin. The chemists recrystallise from hot ethanol-water to obtain $43.7 \, \text{g}$ pure product, yield $87.8\%$. NMR and HPLC confirm purity above $99.5\%$ before release.

Try this

Q1. Write the reaction and state the type for: (a) ethene reacting with bromine; (b) ethanol with acidified potassium dichromate. [3 marks]

• Cue. (a) $\text{CH}_2 = \text{CH}_2 + \text{Br}_2 \to \text{CH}_2 \text{Br CH}_2 \text{Br}$; addition. (b) $\text{CH}_3 \text{CH}_2 \text{OH} \to \text{CH}_3 \text{CHO} \to \text{CH}_3 \text{COOH}$; oxidation.

Q2. Calculate the mass of ethyl ethanoate produced from $4.6 \, \text{g}$ of ethanol reacting with excess ethanoic acid, assuming $65\%$ yield. [4 marks]

• Cue. $n(\text{ethanol}) = 4.6 / 46 = 0.10$ mol; theoretical $n(\text{ester}) = 0.10$ mol; mass $= 0.10 \times 88 = 8.8 \, \text{g}$; actual $= 8.8 \times 0.65 = 5.72 \, \text{g}$.

Q3. Design a 3-step pathway to convert ethane to ethyl ethanoate. (a) State each step's reagents and conditions. (b) Write each balanced equation. (c) Identify the type of each reaction. [2+2+2 marks]

• Cue. (a) Step 1: $\text{Cl}_2 / \text{UV}$ to give chloroethane. Step 2: NaOH(aq) reflux to give ethanol. Step 3: ethanoic acid with conc $\text{H}_2 \text{SO}_4$. (b) Equations as above. (c) Substitution; substitution; condensation (esterification).