← Unit 4: How are organic compounds categorised, analysed and used?
How can the diversity of carbon compounds be explained and categorised?
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
A focused VCE Chemistry Unit 4 answer on organic reactions. Covers substitution of alkanes and alcohols, addition to alkenes, oxidation of primary and secondary alcohols, esterification by condensation, hydrolysis of esters and amides, and the construction of multi-step reaction pathways with reagents and conditions.
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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:
- Substitution with Cl2 or Br2 in UV light.
- Substitution with NaOH (aq) or dilute KOH (aq): R-Br + NaOH (aq) -> R-OH + NaBr.
- Oxidation with acidified K2Cr2O7 / H2SO4 under reflux.
- 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
| Conversion | Reagent | Conditions | Type |
|---|---|---|---|
| Alkane -> haloalkane | Cl2 or Br2 | UV light | Substitution |
| Haloalkane -> alcohol | NaOH(aq) | Reflux in water/ethanol | Substitution |
| Haloalkane -> alkene | NaOH or KOH in ethanol | Reflux in ethanol (concentrated alkali) | Elimination |
| Alkene -> alkane | H2 | Ni/Pt catalyst, heat, pressure | Addition |
| Alkene -> haloalkane | HBr or HCl | Room temperature | Addition (Markovnikov) |
| Alkene -> dihaloalkane | Br2 or Cl2 | Room temperature (CCl4) | Addition |
| Alkene -> alcohol | H2O + dilute H2SO4 | Reflux | Addition (Markovnikov) |
| Primary alcohol -> aldehyde | K2Cr2O7 / H2SO4 (limited) | Gentle heat, distil off aldehyde | Oxidation |
| Primary alcohol -> carboxylic acid | K2Cr2O7 / H2SO4 (excess) | Reflux | Oxidation |
| Secondary alcohol -> ketone | K2Cr2O7 / H2SO4 | Reflux | Oxidation |
| Acid + alcohol -> ester | H2SO4 catalyst | Reflux | Condensation (esterification) |
| Ester + H2O -> acid + alcohol | H2SO4 catalyst (acid) or NaOH (base) | Reflux | Hydrolysis |
| Acid + amine -> amide | Heat (often coupling reagent in practice) | Heat | Condensation |
| Amide + H2O -> acid + amine | H2SO4 or NaOH | Reflux | Hydrolysis |
Worked example
Design a pathway from propan-1-ol to propyl propanoate, an ester used in fruit flavours.
Step 1. Oxidise propan-1-ol to propanoic acid.
CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O
Reagents: K2Cr2O7 / H2SO4 (excess), reflux. Type: oxidation.
Step 2. Esterify propanoic acid with another portion of propan-1-ol.
CH3CH2COOH + CH3CH2CH2OH <=> CH3CH2COOCH2CH2CH3 + H2O
Reagents: concentrated H2SO4 catalyst, heat under reflux. Type: condensation (esterification). Reversible; use excess of one reagent or remove the water to maximise yield.
Total: two steps, both of which can be carried out in the school lab.
Common traps
Forgetting catalyst/conditions. A reagent without conditions is not a complete answer. Reflux is the default for liquid-phase reactions; UV light for the alkane chlorination; H2SO4 for esterification.
Oxidising a tertiary alcohol. Tertiary alcohols do not oxidise under K2Cr2O7 / H2SO4 (no H on the OH carbon to lose). A "no reaction" answer is correct.
Adding HBr to an alkene the wrong way. Markovnikov: H goes to the side with more Hs already (less substituted side); Br goes to the more substituted side.
Forgetting that esterification is reversible. State the equilibrium and how to drive it (excess alcohol or acid, removing water).
Confusing condensation and addition. Condensation produces a small molecule by-product (water). Addition does not (everything ends up in the product).
Using KOH in water for elimination. Elimination requires concentrated alkali in ethanol (and heat). KOH in water gives substitution (haloalkane to alcohol).
In one sentence
The five reactions of organic chemistry are substitution (haloalkanes from alkanes or alcohols), addition (across an alkene C=C, Markovnikov when applicable), oxidation (primary alcohol to aldehyde to acid; secondary to ketone), condensation (esterification: acid + alcohol with H2SO4 catalyst) and hydrolysis (esters and amides back to their parts), and a multi-step pathway is built by chaining these together with the right reagents and conditions at each step.
Past exam questions, worked
Real questions from past VCAA papers on this dot point, with our answer explainer.
2024 VCE5 marksStarting from ethene (C2H4), outline a reaction pathway to produce ethyl ethanoate. Write a balanced equation for each step, name the reagents and conditions, and identify the type of reaction at each step.Show worked answer →
A 5-mark answer needs three steps (ethene to ethanol, ethanol to ethanoic acid, esterification), with reagents and conditions for each.
Step 1: ethene to ethanol (addition / hydration).
CH2=CH2 + H2O -> CH3CH2OH
Conditions: dilute H2SO4 (catalyst), about 300 deg C, high pressure (steam process). Or acid-catalysed hydration of the alkene with dilute H2SO4 at room temperature for lab scale.
Step 2: ethanol to ethanoic acid (oxidation).
CH3CH2OH + 2[O] -> CH3COOH + H2O (overall, with [O] representing the oxidising agent providing two oxygen atoms in stages)
Conditions: heat under reflux with acidified potassium dichromate K2Cr2O7 / H2SO4 (or acidified potassium permanganate KMnO4 / H2SO4). Excess oxidant takes the alcohol all the way to the carboxylic acid.
Step 3: ethanoic acid + ethanol to ethyl ethanoate (esterification / condensation).
CH3COOH + CH3CH2OH <=> CH3COOCH2CH3 + H2O
Conditions: heat under reflux with concentrated H2SO4 (acid catalyst). Reversible reaction; use excess of one reagent or remove the water/ester to push the equilibrium to products.
Markers reward labelling the reaction type at each step (addition, oxidation, condensation), citing the catalyst clearly, and recognising that step 3 is an equilibrium.
2025 VCE3 marks(a) Distinguish between addition and substitution reactions. (b) Predict the major product when 2-methylbut-2-ene reacts with HBr and explain using Markovnikov's rule.Show worked answer →
A 3-mark answer needs the distinction, the product, and Markovnikov reasoning.
(a) Substitution: one atom or group in the molecule is replaced by another. The molecular formula changes by the difference between the two groups but the carbon skeleton remains. Typical for haloalkanes from alkanes and alcohols.
Addition: two atoms or groups add across a multiple bond (e.g. C=C). No atom leaves; the molecular formula gains both new atoms. Typical for alkenes, alkynes and carbonyls.
(b) 2-methylbut-2-ene: (CH3)2C=CH-CH3. Adding HBr across the C=C double bond. The two carbons of the double bond differ in number of attached H atoms (the left C has zero H, the right has one). Markovnikov's rule: the H of HBr adds to the carbon with the more Hs, and the Br adds to the carbon with the fewer Hs.
Major product: 2-bromo-2-methylbutane, (CH3)2CBr-CH2-CH3. The mechanism rationale (not required at VCE depth): the more-substituted carbocation intermediate is more stable, so it forms preferentially.
Related dot points
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