Loav · VCE Chemistry · Unit 4 / Area of Study 1
U4A1-03 · Current 2026 curriculum · Original teaching guide · Study guide
1. Read the change before choosing a reagent
Substitution replaces an atom or group. Addition joins species across a multiple bond. Condensation joins molecules with loss of a small molecule, commonly water. Hydrolysis uses water to split a linkage. Oxidation of an organic molecule often increases its bonding to oxygen or decreases its bonding to hydrogen. These labels identify different changes, so one memorised arrow cannot substitute for inspecting the structure.
Use complete condensed or structural formulas when connectivity matters. Balance all atoms, including halogens and the hydrogens in water. Place catalysts and conditions over an arrow or in accompanying text rather than treating them as net reactants. State symbols depend on the specified medium and temperature; do not invent them if the context does not establish them.
2. Substitution connects haloalkanes, alcohols and amines
An alkane can undergo substitution with chlorine or bromine under ultraviolet light. For example, CH₃CH₃ + Cl₂ → CH₃CH₂Cl + HCl. A hydrogen is replaced by chlorine. Further substitution can occur, so a simple equation does not imply an automatically pure single product. Longer chains may also give positional isomers.
A primary haloalkane can form a primary alcohol with aqueous hydroxide and heating: CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr. Aqueous conditions favour this substitution in the intended pathway. Conversely, a primary alcohol can be converted to a haloalkane using an appropriate hydrogen halide and heating: CH₃CH₂OH + HBr → CH₃CH₂Br + H₂O. The choice of reagent and conditions matters; do not label water alone as a reliable reagent for the reverse conversion.
To make a primary amine, a haloalkane reacts with excess ammonia, commonly in ethanol with heating. An overall equation is CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. One ammonia supplies nitrogen to the amine; another accepts a proton. Excess ammonia reduces further substitution of the amine, although the reaction can still produce mixtures. This is not a home procedure: the chemicals and pressure/temperature controls require appropriate laboratory risk assessment.
3. Addition reactions of alkenes
The C=C becomes C–C while a new bond forms at each of its two carbons. With hydrogen and a nickel catalyst under suitable heating, CH₂=CH₂ + H₂ → CH₃CH₃. With bromine, CH₂=CH₂ + Br₂ → CH₂BrCH₂Br, giving a dihaloalkane; light is not required as it is for alkane radical substitution. With HBr, CH₂=CH₂ + HBr → CH₃CH₂Br. Hydration with steam and an acid catalyst, such as phosphoric acid in an industrial process, gives CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH under suitable elevated temperature and pressure.
For an unsymmetrical alkene and an unsymmetrical reagent, more than one structural product can be conceivable and conditions influence selectivity. Do not claim all hydration reactions give primary alcohols: hydration of propene commonly favours propan-2-ol. Ethene avoids this positional ambiguity in introductory pathway calculations.
4. Oxidation provides a route to carboxylic acids
A primary alcohol can be oxidised first to an aldehyde and then to a carboxylic acid. Acidified dichromate or another suitable oxidising agent provides the oxidation conditions. Distilling an aldehyde as it forms can limit further oxidation; heating under reflux with excess oxidant favours the carboxylic acid. Reflux returns condensed vapour to the reaction mixture, allowing continued heating without deliberately collecting a volatile product.
CH₃CH₂OH + [O] → CH₃CHO + H₂O
CH₃CHO + [O] → CH₃COOH
Overall: CH₃CH₂OH + 2[O] → CH₃COOH + H₂O
[O] is a shorthand for oxygen supplied by an oxidising agent, not an instruction to use oxygen gas and not a complete redox equation for a named reagent. If a question asks for a fully balanced ionic redox equation, include the oxidant’s reduction half-equation. Secondary alcohols oxidise to ketones, whereas tertiary alcohols do not undergo the analogous simple oxidation under ordinary school-level conditions.
5. Esterification and two kinds of ester hydrolysis
Heating a carboxylic acid with an alcohol and an acid catalyst establishes an esterification equilibrium. For ethyl ethanoate:
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O
A suitable acid catalyst, commonly concentrated sulfuric acid in the teaching model, speeds the forward and reverse reactions. It does not change the equilibrium constant at a fixed temperature. Acid-catalysed hydrolysis with water reverses esterification, forming the carboxylic acid and alcohol. Excess water can favour hydrolysis. Alkaline hydrolysis has a different immediate product:
CH₃COOCH₂CH₃ + NaOH → CH₃COONa + CH₃CH₂OH
The product is a carboxylate salt, not neutral ethanoic acid. An additional acidification step is needed to produce CH₃COOH from CH₃COO⁻. Specify the medium rather than writing “hydrolysis” without a product distinction.
6. Biodiesel: replace an ester’s alcohol portion
A plant triglyceride is a triester of glycerol and fatty acids. Transesterification with methanol produces three fatty-acid methyl ester molecules and glycerol, commonly using a suitable base catalyst such as sodium hydroxide. With ethanol, the products are ethyl esters. Heating and a catalyst increase reaction rate; feedstock impurities and water can complicate conversion. The core stoichiometry is one triglyceride plus three alcohol molecules.
Triglyceride + 3CH₃OH ⇌ glycerol + R¹COOCH₃ + R²COOCH₃ + R³COOCH₃
The three R groups can differ. This reaction exchanges the ester-linked alcohol component; it is not hydrolysis and it does not produce water in its net equation. Biodiesel is the fatty-acid ester mixture, not the glycerol coproduct or the unmodified plant oil.
Worked example: build an ester from a haloalkane
Target ethyl ethanoate from bromoethane, with other reagents available. First use aqueous hydroxide and heat to form ethanol. Oxidise a portion of this ethanol under reflux with excess acidified oxidant to make ethanoic acid. React that acid with another portion of ethanol under acid-catalysed esterification conditions. Check both carbon fragments in CH₃COOCH₂CH₃: one comes from the acid and one from the alcohol. Simply oxidising all the ethanol would leave no ethanol for the final step unless more were supplied.
Practice
- Write the balanced overall equation for making propan-1-amine from 1-bromopropane and ammonia.
- Give the product of CH₃CH=CHCH₃ + H₂ with a suitable catalyst.
- Identify the immediate organic products when methyl propanoate undergoes alkaline hydrolysis.
- How many moles of ethanol are required stoichiometrically for 0.250 mol of triglyceride transesterification? Name the ester type.
Answers with reasoning
- CH₃CH₂CH₂Br + 2NH₃ → CH₃CH₂CH₂NH₂ + NH₄Br. Excess ammonia, an appropriate solvent and heating favour primary amine formation. Both sides have three carbons, thirteen hydrogens, two nitrogens and one bromine.
- Butane, CH₃CH₂CH₂CH₃. One H adds to each former double-bond carbon, consuming one H₂ per C=C; nickel is a suitable catalyst.
- Propanoate ions, commonly a sodium salt with NaOH, and methanol. CH₃CH₂COOCH₃ + NaOH → CH₃CH₂COONa + CH₃OH. Acidification would be a separate step.
- 0.750 mol ethanol from the 3:1 ratio. Fatty-acid ethyl esters form, along with 0.250 mol glycerol. This is the theoretical stoichiometric requirement, not a practical recipe or claim of complete conversion.
Go deeper: plan backwards, then verify forwards
Work backwards from the target functional group. For an ester, split the C(=O)–O connection conceptually into an acid-derived portion and an alcohol-derived portion. Ask which allowed transformations can supply those two molecules. Then run the proposed route forwards, checking every carbon atom and functional group after each arrow. A route is not complete if it names reagents but loses track of which product they make.
Worked example 2 · Develop: an oxidation that must stop early
You have butan-1-ol, CH₃CH₂CH₂CH₂OH, and want butanal, CH₃CH₂CH₂CHO, rather than butanoic acid. The first oxidation is CH₃CH₂CH₂CH₂OH + [O] → CH₃CH₂CH₂CHO + H₂O. Use suitable controlled oxidation conditions and remove the aldehyde by distillation as it forms to reduce its exposure to oxidant. Simply writing “excess acidified oxidant, reflux” would instead favour further oxidation to CH₃CH₂CH₂COOH.
The carbon count remains four. Check the atom balance: C₄H₁₀O + O → C₄H₈O + H₂O. Distillation does not change an alcohol into an aldehyde by itself; the oxidant makes the chemical change, while product removal supports selectivity. These are conceptual conditions for a supervised laboratory, not an executable preparation.
Worked example 3 · Apply: use symmetry to avoid false products
Hydration of but-2-ene can attach the OH component at either carbon of the original double bond. Sketch the two possibilities: CH₃CH(OH)CH₂CH₃ and CH₃CH₂CH(OH)CH₃. Reverse the second chain and they have identical connectivity. Thus there is one constitutional alcohol product, butan-2-ol, in this comparison. The balanced net equation is C₄H₈ + H₂O ⇌ C₄H₁₀O, under suitable acid-catalysed hydration conditions.
This exercise counts structural products, not stereoisomers. It does not mean every alkene gives one structural alcohol. For an unsymmetrical alkene, draw both placements and check whether they are actually different; then use supplied selectivity information rather than assuming equal yields or a convenient primary alcohol.
In pathway explanations, separate three ideas. A catalyst affects rate. Reflux or distillation controls how the reaction mixture is heated and whether volatile material returns or is collected. Reactant proportions and removal of a product may influence equilibrium composition or practical recovery. Saying “the catalyst makes more ester at equilibrium” collapses these different roles and is incorrect at fixed temperature.
Application practice: defend each arrow
- Write the addition product of CH₃CH=CHCH₂CH₃ with Br₂. Explain why putting both bromines on the first two chain carbons is wrong.
- A route proposes oxidising butan-2-ol directly to butanoic acid using ordinary alcohol-oxidation conditions. Identify the expected product and repair the starting-material choice for a route to the acid.
- Compare complete acidic hydrolysis and methanol transesterification of one triglyceride molecule. Give the small-molecule reagent counts and products for each.
Application answers and checks
- CH₃CHBrCHBrCH₂CH₃, 2,3-dibromopentane. Each original double-bond carbon receives one Br. The C=C was between carbons 2 and 3, so shifting addition to carbons 1 and 2 would change the required connectivity. Formula check: C₅H₁₀ + Br₂ → C₅H₁₀Br₂.
- Butan-2-one, CH₃COCH₂CH₃, forms from the secondary alcohol. Use butan-1-ol instead for oxidation via butanal to butanoic acid with excess suitable oxidant under reflux. Do not turn a ketone into a same-chain carboxylic acid with an unjustified one-arrow rule.
- Hydrolysis consumes 3H₂O and yields one glycerol plus three fatty acids. Transesterification consumes three methanol molecules and gives one glycerol plus three fatty-acid methyl esters. Neither process removes the fatty-acid chain carbons. Their coproducts and reagents differ even though both alter ester connections.
