Loav · VCE Chemistry · Unit 4 / Area of Study 1
U4A1-04 · Current 2026 curriculum · Original teaching guide · Study guide
1. Join by condensation; separate by hydrolysis
In a condensation model, two molecules join and a small molecule is eliminated. For the biological linkages studied here, that small molecule is water. In hydrolysis, water is a reactant: its H and OH end up on the fragments produced when a linkage breaks. Condensation and hydrolysis therefore reverse the connectivity change, although actual biological synthesis often uses activated intermediates and enzyme-coupled steps rather than direct mixing of free monomers.
Write a net structural equation to keep track of the atoms. Do not say that water “breaks a bond” and then forget to incorporate its atoms into the products. Each hydrolysed linkage consumes one water molecule in the simple net accounting used here. Specific enzyme names are not required by this area of study, but suitable enzymes and conditions are needed in biological systems. Laboratory hydrolysis can instead use appropriate acid or base and heating under supervision.
2. Proteins: amino acids joined by peptide bonds
An α-amino acid has an amino group and a carboxyl group associated with the same central carbon, together with H and a variable side group R. A common neutral structural template is H₂N–CH(R)–COOH. Proline is an exception: its nitrogen is part of a ring and its amino group is secondary. Actual charge state depends on pH; the neutral template is useful for tracing condensation, not a claim that all amino acids remain neutral in water.
The carboxyl group of one amino acid joins to the amino group of another. A water molecule is eliminated, and the resulting –C(=O)–NH– linkage is a peptide bond, an amide linkage. The original amino acids become residues within the chain. A chain has an amino end and a carboxyl end; sequence matters because exchanging the order of different residues changes the molecule.
H₂NCH₂COOH + H₂NCH(CH₃)COOH → H₂NCH₂CONHCH(CH₃)COOH + H₂O
This net condensation joins glycine to alanine. The product contains one peptide linkage, not two. Hydrolysis adds water across that linkage and restores the two amino-acid units; their protonation state depends on solution pH. Complete hydrolysis of a single unbranched chain containing n amino-acid residues requires n − 1 water molecules because it contains n − 1 peptide bonds. Disulfide links, if present in a protein, are not peptide bonds and are not counted as peptide hydrolysis sites.
3. Starch and glycogen: glucose storage polymers
Glucose units can join through glycosidic linkages. In a simple condensation representation, an OH from one unit and H from another produce water while an oxygen-containing bridge connects the carbon frameworks. Hydrolysis reverses this linkage formation. Starch in plants and glycogen in animals store glucose units; digestion progressively breaks glycosidic linkages to produce smaller carbohydrates and ultimately glucose.
Starch includes amylose, principally an unbranched chain with α(1→4) links, and amylopectin, with α(1→4) chain links and α(1→6) branch points. Glycogen is also branched and has frequent α(1→6) branching. These labels describe which glucose carbons are connected and the orientation at the anomeric carbon. You need to recognise the oxygen bridge and glucose building blocks rather than mistake the ring oxygen already inside each glucose unit for a linkage between units.
The repeated glucose-derived unit is commonly represented as C₆H₁₀O₅. This repeat-unit shorthand omits end-group detail. For an exact finite molecule made from n glucose molecules joined by n − 1 links, condensation releases n − 1 H₂O, giving C₆ₙH₁₀ₙ₊₂O₅ₙ₊₁. Use the convention specified in a question. A branched connected molecule with no closed loop still has n − 1 links: each newly attached glucose adds one link, even at a branch.
4. Fats and oils: three fatty acids plus glycerol
Glycerol has three hydroxyl groups: HOCH₂CH(OH)CH₂OH. A fatty acid has a long hydrocarbon chain attached to COOH. Each glycerol OH can form an ester linkage with a fatty-acid carboxyl group. Joining three fatty acids to one glycerol forms a triglyceride and three water molecules. The fatty acids need not be identical.
Glycerol + 3 fatty acids ⇌ triglyceride + 3H₂O
A triglyceride is a triester, not a chain polymer made from endlessly repeated glycerol units. Hydrolysis breaks its three ester linkages. Acidic or enzyme-catalysed hydrolysis is represented as producing glycerol and fatty acids. Alkaline hydrolysis produces glycerol and fatty-acid carboxylate salts, the chemical basis of soap formation. An additional acidification step would convert those salts to neutral fatty acids.
“Fat” and “oil” usually describe physical state near room temperature. Many oils contain a larger proportion of unsaturated chains, but they are mixtures, not single molecules with one universal structure. A saturated fatty-acid chain contains no C=C bonds; a monounsaturated chain has one, and a polyunsaturated chain has more than one. Ester carbonyl C=O bonds do not count as carbon–carbon unsaturation. Cis C=C bonds can introduce bends that impede packing and help explain lower melting behaviour; chain length and composition also matter.
Worked example 1: peptide water accounting
A single unbranched peptide contains 8 amino-acid residues. It has 7 peptide bonds. Making it from eight free amino acids releases 7 water molecules per peptide; complete hydrolysis consumes 7. For 0.0200 mol of this peptide, hydrolysis requires 7 × 0.0200 = 0.140 mol H₂O and yields 8 × 0.0200 = 0.160 mol total amino acids. These are illustrative stoichiometric quantities, assuming complete hydrolysis.
Worked example 2: exact carbohydrate formula
Join three glucose molecules by two glycosidic links. Before condensation the total is C₁₈H₃₆O₁₈. Subtract 2H₂O to obtain C₁₈H₃₂O₁₆. Complete hydrolysis reverses this: C₁₈H₃₂O₁₆ + 2H₂O → 3C₆H₁₂O₆. The common repeat unit alone would miss the end-group atoms in this small, finite example.
Practice
- Name the linkage broken in protein digestion and the small building blocks ultimately formed.
- How much water is consumed by complete ester hydrolysis of 0.150 mol triglyceride? How much glycerol forms?
- A connected, branched glucose molecule has 20 glucose residues and no closed loops. How many links connect them?
- One triglyceride contains fatty-acid chains with zero, one and two C=C bonds respectively. How many chain C=C bonds are present? Do its carbonyls change that count?
Answers with reasoning
- Peptide (amide) bonds; amino acids. Water contributes atoms to the newly formed amino and carboxyl ends, with ionisation dependent on pH.
- 0.450 mol water and 0.150 mol glycerol. There are three ester groups per triglyceride but only one glycerol backbone. In alkaline hydrolysis, express the net reagent as hydroxide and the acid-derived products as carboxylate salts instead.
- 19 links. Begin with one glucose and attach each of the other 19 by one link. Branching changes arrangement but not this count unless extra connections create loops.
- Three chain C=C bonds. Its three ester C=O groups are not carbon–carbon double bonds, so they do not increase this count.
Go deeper: locate the linkage, not just the element
A protein can contain oxygen in a side-chain OH as well as in its peptide backbone. A polysaccharide has oxygen atoms within glucose rings as well as between glucose units. A triglyceride has three carbonyl groups, but all three belong to esters. Counting O or N atoms alone therefore does not identify the bonds that hydrolyse. Trace the covalent backbone, mark the characteristic linkage, and only then add water at that location.
For a peptide, follow the repeated backbone sequence N–Cα–C(=O). The side group branches from Cα; it is not the next residue. Cutting a peptide linkage conceptually means separating the carbonyl carbon from the next nitrogen and restoring the appropriate OH and H. Leave the side groups intact. Counting all carbonyls as peptide bonds fails at the terminal carboxyl group and can fail when side chains also contain carbonyls.
Condensation changes molar mass predictably. For an unbranched peptide assembled from n free amino acids, add their molar masses, then subtract (n − 1) × M(H₂O). Hydrolysis reverses this mass change. The combined mass of the released amino acids is greater than the initial peptide mass because water contributes atoms; this obeys conservation of mass when the water is included. The same principle applies to a triglyceride, but its fixed three ester links mean subtracting three waters during formation, rather than a number based on its total carbon count.
Worked example 3 · Apply: infer a triglyceride’s building block
An original model triglyceride consists of glycerol and three identical saturated fatty acids. Its formula is C₄₅H₈₆O₆. Use the exercise atomic masses C = 12.0, H = 1.00 and O = 16.0. M(triglyceride) = 45 × 12.0 + 86 × 1.00 + 6 × 16.0 = 722 g mol⁻¹. Glycerol is C₃H₈O₃, 92.0 g mol⁻¹; water is 18.0 g mol⁻¹.
Complete hydrolysis gives 722 + 3 × 18.0 = 776 g of combined products per mole of triglyceride. Subtract the 92.0 g glycerol to leave 684 g for three moles of fatty acid. Each fatty acid therefore has M = 228 g mol⁻¹. Formula accounting independently gives (C₄₅H₈₆O₆ + 3H₂O − C₃H₈O₃)/3 = C₁₄H₂₈O₂. Its molar mass is indeed 228 g mol⁻¹.
For a real oil containing different fatty-acid chains, that calculation could give an average molar mass rather than identify a unique acid. The assumption of three identical chains is essential evidence, not an optional detail. No systematic naming of this long-chain molecule is required here.
Application practice: account for every water
- A peptide consists of two glycine residues and one alanine residue in one unbranched chain. Using free amino-acid formulas C₂H₅NO₂ and C₃H₇NO₂, find its molecular formula. Does the formula identify the sequence?
- Complete hydrolysis of 0.0300 mol of a five-residue unbranched peptide produces how many moles of total amino acids and consumes how many moles of water?
- One mole of a triglyceride gives one mole glycerol and three moles fatty-acid salts in aqueous NaOH. Write the reagent coefficient and explain why neutral fatty acids are not the immediate products.
Application answers and checks
- Add 2C₂H₅NO₂ + C₃H₇NO₂ to obtain C₇H₁₇N₃O₆, then subtract two waters because three residues require two links: C₇H₁₃N₃O₄. The formula does not identify sequence: changing the order of these same residues changes connectivity without changing total atoms.
- There are four peptide links per chain. Water consumed = 4 × 0.0300 = 0.120 mol; total amino acids = 5 × 0.0300 = 0.150 mol. Do not use the same coefficient for links and residues.
- Three NaOH per triglyceride. The three ester linkages yield three carboxylate groups in the basic solution, each paired with Na⁺, plus glycerol. The overall model is triglyceride + 3NaOH → glycerol + three sodium fatty-acid salts. Adding acid afterwards is a separate reaction if neutral fatty acids are required.
