CLASS 12 CHEMISTRY · CHAPTER 10 · STUDY NOTES
Biomolecules
An exploration of the complex organic compounds that sustain life, including carbohydrates, proteins, enzymes, vitamins, nucleic acids, and hormones, focusing on their chemical structures and biological roles.
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Introduction to Carbohydrates
Identify the chemistry, rather than treating a sugar as carbon plus trapped water.Recognise a carbohydrate from its groups
- Carbohydrate
- A polyhydroxy aldehyde or ketone, or a substance that yields such units on hydrolysis.
Glucose fits C₆(H₂O)₆, but that formula alone is not a definition. Acetic acid fits a similar carbon/water ratio and is not a carbohydrate; rhamnose is a carbohydrate that does not fit that ratio. Look for the relevant functional groups and hydrolysis products.
Carbohydrates serve as energy sources, storage substances and structural materials. Glucose and fructose are individual sugar units; starch and cellulose join many units. Their different structures matter as much as their elemental composition.
Let’s look a little closer.
Go deeper: Polar groups and living molecules
Multiple hydroxyl groups can form hydrogen-bond contacts with water, helping explain the water solubility of many small sugars. A contact diagram does not count a fixed number of bound water molecules. Large polymers can behave differently because structure and intermolecular association also matter.
Biomolecules include carbohydrates, proteins, nucleic acids and other biologically important compounds. These categories connect molecular structure to function; they are not separate ingredients that each do only one job. This chapter also introduces vitamins, enzymes and hormones.
Take a moment. What do you think?
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Question
A compound has a carbon-to-water formula pattern but only one COOH group and no series of hydroxyl groups. Is that pattern sufficient to classify it as a carbohydrate?
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Classification of Carbohydrates
Use two independent classifications: number of sugar units and position of the carbonyl group.Classify by hydrolysis and carbonyl position
- Monosaccharide
- A carbohydrate unit that cannot be hydrolysed to a simpler carbohydrate.
| Classification | Question answered |
|---|---|
| Mono/di/poly | How many simpler sugar units can hydrolysis yield? |
| Aldose/ketose | Where is the open-chain carbonyl? |
| Pentose/hexose | How many carbons are in one unit? |
| Reducing/non-reducing | What reaction occurs in the specified reagent conditions? |
Hydrolysis of a disaccharide consumes water and yields two monosaccharides, which can be the same or different. An oligosaccharide yields a small number of units, conventionally 2–10 in this chapter; a polysaccharide contains many linked units. These counts refer to units, not to carbon atoms.
Glucose is an aldohexose: its open-chain form has six carbons and an aldehyde at C1. Fructose is a ketohexose: six carbons with a ketone at C2. Triose, tetrose, pentose and hexose specify the carbon count; aldose/ketose specify the carbonyl type.
Go deeper: Reducing property is a reaction, not a taste
Glucose can expose an aldehyde by ring opening and reduces Tollens/Fehling reagents under their specified conditions. Fructose also gives a reducing response because alkaline conditions allow rearrangement to aldose forms. A ketose is therefore not automatically non-reducing.
In sucrose both anomeric centres participate in its glycosidic linkage, preventing the ordinary free reducing-end response. Maltose and lactose retain a free anomeric centre. Classifying a sugar by unit count does not itself determine whether it is reducing.
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Question
One sugar gives two smaller sugar molecules on hydrolysis. A second six-carbon sugar has an internal carbonyl and cannot be hydrolysed to simpler carbohydrate units. Classify the second and explain why carbon count cannot classify the first.
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Structure and Evidence of Glucose
Build the glucose structure by combining separate chemical observations.Build a structure from independent evidence
- Fischer configuration
- A projection convention in which horizontal bonds point towards the viewer and vertical bonds point away.
Glucose has formula C₆H₁₂O₆. Prolonged HI reduction gives n-hexane, supporting six carbons in an unbranched skeleton. Oxime and cyanohydrin formation support a reactive carbonyl. Mild bromine-water oxidation gives gluconic acid, supporting an aldehyde end.
Acetylation gives a pentaacetate, supporting five hydroxyl groups. Strong nitric-acid oxidation produces a dicarboxylic acid, showing that the other end contains a primary alcohol. Together these observations support the open-chain CHO–(CHOH)₄–CH₂OH structure; they do not independently assign every OH configuration.
These reactions establish glucose as a straight-chain pentahydroxy aldehyde.
Go deeper: Configuration and the open-chain limitation
For D-glucose, the C2→C5 OH pattern in a Fischer projection is right–left–right–right. The OH at the lowest stereocentre establishes the D relationship. L-glucose is its mirror. Rotating a projection carelessly can change what it means; D/L is not simply the direction of light rotation.
The simple open-chain drawing does not explain all observations. Glucose does not give the ordinary Schiff response or sodium hydrogensulphite addition product, and glucose pentaacetate does not react with hydroxylamine as a freely exposed aldehyde would. Cyclic hemiacetal forms explain why an aldehyde classification and predominant ring structures can coexist.
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Question
Mild oxidation changes one end of glucose to COOH; strong oxidation changes both ends. What additional evidence does the strong oxidation provide?
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Cyclic Structure of Glucose and Fructose
Count ring atoms and locate the new stereocentre before using alpha or beta.Locate the anomeric carbon in a real ring
- Anomers
- Stereoisomers differing at the anomeric carbon created when the carbonyl group forms a cyclic hemiacetal or hemiketal.
Glucose C5 hydroxyl oxygen adds to the C1 aldehyde, making a six-membered ring with five carbons and one oxygen. C6 remains outside. For the shown D-glucopyranose projection, alpha C1 OH is opposite the C5-linked CH₂OH reference; beta is on the same side.
Fructose C5 hydroxyl oxygen can add to its C2 ketone, giving the chapter’s five-membered fructofuranose ring. C1 and C6 remain outside, and C2 is the anomeric carbon. Changing alpha to beta switches both OH and the opposing C1 substituent at C2. Other ring forms are also possible.
Go deeper: Addition is not dehydration; D is not plus
Ring closure is intramolecular addition: no water molecule is lost. The carbonyl oxygen becomes an OH and the C5 hydroxyl oxygen becomes the ring oxygen. A Haworth projection indicates stereochemical relationships; it does not assert that an actual ring is perfectly flat.
D/L describes configuration relative to glyceraldehyde, while +/− records the observed direction of optical rotation under measurement conditions. D-glucose is (+), while D-fructose is (−). Alpha/beta is a third distinction, concerning the anomeric configuration rather than changing every stereocentre.
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Question
A proposed fructofuranose contains C1–C6 and one oxygen in its ring and labels C1 as anomeric. Repair both counts.
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Disaccharides and Polysaccharides
Find the linked carbons to predict hydrolysis products and reducing ends.Which carbons does the oxygen bridge join?
- Glycosidic linkage
- A covalent connection through oxygen that joins sugar units, with the linked carbon positions and configurations specified.
| Disaccharide | Link | Hydrolysis products | Reducing response |
|---|---|---|---|
| Sucrose | Glucose C1 → fructose C2 | Glucose + fructose | No ordinary response |
| Maltose | Glucose C1 → glucose C4 | Glucose + glucose | Yes |
| Lactose | Galactose C1 → glucose C4 | Galactose + glucose | Yes |
Sucrose joins alpha-D-glucose C1 to beta-D-fructose C2. Both anomeric centres are bound, so sucrose is non-reducing in the ordinary tests. Hydrolysis consumes water and gives one glucose plus one fructose.
Maltose has an alpha(1→4) glucose link; lactose has a beta(1→4) galactose-to-glucose link. Each leaves the receiving glucose C1 able to open, so each is reducing. Their hydrolysis gives glucose+glucose and galactose+glucose respectively.
Go deeper: Storage, structure and inversion
Amylose contains predominantly unbranched alpha(1→4) glucose chains; amylopectin adds alpha(1→6) branches. Glycogen is an animal storage polymer with more branching. Cellulose uses beta(1→4) glucose and forms structural chains in plant cell walls. The source’s broad bacterial-cell-wall claim is corrected: most bacterial walls instead have peptidoglycan.
Sucrose hydrolysis changes the source mixture’s optical-rotation sign from positive to negative; this is inversion. Fructose’s negative contribution outweighs glucose’s positive contribution under the stated comparison. Source specific rotations +52.5° and −92.4° are not raw angles to add without accounting for concentration and path length.
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Question
A disaccharide joins glucose C1 to another glucose C4. Explain why linking one anomeric carbon does not necessarily make the whole molecule non-reducing.
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Amino Acids and Proteins
Connect side-chain identity, acid–base form and levels of protein structure.From an amino acid to a folded protein
- Peptide bond
- The C(=O)–NH amide linkage between amino-acid residues in a peptide chain.
An alpha-amino acid has amino and carboxyl groups attached to the same alpha carbon, together with H and side chain R. Glycine has two H at that carbon and is achiral; most other common alpha-amino acids have a stereocentre. Acidic/basic side chains affect overall charge.
In the glycine model, acidic medium favours NH₃⁺/COOH, a zwitterion has NH₃⁺/COO⁻, and basic medium favours NH₂/COO⁻. A zwitterion has internal positive and negative charges while its net charge can be zero. Carboxyl and amino groups exchange protons under different conditions.
Go deeper: From a sequence to a functional shape
Net peptide formation removes carboxyl OH plus one amino H as water, retaining C=O and creating C–N. Two residues give a dipeptide; three give two peptide links. Essential amino acids must be supplied in the diet when synthesis cannot meet need; requirements depend on biological context, rather than assuming every amino acid is synthesised adequately.
Primary structure is sequence; secondary includes alpha helices and beta sheets stabilised by backbone hydrogen bonds. Tertiary describes one chain’s overall fold; quaternary describes multiple subunits. H bonds, disulphide links and other interactions contribute to structure. Fibrous and globular describe broad shapes. A rigid 100-residue protein cutoff fails even the source insulin example of 51 residues.
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Question
A linear peptide has five residues and no ring closure. How many peptide bonds connect them, and how many waters appear in the net condensation accounting?
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Denaturation and Enzymes
Separate a change in protein shape from a change in reaction energy barrier.Shape controls activity; catalysts change the barrier
- Denaturation
- Disturbance of a protein’s native higher structure, often reducing its activity without hydrolysing its peptide backbone.
Heat or a chemical/pH change can disrupt interactions maintaining native protein shape. Ordinary denaturation alters secondary/tertiary organisation while retaining primary sequence; boiling egg white illustrates the resulting change. Severe treatment can also cause chemical damage, and reversal is not guaranteed.
Enzymes are biological catalysts, mostly proteins in this chapter’s treatment. An active site binds suitable substrate and enables a lower-barrier route. Products leave, making enzyme available for another cycle; a temporary enzyme–substrate complex is not permanent consumption of the catalyst.
Go deeper: Specificity and energy accounting
Names often reflect substrate or reaction, such as maltase and oxidoreductase. Maltase hydrolyses maltose: C₁₂H₂₂O₁₁ + H₂O → 2C₆H₁₂O₆. Water is necessary to balance H/O and is restored here despite its omission from the source’s displayed equation.
A catalyst lowers activation barrier without changing reactant/product energy difference or the equilibrium constant for the same overall reaction. The source reports 6.22 versus 2.15 kJ mol⁻¹ for its compared acid/sucrase routes; these are context-specific reported values. A schematic energy profile is not a time trace, measured potential surface or independent rate calculation.
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Question
A sketch lowers the enzyme-route peak but also makes its products a different energy from the uncatalysed products. Which change is justified?
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Vitamins and Deficiency
Learn the class, a biological role and a deficiency association together.Connect vitamin class with biological function
- Vitamin
- An organic nutrient needed in small amounts for normal biological functions, with requirements varying by organism and context.
| Vitamin class | Examples | Storage distinction |
|---|---|---|
| Fat soluble | A, D, E, K | Can accumulate in body stores |
| Water soluble | B group, C | Storage varies; B12 is an important exception |
Vitamins A, D, E and K are fat-soluble; the B group and C are water-soluble. Solubility helps explain handling and storage but is not an absolute no-storage rule: B12 has substantial body stores, particularly in liver.
A supports vision; D supports bone mineralisation; E contributes antioxidant protection; K supports clotting-protein function. B vitamins have important metabolic roles; B12 supports blood-cell and nerve function, and C supports connective tissue. Select each vitamin for source examples and deficiency associations.
Go deeper: Deficiency associations need their causal direction
The chapter links A deficiency with night blindness/xerophthalmia, B1 with beri beri, B2 with cheilosis, C with scurvy, D with rickets/osteomalacia and K with longer clotting time. These are learning associations: a single symptom is not a diagnosis, and other causes can give similar signs.
B12 deficiency can cause megaloblastic anaemia. Pernicious anaemia is an absorption disorder that can cause B12 deficiency, rather than the inevitable result of every low intake. Vitamin requirements, storage, absorption and deficiency are different ideas; the notes do not supply a dose or treatment plan.
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Question
A learner says all water-soluble vitamins have no body stores. Use one chapter example to repair the rule.
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Nucleic Acids and Heredity
Follow the covalent backbone separately from the hydrogen bonds joining bases.Follow the sugar, backbone and paired bases
- Nucleotide
- A unit containing a nitrogenous base, a pentose sugar and phosphate; a nucleoside contains base plus sugar without that phosphate.
DNA has 2-deoxyribose, with H at sugar C2′; RNA has ribose, with OH there. DNA bases are A,G,C,T; RNA uses U in place of T. A/G are purines and C/T/U are pyrimidines. Primes identify sugar carbons, keeping them distinct from base positions.
A base attaches at sugar C1′ to form a nucleoside. Adding phosphate, shown at C5′, forms a nucleotide. Adjacent units connect through a 3′–O–P–O–5′ phosphodiester; the covalent sugar–phosphate backbone is not held together by base-pair hydrogen bonds.
Go deeper: Information, pairing and profiles
Ordinary DNA has complementary antiparallel strands: A–T gives two H-bond contacts and G–C three. Complementarity supports copying genetic information. DNA stores hereditary information; mRNA carries coding information, tRNA delivers amino acids via anticodon matching, and rRNA contributes ribosome structure and catalysis during protein synthesis.
DNA profiling compares selected variable regions rather than showing every base. Most nucleated cells share the individual’s genomic pattern, but mature red blood cells lack a nucleus. Ordinary profiles may not distinguish identical twins; mutation and technical sampling matter. A profile comparison is not proof of guilt or a biological racial classification.
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Question
A proposed nucleotide puts the base at C5′ and phosphate at C1′, then calls the link between paired bases phosphodiester. Repair the attachments and the inter-strand link.
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Hormones and Biological Regulation
Distinguish the signal, its transport and the target response.A signal acts only where its receptor can respond
- Hormone
- A chemical signal released by endocrine cells and acting on responsive target cells through appropriate receptors.
Blood carries an endocrine signal to many locations, but only cells able to respond through appropriate receptors show the relevant response. Hormones can be steroids, peptides/proteins or amino-acid derivatives. Their structure affects how they interact with cells; not all receptors are on the outer membrane.
Insulin helps lower blood glucose through uptake/storage responses, while glucagon helps raise it through liver glucose release. These are opposing physiological roles, not instructions to alter hormone levels. Epinephrine/norepinephrine participate in stress responses; thyroxine supports metabolic regulation.
Go deeper: Examples connect chemistry and physiology
Testosterone, estradiol and progesterone are steroid examples with differing biological roles; sharing a fused-ring nucleus does not give identical action. Insulin is a peptide hormone. Thyroxine and epinephrine are amino-acid-derived examples, not amino acids used directly as protein monomers.
Iodine is needed for thyroid hormone synthesis. Deficiency can impair synthesis and contribute to thyroid enlargement, but thyroid conditions have multiple causes. The chapter links iodised salt with iodine provision; these notes explain the chemical/physiological relationship without diagnosing a learner or setting intake/treatment advice.
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Question
Two cells encounter the same circulating hormone, but only one responds. Does exposure alone determine the response?
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