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State Board · Class 12 · Chemistry

Lesson 14: Biomolecules

7 concepts · 56 practice questions with answers · Free to read

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Carbohydrates: classification and preparation of glucose

From daily life

Rice, bread, sugar and paper are all carbohydrates.

Simple definition

Carbohydrates are polyhydroxy aldehydes/ketones or compounds yielding them on hydrolysis: mono-, oligo-, polysaccharides.

ClassificationCarbohydrates: classificationMonosaccharideno hydrolysis: glucose, fructoseOligosaccharide2–10 units: sucrose, maltose, lactosePolysaccharidemany units: starch, cellulose, glycogenAldose / ketose–CHO (glucose) / >C=O (fructose)Triose … hexose3 … 6 carbons (glucose = aldohexose)Preparation: sucrose + H₂O (H⁺) ⇒ glucose + fructosestarch + H₂O (H⁺, 393 K, pressure) ⇒ glucose

In the figure : Classification.

Rule 1: Mono/oligo/poly — Monosaccharides do not hydrolyse; disaccharides have two units; polysaccharides many.

  • Easy : Glucose = mono.
  • Medium : Sucrose = di.
  • Exam-type : Starch = poly.

Rule 2: Aldose/ketose — –CHO ⇒ aldose; >C=O ⇒ ketose; triose … hexose by carbon count.

  • Easy : Glucose = aldohexose.
  • Medium : Fructose = ketohexose.
  • Exam-type : Ribose = aldopentose.

Rule 3: Making glucose — Sucrose + H₂O (dil. HCl/H₂SO₄) ⇒ glucose + fructose. Starch + H₂O (H⁺, 393 K, pressure) ⇒ glucose.

  • Easy : 34.2 g sucrose ⇒ 18.0 g glucose + 18.0 g fructose.
  • Medium : Mass 34.2 + 1.8 = 36.0 g.
  • Exam-type : Industrial: from starch.

Calculation: glucose from 34.2 g sucrose

  • Step 1: n = 34.2 ÷ 342.3 = 0.1 mol.
  • Step 2: 1 mol sucrose ⇒ 1 mol glucose + 1 mol fructose.
  • Step 3: 0.1 × 180.16 = 18.0 g of each.

CET level (MCQ)

  • CET 1) Glucose is a(n)? 1) aldohexose 2) ketohexose 3) aldopentose 4) disaccharide — Answer: 1).
  • CET 2) Hydrolysis of sucrose gives? 1) glucose + fructose 2) 2 glucose 3) glucose + galactose 4) 2 fructose — Answer: 1).

Real-life use: Food, energy, fuel (ethanol).

Remember

Monosaccharides do not hydrolyse; glucose = aldohexose; fructose = ketohexose.

Common mistake

Calling fructose an aldose.

CET trap

Aldose = –CHO; ketose = C=O.

Shortcut

Memory clue: "Glucose has the aldehyde; fructose has the ketone".

Spot the difference

SugarGroupClass
Glucose–CHOaldohexose
Fructose>C=Oketohexose
Sucrose—disaccharide

Practice questions

Easy
  1. 1What are carbohydrates?
  2. 2Mono vs oligo vs poly.
  3. 3Aldose vs ketose.
Medium
  1. 4How is glucose prepared?
  2. 534.2 g sucrose ⇒ ?
  3. 6Type of ribose?
Challenging
  1. 7Starch hydrolysis conditions.
  2. 8What is a hexose?
Show answers (try them yourself first)
  1. 1Polyhydroxy carbonyl compounds.
  2. 2Number of hydrolysis products.
  3. 3–CHO vs C=O.
  4. 4Sucrose/starch + H⁺.
  5. 518.0 g glucose + 18.0 g fructose.
  6. 6Aldopentose.
  7. 7H⁺, 393 K, pressure.
  8. 8A six-carbon sugar.

Structure and stereochemistry of glucose

From daily life

Doctors test for glucose with Benedict/Fehling because it carries –CHO.

Simple definition

Glucose = HOCH₂–(CHOH)₄–CHO. Four chiral carbons ⇒ 2⁴ = 16 stereoisomers; natural glucose is D(+).

Glucose evidenceStructure of glucose: evidenceHI, heat⇒ n-hexane ⇒ straight 6-C chainBr₂ water⇒ gluconic acid ⇒ –CHO presentconc. HNO₃⇒ saccharic acid ⇒ a 1° OH present(CH₃CO)₂O⇒ pentaacetate ⇒ five –OHNH₂OH / HCN⇒ oxime / cyanohydrin ⇒ C=ONaHSO₃, Schiff⇒ no reaction ⇒ ring form (hemiacetal)4 chiral C ⇒ 2⁴ = 16 stereoisomers (8 D + 8 L); natural = D

In the figure : Structural evidence.

Rule 1: Open-chain evidence — HI ⇒ n-hexane (straight 6 C). Br₂ water ⇒ gluconic acid (–CHO). HNO₃ ⇒ saccharic acid (terminal 1° OH).

  • Easy : Pentaacetate ⇒ five OH.
  • Medium : 18.0 g glucose ⇒ 39.0 g pentaacetate.
  • Exam-type : NH₂OH ⇒ oxime ⇒ C=O.

Rule 2: D and L — OH on the chiral carbon farthest from CHO (C-5) on the right ⇒ D; on the left ⇒ L.

  • Easy : The mirror image of D is L.
  • Medium : D(+) rotates plane light to the right.
  • Exam-type : D/L ≠ d/l.

Rule 3: Limitations — Glucose fails the NaHSO₃ and Schiff tests and its pentaacetate does not react with NH₂OH, so a ring form is needed.

  • Easy : Schiff ⇒ no.
  • Medium : Mutarotation.
  • Exam-type : Two anomers.

Reasoning: stereoisomers of an aldohexose

  • Step 1: Chiral carbons = 4 (C2–C5).
  • Step 2: 2⁴ = 16.
  • Step 3: 8 D + 8 L.

CET level (MCQ)

  • CET 1) Chiral carbons in glucose? 1) 4 2) 3 3) 5 4) 6 — Answer: 1).
  • CET 2) Bromine water with glucose ⇒ ? 1) gluconic acid 2) saccharic acid 3) n-hexane 4) sorbitol — Answer: 1).

Real-life use: Diabetes tests, IV fluids.

Remember

HI ⇒ hexane; Br₂ water ⇒ gluconic; HNO₃ ⇒ saccharic; 4 C* ⇒ 16.

Common mistake

Mistaking D/L for d/l (rotation).

CET trap

D/L = configuration; (+)/(–) = rotation direction.

Shortcut

Memory clue: "C-5 OH on the right = D".

Spot the difference

ReagentProduct
HIn-hexane
Br₂/H₂Ogluconic acid
HNO₃saccharic acid

Practice questions

Easy
  1. 1How is the open chain proved?
  2. 2How 2⁴ = 16?
  3. 3D vs L.
Medium
  1. 4What does pentaacetate show?
  2. 5Why no Schiff test?
  3. 6Product with HNO₃.
Challenging
  1. 7C* in a ketohexose?
  2. 8Pentaacetate from 18.0 g?
Show answers (try them yourself first)
  1. 1HI ⇒ hexane, Br₂ ⇒ acid.
  2. 24 chiral carbons.
  3. 3C-5 OH right/left.
  4. 4Five OH groups.
  5. 5A ring (hemiacetal) form.
  6. 6Saccharic acid.
  7. 73 ⇒ 8 isomers.
  8. 839.0 g.

Ring structures, anomers and fructose

From daily life

A fresh glucose solution slowly changes its optical rotation — mutarotation.

Simple definition

C-1 aldehyde reacts with the C-5 OH forming a hemiacetal ⇒ six-membered pyranose ring. C-1 OH down ⇒ α, up ⇒ β (anomers).

Ring formsGlucose: pyranose ring, α and β anomersOCH₂OHC5OHC1C2C3C4α-D-glucoseC1–OH downOCH₂OHC5OHC1C2C3C4β-D-glucoseC1–OH up[α]D = +112°[α]D = +18.7°Equilibrium (mutarotation): +52.7° ⇒ ≈ 36 % α : 64 % βFructose: five-membered furanose ring

In the figure : Haworth structures.

Rule 1: Anomers — α-D-glucose [α]D = +112°; β-D-glucose +18.7°; they differ only at C-1 (the anomeric carbon).

  • Easy : α: OH down.
  • Medium : β: OH up.
  • Exam-type : Both are diastereomers.

Rule 2: Mutarotation — In solution α ⇌ β via the open chain; rotation settles at +52.7° ⇒ ≈ 36 % α : 64 % β.

  • Easy : 112x + 18.7(1–x) = 52.7.
  • Medium : x ≈ 0.364.
  • Exam-type : β is more stable.

Rule 3: Fructose — A ketohexose; ring is five-membered furanose (O between C-2 and C-5). D-fructose [α]D = –92.4°, hence 'laevulose'.

  • Easy : The sweetest sugar.
  • Medium : Fructose gives Fehling's test.
  • Exam-type : Sweetest of the hexoses.

Calculation: α fraction at equilibrium

  • Step 1: 112x + 18.7(1 – x) = 52.7.
  • Step 2: 93.3x = 34.0 ⇒ x ≈ 0.364.
  • Step 3: α ≈ 36 %, β ≈ 64 %.

CET level (MCQ)

  • CET 1) Anomers differ at which carbon? 1) C-1 2) C-2 3) C-5 4) C-6 — Answer: 1).
  • CET 2) Members in the glucose pyranose ring? 1) 6 2) 5 3) 4 4) 7 — Answer: 1).

Real-life use: Honey, sweeteners, energy drinks.

Remember

α = OH down; β = up; equilibrium +52.7°; fructose five-membered.

Common mistake

Confusing α/β with D/L.

CET trap

α/β concern the C-1 OH; D/L concerns C-5.

Shortcut

Memory clue: "α = down (alpha-low)".

Spot the difference

Form[α]D
α-D-glucose+112°
β-D-glucose+18.7°
Equilibrium+52.7°

Practice questions

Easy
  1. 1What are anomers?
  2. 2What is mutarotation?
  3. 3Find the α fraction.
Medium
  1. 4Ring of fructose.
  2. 5Pyranose vs furanose.
  3. 6Why is fructose laevulose?
Challenging
  1. 7What is a hemiacetal?
  2. 8Difference between α and β.
Show answers (try them yourself first)
  1. 1Isomers differing at C-1.
  2. 2Rotation changes through α ⇌ β.
  3. 336 %.
  4. 4Five-membered.
  5. 56 vs 5.
  6. 6[α]D –92.4°.
  7. 7C(OH)(OR).
  8. 8C-1 OH down/up.

Disaccharides and polysaccharides

From daily life

The starch in flour, the sugar in honey and the lactose in milk are all here.

Simple definition

A glycosidic bond is the C–O–C link joining two sugars: sucrose (1→2), maltose and lactose (1→4).

Di/polyDisaccharides and polysaccharidesSucroseα-Glu(1→2)β-Fru · non-reducingMaltoseα-Glu(1→4)Glu · reducingLactoseβ-Gal(1→4)Glu · reducingStarchamylose 15–20 % (linear) + amylopectin 80–85 % (branched)Celluloseβ-1,4 glucose, linear; not digested by humansGlycogenα-1,4 + α-1,6; highly branched; animal storeSucrose [α]D +66.5° ⇒ hydrolysis ⇒ –20° (invert sugar)

In the figure : Di- and polysaccharides.

Rule 1: Disaccharides — Sucrose = α-glucose + β-fructose (1→2), non-reducing. Maltose = glucose + glucose (α-1,4), reducing. Lactose = galactose + glucose (β-1,4), reducing.

  • Easy : Sucrose with Fehling ⇒ none.
  • Medium : Maltose ⇒ red Cu₂O.
  • Exam-type : Lactose = milk sugar.

Rule 2: Invert sugar — Sucrose [α]D = +66.5°; after hydrolysis the glucose + fructose mixture is ≈ –20°, so rotation inverts.

  • Easy : Fructose (–92.4°) dominates.
  • Medium : Hence 'invert'.
  • Exam-type : Honey contains invert sugar.

Rule 3: Polysaccharides — Starch: amylose (α-1,4 linear, 15–20 %) + amylopectin (α-1,4 and α-1,6 branched, 80–85 %). Cellulose: β-1,4. Glycogen: highly branched.

  • Easy : Starch + I₂ ⇒ blue.
  • Medium : Humans cannot digest cellulose.
  • Exam-type : Glycogen = animal starch.

Reasoning: why no Fehling result for sucrose?

  • Step 1: Glucose C-1 and fructose C-2 are joined together.
  • Step 2: No free hemiacetal OH remains.
  • Step 3: So it is non-reducing.

CET level (MCQ)

  • CET 1) Non-reducing disaccharide? 1) sucrose 2) maltose 3) lactose 4) cellobiose — Answer: 1).
  • CET 2) Starch + iodine ⇒ ? 1) blue 2) red 3) yellow 4) green — Answer: 1).

Real-life use: Food, cloth (cotton), paper.

Remember

Sucrose non-reducing; cellulose β; starch α; I₂ ⇒ blue.

Common mistake

Calling sucrose reducing.

CET trap

Starch is α-1,4, cellulose β-1,4 — only the link direction differs.

Shortcut

Memory clue: "sucrose = safe (non-reducing)".

Spot the difference

SugarUnitsBond
SucroseGlu + Fruα-1,2-β
MaltoseGlu + Gluα-1,4
LactoseGal + Gluβ-1,4

Practice questions

Easy
  1. 1Structure of sucrose.
  2. 2What is invert sugar?
  3. 3Amylose vs amylopectin.
Medium
  1. 4Why isn't cellulose digested?
  2. 5What is a glycosidic bond?
  3. 6Units of lactose.
Challenging
  1. 7Iodine test of starch.
  2. 8Where is glycogen stored?
Show answers (try them yourself first)
  1. 1α-Glu(1→2)β-Fru.
  2. 2+66.5° ⇒ –20°.
  3. 3Linear vs branched.
  4. 4No enzyme to cleave β links.
  5. 5A C–O–C link.
  6. 6Galactose + glucose.
  7. 7A blue colour.
  8. 8Liver and muscles.

Amino acids and peptides

From daily life

Egg, pulses and milk contain proteins that are chains of amino acids.

Simple definition

α-Amino acids are H₂N–CHR–COOH and exist as zwitterions. Two amino acids lose H₂O to form a peptide bond.

Amino acidsAmino acids and the peptide bondH₃N⁺–CHR–COO⁻zwitterion (glycine pI = 5.97)20 standard α-amino acids10 essential (not made in body)H₂N–CHR–COOH + H₂N–CHR′–COOH– H₂OH₂N–CHR–CO–NH–CHR′–COOH–CO–NH– = peptide (amide) bondn amino acids ⇒ (n – 1) peptide bonds + (n – 1) H₂OpH < pI ⇒ cation; pH > pI ⇒ anion; pH = pI ⇒ net charge 0

In the figure : Amino acids and the peptide bond.

Rule 1: Classification — 20 standard amino acids; essential = the body cannot make them (valine, leucine, lysine); non-essential = it can (glycine, alanine).

  • Easy : Glycine (R = H) is achiral.
  • Medium : Others are mainly L.
  • Exam-type : Essential ones come from diet.

Rule 2: Zwitterion and pI — H₃N⁺–CHR–COO⁻. pH < pI ⇒ cation; pH > pI ⇒ anion; pH = pI ⇒ net charge 0 (lowest solubility). Glycine pI = 5.97.

  • Easy : In acid: H₃N⁺CH₂COOH.
  • Medium : In base: H₂NCH₂COO⁻.
  • Exam-type : Precipitation is easiest at the pI.

Rule 3: Peptides — n amino acids ⇒ (n – 1) peptide bonds + (n – 1) H₂O; di-, tri-, polypeptide (above ≈ 100 units = protein).

  • Easy : 2 Gly ⇒ Gly-Gly + H₂O.
  • Medium : 15.0 g Gly ⇒ 13.2 g dipeptide + 1.8 g H₂O.
  • Exam-type : A tetrapeptide has 3 bonds.

Calculation: Gly-Gly from 15.0 g glycine

  • Step 1: n = 15.0 ÷ 75.07 = 0.2 mol.
  • Step 2: 2 Gly ⇒ 1 Gly-Gly ⇒ 0.1 mol.
  • Step 3: M = 2×75.07 – 18.02 = 132.1 ⇒ 13.2 g.

CET level (MCQ)

  • CET 1) Peptide bonds in a tetrapeptide? 1) 3 2) 4 3) 2 4) 5 — Answer: 1).
  • CET 2) Achiral amino acid? 1) glycine 2) alanine 3) valine 4) lysine — Answer: 1).

Real-life use: Diet, drugs, supplements.

Remember

n units ⇒ (n–1) bonds; charge 0 at pI; glycine is achiral.

Common mistake

Writing n peptide bonds for n amino acids.

CET trap

A linear chain has (n–1); a cyclic peptide has n.

Shortcut

Memory clue: "one more acid than bonds".

Spot the difference

nPeptide bondsH₂O
211
322
1099

Practice questions

Easy
  1. 1General structure of an amino acid.
  2. 2What is a zwitterion?
  3. 3Write a peptide bond.
Medium
  1. 4Essential amino acids.
  2. 5Meaning of pI.
  3. 6Bonds in a pentapeptide?
Challenging
  1. 7Why is glycine achiral?
  2. 815.0 g Gly ⇒ ?
Show answers (try them yourself first)
  1. 1H₂N–CHR–COOH.
  2. 2H₃N⁺ and COO⁻ in one molecule.
  3. 3–CO–NH–.
  4. 4Those the body cannot make.
  5. 5pH of zero net charge.
  6. 64.
  7. 7R = H.
  8. 813.2 g.

Protein structure, denaturation and enzymes

From daily life

Cooking an egg turns the clear white opaque — protein denaturation.

Simple definition

Proteins have 1° (sequence), 2° (α-helix, β-sheet via H-bonds), 3° (3-D folding) and 4° (several chains) structure.

Protein structureLevels of protein structure1°amino-acid sequence2°α-helix / β-pleated sheet3°3-D folding4°several chainsDenaturation (heat, acid, Hg²⁺)2°/3° lost, 1° intact; egg cookingEnzymelowers Ea; specific; 310 K, pH ≈ 7

In the figure : Structure levels.

Rule 1: Structure levels — 1°: peptide bonds. 2°: C=O···H–N H-bonds. 3°: H-bonds, S–S, ionic, hydrophobic. 4°: assembly of subunits (e.g. haemoglobin).

  • Easy : Keratin is α-helix.
  • Medium : Silk is β-sheet.
  • Exam-type : Haemoglobin has 4°.

Rule 2: Denaturation — Heat, acid/base, heavy metals (Hg²⁺), urea destroy 2° and 3°; 1° stays; activity is lost.

  • Easy : Cooking an egg.
  • Medium : Curdling of milk.
  • Exam-type : Usually irreversible.

Rule 3: Enzymes — Biological catalysts (proteins): lower activation energy; specific (lock and key); optimum ≈ 310 K, pH ≈ 7 (pepsin ≈ 2).

  • Easy : Amylase breaks starch.
  • Medium : Urease breaks urea.
  • Exam-type : Heating denatures enzymes.

Reasoning: why does heat stop an enzyme?

  • Step 1: Heat breaks H-bonds.
  • Step 2: 3° folding fails and the active site changes.
  • Step 3: The substrate no longer fits so activity stops.

CET level (MCQ)

  • CET 1) 2° structure arises from? 1) H-bonds 2) S–S 3) ionic 4) peptide — Answer: 1).
  • CET 2) Which structure survives denaturation? 1) 1° 2) 2° 3) 3° 4) 4° — Answer: 1).

Real-life use: Digestion, drugs, detergents, cheese.

Remember

2° = H-bonds; denaturation keeps 1°; enzymes lower Ea, 310 K.

Common mistake

Thinking denaturation breaks peptide bonds.

CET trap

Denaturation destroys 2°, 3°, but 1° remains.

Shortcut

Memory clue: "1° order, 2° fold, 3° shape, 4° team".

Spot the difference

LevelHeld by
1°peptide
2°H-bonds
3°S–S, ionic
4°subunits

Practice questions

Easy
  1. 1Four levels of protein.
  2. 2α-helix and β-sheet.
  3. 3What is denaturation?
Medium
  1. 4How do enzymes work?
  2. 5Optimum temperature.
  3. 6Haemoglobin's level?
Challenging
  1. 7Level with S–S bonds?
  2. 8Lock-and-key model.
Show answers (try them yourself first)
  1. 11° 2° 3° 4°.
  2. 2Two kinds of 2°.
  3. 3Loss of 2°/3°.
  4. 4By lowering Ea.
  5. 5≈ 310 K.
  6. 64°.
  7. 73°.
  8. 8Substrate fits the active site.

Nucleic acids: nucleotides, DNA and RNA

From daily life

Children inherit parental traits because of information stored in DNA.

Simple definition

Nucleotide = base + pentose sugar + phosphate. DNA: 2-deoxyribose, A G C T. RNA: ribose, A G C U.

DNA / RNANucleic acids: DNA and RNANucleotide = base + pentose sugar + phosphateATGCTACGA=T: 2 H-bonds · G≡C: 3 H-bondsDNA: 2-deoxyribose, A G C T, double helixRNA: ribose, A G C U, single strandPurines: A, G (two rings) · Pyrimidines: C, T, U (one ring)

In the figure : DNA and RNA.

Rule 1: Components — Purines: A, G. Pyrimidines: C, T, U. Nucleoside = base + sugar; nucleotide = + phosphate; linkage: phosphodiester.

  • Easy : T in DNA; U in RNA.
  • Medium : Two rings: A, G.
  • Exam-type : One ring: C, T, U.

Rule 2: DNA double helix — Two complementary strands: A=T (2 H-bonds), G≡C (3 H-bonds); %A = %T and %G = %C (Chargaff).

  • Easy : A = 30 % ⇒ T = 30 %, G = C = 20 %.
  • Medium : 100 pairs (30 A–T + 70 G–C) ⇒ 270 H-bonds.
  • Exam-type : G–C pairs are more stable.

Rule 3: Types of RNA — mRNA (message), tRNA (carries amino acids), rRNA (ribosomes). DNA ⇒ RNA ⇒ protein (transcription, translation).

  • Easy : DNA replication.
  • Medium : tRNA anticodon.
  • Exam-type : Three bases = one codon.

Calculation: H-bonds in a 100-pair DNA piece

  • Step 1: 30 A–T pairs × 2 = 60.
  • Step 2: 70 G–C pairs × 3 = 210.
  • Step 3: Total 60 + 210 = 270.

CET level (MCQ)

  • CET 1) Base absent from RNA? 1) thymine 2) uracil 3) adenine 4) guanine — Answer: 1).
  • CET 2) H-bonds in G≡C? 1) 3 2) 2 3) 1 4) 4 — Answer: 1).

Real-life use: DNA fingerprinting, drugs, vaccines.

Remember

A=T 2; G≡C 3; DNA has T, RNA has U; deoxyribose.

Common mistake

Writing thymine for RNA.

CET trap

DNA = 2-deoxyribose; RNA = ribose (OH at 2').

Shortcut

Memory clue: "A-T two, G-C three".

Spot the difference

FeatureDNARNA
Sugardeoxyriboseribose
BasesA G C TA G C U
Strandstwoone

Practice questions

Easy
  1. 1What is a nucleotide?
  2. 2DNA vs RNA.
  3. 3Chargaff's rule.
Medium
  1. 4A = 30 % ⇒ G?
  2. 5Purines and pyrimidines.
  3. 6Types of RNA.
Challenging
  1. 7Phosphodiester linkage.
  2. 8Why can DNA replicate easily?
Show answers (try them yourself first)
  1. 1Base + sugar + phosphate.
  2. 2Sugar, T/U, strands.
  3. 3A = T, G = C.
  4. 420 %.
  5. 5Two rings vs one ring.
  6. 6m, t, r.
  7. 7Joins sugar and phosphate.
  8. 8Complementary pairing.

Quick revision of the whole lesson

ConceptKey formula / fact
Glucosealdohexose, 16 isomers
Anomersdiffer at C-1; +52.7°
Sucrosenon-reducing; +66.5° ⇒ –20°
Peptiden ⇒ (n–1) bonds
pInet charge 0
Denaturation2°, 3° lost; 1° stays
DNA pairsA=T (2), G≡C (3)
RNAribose, U, single strand
Live figure

🚀 Quick revision of the whole lesson

In the app you can move this figure with your finger.

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🎯 Most important points for the exam

  • Classification.
  • Glucose structure.
  • Ring, anomers.
  • Di/polysaccharides.
  • Amino acids, peptides.
  • Proteins, enzymes.
  • DNA, RNA.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.