Class 12 Biomolecules — The Structures Worth Knowing
Biomolecules looks like a reading chapter, and that is exactly why students lose marks in it. The questions are precise: which sugar is reducing, which linkage joins starch, what survives denaturation, the difference between a nucleoside and a nucleotide. This guide takes the structures that actually get asked, states each one carefully in words and tables, and works the numerical checks that prove you have the formula right.
Carbohydrates — the classification you are graded on
| Class | Meaning | Examples |
|---|---|---|
| Monosaccharide | Cannot be hydrolysed to a simpler sugar | Glucose, fructose, ribose, galactose |
| Oligosaccharide | Gives 2–10 monosaccharide units on hydrolysis | Sucrose, maltose, lactose (all disaccharides) |
| Polysaccharide | Gives a large number of monosaccharide units | Starch, cellulose, glycogen |
| Reducing sugar | Reduces Tollens' and Fehling's reagents — has a free aldehyde or ketone group | All monosaccharides, plus maltose and lactose |
| Non-reducing sugar | Does not reduce those reagents | Sucrose |
Glucose — an aldohexose, described carbon by carbon
Glucose is C₆H₁₂O₆. Its open-chain structure is a straight six-carbon chain:
| Carbon | Group | Evidence in the chapter |
|---|---|---|
| C1 | –CHO (aldehyde) | Gives silver mirror with Tollens', red Cu₂O with Fehling's; oxidised to gluconic acid by mild bromine water |
| C2–C5 | –CHOH (four secondary alcohols) | Forms a pentaacetate with acetic anhydride — proving five –OH groups in total |
| C6 | –CH₂OH (primary alcohol) | With C1, gives saccharic acid on oxidation with nitric acid, showing both ends can be oxidised |
| Chain | Straight, unbranched | On prolonged heating with HI it gives n-hexane |
Why the open chain is not the whole story. Glucose fails some standard aldehyde tests: it gives no reaction with sodium hydrogensulphite or ammonia, it does not form the expected Schiff's test result, and its pentaacetate does not react with hydroxylamine. These observations show that the aldehyde group is not freely available — glucose exists mostly as a six-membered pyranose ring, formed when the –OH on C5 adds across the C1 carbonyl.
Ring closure creates a new stereocentre at C1, giving two forms called α-D-glucose and β-D-glucose. In solution the two interconvert through the open-chain form, so the optical rotation of a freshly prepared solution changes until it settles at an equilibrium value. This is mutarotation: pure α-D-glucose starts at about +111°, pure β-D-glucose at about +19.2°, and both drift to the equilibrium value of about +52.5°.
Fructose, also C₆H₁₂O₆, is a ketohexose — the carbonyl is at C2, not C1 — and it closes into a five-membered furanose ring. Even though it is a ketone, fructose still reduces Tollens' and Fehling's reagents, because in alkaline solution it rearranges to an aldose. That single sentence answers a very common two-mark question.
Disaccharides — the linkage decides everything
| Sugar | Units | Linkage | Reducing? |
|---|---|---|---|
| Sucrose | α-D-glucose + β-D-fructose | C1 of glucose to C2 of fructose | No — both anomeric carbons are used up in the link |
| Maltose | Two α-D-glucose | α-1,4 glycosidic | Yes — the second unit keeps a free anomeric carbon |
| Lactose | β-D-galactose + β-D-glucose | β-1,4 glycosidic | Yes |
Invert sugar. Sucrose is dextrorotatory, about +66.5°. Hydrolysis gives an equimolar mixture of glucose (+52.5°) and fructose (−92.4°). The sign of the mixture flips to negative, which is why hydrolysed sucrose is called invert sugar.
Compute the rotation of the mixture. An equimolar mixture rotates by the average of the two values:
(+52.5) + (−92.4) = −39.9
−39.9 ÷ 2 = −19.95 ≈ −20°
Positive before hydrolysis, negative after — the inversion is a straightforward arithmetic consequence of fructose being much more strongly laevorotatory than glucose is dextrorotatory.
Polysaccharides — same monomer, different linkage
| Polysaccharide | Monomer | Linkage and shape | Role |
|---|---|---|---|
| Starch — amylose | α-D-glucose | α-1,4 only; long unbranched coiled chain; water soluble | Plant food store; the fraction that gives the blue colour with iodine |
| Starch — amylopectin | α-D-glucose | α-1,4 chain with α-1,6 branch points; water insoluble | The branched, major fraction of starch |
| Cellulose | β-D-glucose | β-1,4 only; straight, rigid chains that pack into fibres | Plant cell wall — structural, not food |
| Glycogen | α-D-glucose | Like amylopectin but more highly branched | Animal food store — "animal starch" |
Starch and cellulose are built from the same sugar. The only difference is α versus β at the linking carbon — and that one difference is why humans can digest starch but not cellulose.
Worked example 1 — glucose, sucrose, and a hydrolysis cross-check
Atomic masses: C = 12.011, H = 1.008, O = 15.999.
Glucose, C₆H₁₂O₆
C: 6 × 12.011 = 72.066
H: 12 × 1.008 = 12.096
O: 6 × 15.999 = 95.994
M = 72.066 + 12.096 = 84.162; 84.162 + 95.994 = 180.156 ≈ 180.16 g/mol
Sucrose, C₁₂H₂₂O₁₁
C: 12 × 12.011 = 144.132
H: 22 × 1.008 = 22.176
O: 11 × 15.999 = 175.989
M = 144.132 + 22.176 = 166.308; 166.308 + 175.989 = 342.297 ≈ 342.30 g/mol
Cross-check using the hydrolysis reaction. One sucrose plus one water gives one glucose plus one fructose, and glucose and fructose have the same formula:
Left-hand side: 342.297 + 18.015 = 360.312
Right-hand side: 2 × 180.156 = 360.312 ✓
The two sides match exactly, which confirms both molar masses and the reaction stoichiometry at the same time.
Proteins — from one amino acid to four levels of structure
An α-amino acid carries both an amino group and a carboxyl group on the same carbon:
There are 20 α-amino acids in proteins; about 10 of them cannot be made by the human body and must come from the diet, so they are called essential. In the solid state and in neutral solution an amino acid exists as a zwitterion, ⁺H₃N–CH(R)–COO⁻, which is why amino acids are high-melting, water-soluble solids rather than ordinary organic liquids. At the isoelectric point the zwitterion carries no net charge and does not migrate in an electric field.
Two amino acids join by losing a water molecule between the –COOH of one and the –NH₂ of the other. The resulting –CO–NH– link is the peptide bond.
| Level | What it describes | Held together by |
|---|---|---|
| Primary | The exact sequence of amino acids in the chain | Peptide (covalent) bonds |
| Secondary | Local shape: the right-handed α-helix or the β-pleated sheet | Hydrogen bonds between C=O and N–H groups of the backbone |
| Tertiary | The overall three-dimensional folding of one chain | Hydrogen bonds, disulphide bridges, electrostatic attraction, van der Waals forces |
| Quaternary | How two or more folded chains (subunits) assemble | The same non-covalent forces, acting between subunits |
Denaturation — heating, or adding acid — destroys the secondary and tertiary structure while leaving the primary structure intact, so the protein loses its biological activity. Boiling an egg and curdling milk are the two standard examples.
Worked example 2 — a dipeptide, checked two ways
Glycine, C₂H₅NO₂ (R = H)
C: 2 × 12.011 = 24.022; H: 5 × 1.008 = 5.040; N: 14.007; O: 2 × 15.999 = 31.998
M = 24.022 + 5.040 = 29.062; + 14.007 = 43.069; + 31.998 = 75.067 ≈ 75.07 g/mol
Alanine, C₃H₇NO₂ (R = CH₃)
C: 3 × 12.011 = 36.033; H: 7 × 1.008 = 7.056; N: 14.007; O: 31.998
M = 36.033 + 7.056 = 43.089; + 14.007 = 57.096; + 31.998 = 89.094 ≈ 89.09 g/mol
Route 1 — subtract the water lost. Forming one peptide bond releases one
H₂O:
75.067 + 89.094 = 164.161
164.161 − 18.015 = 146.146 g/mol
Route 2 — build the formula and add it up. The dipeptide glycylalanine
is C₅H₁₀N₂O₃:
C: 5 × 12.011 = 60.055; H: 10 × 1.008 = 10.080; N: 2 × 14.007 = 28.014;
O: 3 × 15.999 = 47.997
60.055 + 10.080 = 70.135; + 28.014 = 98.149; + 47.997 = 146.146 g/mol ✓
Both routes give 146.146 g/mol. When two independent methods agree, the formula is right.
Enzymes and vitamins — the exam-relevant facts
Enzymes are biocatalysts, and almost all of them are proteins. They are extremely specific — one enzyme typically catalyses one reaction — and they work by lowering the activation energy of that reaction, so it proceeds far faster at body temperature than it otherwise would. Because they are proteins, they are destroyed by the same conditions that denature any protein.
Vitamins divide into two groups by solubility, and the grouping explains the storage behaviour:
| Group | Members | Stored in the body? | Named deficiency disease (syllabus) |
|---|---|---|---|
| Fat-soluble | A, D, E, K | Yes — stored in liver and fatty tissue | A: night blindness / xerophthalmia · D: rickets and osteomalacia · K: increased blood-clotting time |
| Water-soluble | B group and C | No — excreted, so a regular supply is needed | B₁: beriberi · B₁₂: pernicious anaemia · C: scurvy |
These deficiency links are syllabus content for a chemistry paper, not medical guidance. Nutrition and any supplement decision belongs to a qualified doctor, never to a textbook list.
Nucleic acids — get these two definitions exactly right
Nucleotide = nitrogenous base + pentose sugar + phosphate group
| DNA | RNA | |
|---|---|---|
| Sugar | 2-deoxy-D-ribose | D-ribose |
| Bases | Adenine, Guanine, Cytosine, Thymine | Adenine, Guanine, Cytosine, Uracil |
| Strands | Double helix, two complementary strands | Usually a single strand |
| Function | Stores and transmits hereditary information | Mainly protein synthesis |
Adenine and guanine are purines (two fused rings); cytosine, thymine and uracil are pyrimidines (one ring). In the double helix the bases pair specifically: A with T through two hydrogen bonds, G with C through three. Because G–C pairs have an extra hydrogen bond, a region rich in G and C is harder to separate — a fact worth remembering as the reason the pairing is not arbitrary.
Mistakes that lose marks in this chapter
- Calling sucrose a reducing sugar. Both anomeric carbons are locked into the glycosidic bond, so there is no free carbonyl. Maltose and lactose are reducing.
- Saying starch and cellulose have different monomers. Both are made of glucose. The difference is α-1,4 versus β-1,4 linkage.
- Swapping nucleoside and nucleotide. The phosphate is what makes it a nucleotide.
- Saying denaturation breaks peptide bonds. It destroys secondary and tertiary structure only; the primary sequence survives.
- Writing that fructose does not reduce Fehling's solution. It does, because it isomerises to an aldose in alkaline medium.
- Forgetting why the cyclic structure was proposed. The evidence is the set of aldehyde tests glucose fails, plus mutarotation.
- Claiming all vitamins are stored in the body. Only the fat-soluble ones are.
Where biomolecules is examined
| Exam | Typical question |
|---|---|
| CBSE Class 12 | Why glucose does not give certain aldehyde tests; reducing vs non-reducing sugars; the four levels of protein structure; nucleoside vs nucleotide |
| ISC Class 12 | Structures of glucose and fructose; hydrolysis products of disaccharides; zwitterion and isoelectric point |
| NEET | High-frequency factual questions on linkages, base pairing and vitamin deficiency |
| IIT-JAM / CUET-PG | Stereochemistry of sugars, optical rotation and peptide sequencing |
Verify every formula in this chapter numerically. Type C6H12O6, C12H22O11, C2H5NO2 or C3H7NO2 into the Molar Mass & Composition calculator and check the values worked out above — then confirm the sucrose hydrolysis balance for yourself: sucrose plus water must equal two glucose units by mass.
Open the Molar Mass & Composition Calculator →Revising the Class 12 organic chapters for boards? ABC Chemistry runs Class 11–12 chemistry coaching at its Gurugram centre and online classes across India — details at abcchemistry.in.