Biomolecules are the chemicals of life. Carbohydrates, proteins, and fats make up the bulk of every living cell, and understanding their chemistry — their structure, classification, and the tests used to identify them — is central to CBSE Class 12, JEE Main, and NEET. This guide builds from structural basics to exam-level analytical chemistry.
Key Terms
Monosaccharide: The simplest carbohydrate unit; cannot be hydrolysed further. Examples: glucose, fructose, galactose.
Disaccharide: Two monosaccharides joined by a glycosidic bond. Examples: sucrose, maltose, lactose.
Polysaccharide: Many monosaccharides linked in long chains. Examples: starch, glycogen, cellulose.
Amino acid: The monomer of proteins. Contains an amino group (−NH₂) and a carboxyl group (−COOH) attached to the same carbon (α-carbon).
Peptide bond: The amide bond (−CO−NH−) formed between the carboxyl group of one amino acid and the amino group of another, with loss of water.
Fatty acid: A long hydrocarbon chain with a carboxyl group (−COOH) at one end. Can be saturated (no double bonds) or unsaturated (one or more double bonds).
Triglyceride: A glycerol molecule esterified with three fatty acids. The main form of stored fat.
Saponification: Hydrolysis of ester bonds in fats with alkali (NaOH), yielding glycerol and soap (fatty acid salts).
Carbohydrates
Classification
Monosaccharides:
- Glucose (C₆H₁₂O₆): Most important monosaccharide; primary fuel for cellular respiration. Exists in open-chain (Fischer projection) and ring (Haworth projection) forms.
- Fructose: Isomer of glucose; sweeter taste; found in fruit and honey.
- Ribose/Deoxyribose: 5-carbon sugars (pentoses); components of RNA and DNA respectively.
Classification by carbonyl group:
- Aldoses: aldehyde group (−CHO) at C1. Example: glucose (aldohexose).
- Ketoses: ketone group at C2. Example: fructose (ketohexose).
Disaccharides:
| Disaccharide | Monomers | Glycosidic bond | Reducing? |
|---|---|---|---|
| Maltose | Glucose + Glucose | α(1→4) | Yes |
| Lactose | Glucose + Galactose | β(1→4) | Yes |
| Sucrose | Glucose + Fructose | α,β(1→2) | No |
Sucrose is a non-reducing sugar because its glycosidic bond involves both anomeric carbons (the C1 of glucose and C2 of fructose), leaving no free aldehyde or ketone group.
Polysaccharides:
- Starch: Storage polysaccharide in plants. Two components: amylose (linear, α(1→4) linkages) and amylopectin (branched, α(1→4) main chain with α(1→6) branch points every ~25 glucose units).
- Glycogen: Storage polysaccharide in animals (liver, muscles). More branched than amylopectin.
- Cellulose: Structural polysaccharide in plant cell walls. β(1→4) linkages create rigid, unbranched chains that humans cannot digest (no β-glucosidase).
Chemical Tests for Carbohydrates
Fehling’s test / Benedict’s test: Detects reducing sugars (free aldehyde or ketone). Reducing sugars reduce Cu²⁺ (blue) to Cu₂O (brick-red precipitate). Positive: glucose, fructose, maltose, lactose. Negative: sucrose (non-reducing).
Tollens’ test (silver mirror test): Aldehyde reduces silver ions to metallic silver (mirror). Positive for aldoses; fructose gives a positive test too (mild alkaline conditions cause rearrangement to aldose).
Iodine test: Starch gives a characteristic blue-black colour with iodine (I₂/KI). Glycogen gives red-brown; cellulose gives no colour change. Used to distinguish starch from other polysaccharides.
Molisch’s test: General test for carbohydrates. All carbohydrates give a purple ring at the interface of the sample and H₂SO₄ (due to furfural/hydroxymethylfurfural condensation with α-naphthol).
Proteins
Amino Acids — The Building Blocks
Twenty standard amino acids are encoded by the genetic code. Each has:
- A central α-carbon
- An amino group (−NH₂)
- A carboxyl group (−COOH)
- A hydrogen atom
- A side chain (R group) — this is what makes each amino acid unique
Classification by R group:
- Non-polar/hydrophobic (Gly, Ala, Val, Leu, Ile, Pro, Phe, Trp, Met): tend to cluster in protein interior
- Polar/uncharged (Ser, Thr, Cys, Asn, Gln, Tyr): form hydrogen bonds
- Acidic (Asp, Glu): negatively charged at physiological pH
- Basic (Lys, Arg, His): positively charged at physiological pH
Essential amino acids (9, must come from diet): Histidine, Isoleucine, Leucine, Lysine, Methionine, Phenylalanine, Threonine, Tryptophan, Valine. Mnemonic: “His Ile Leu Lys Met Phe Thr Trp Val.”
Protein Structure Levels
Primary structure: The sequence of amino acids linked by peptide bonds. Determined by DNA sequence. The complete amino acid sequence of a protein is its “primary structure.”
Secondary structure: Regular local conformations stabilised by hydrogen bonds between backbone NH and C=O groups. Two types:
- α-helix: Coiled structure; H-bonds between residue and
- β-pleated sheet: Extended strands hydrogen-bonded side by side (parallel or antiparallel)
Tertiary structure: Overall 3D folding of the entire polypeptide chain. Stabilised by:
- Disulphide bonds (−S−S−, between cysteine residues) — covalent
- Hydrophobic interactions
- Hydrogen bonds
- Ionic bonds (between acidic and basic side chains)
Quaternary structure: Arrangement of multiple polypeptide chains (subunits) into a functional complex. Example: haemoglobin (2α + 2β subunits).
Denaturation
Denaturation is the disruption of protein secondary, tertiary, or quaternary structure without breaking peptide bonds (primary structure is preserved). Caused by heat, extremes of pH, heavy metal ions, organic solvents.
Example: Cooking an egg denatures the albumin (egg white protein) from soluble and transparent to solid and white. The primary amino acid sequence is unchanged; the folding is destroyed.
Denaturation is usually irreversible (egg white can’t be “uncooked”).
Chemical Tests for Proteins
Biuret test: Detects peptide bonds (≥2 peptide bonds needed). Protein + NaOH + CuSO₄ (dilute) → violet/purple colour (Cu²⁺ coordinates with N atoms of peptide bonds). Used for quantitative protein determination.
Ninhydrin test: Detects free amino groups (α-amino acids and proteins). Amino acid + ninhydrin → purple colour (Ruhemann’s purple). Proline gives yellow-orange. Used in chromatography to detect amino acids.
Xanthoproteic test: Detects aromatic amino acids (Phe, Tyr, Trp). Protein + conc. HNO₃ → yellow colour (nitration of aromatic ring). On adding NaOH → orange colour. This is why skin turns yellow when it contacts concentrated nitric acid.
Fats (Lipids)
Structure of Fats
Fats are esters of glycerol with fatty acids (triglycerides):
Glycerol (3-OH groups) + 3 Fatty acids → Triglyceride + 3 H₂O
Fatty acids:
- Saturated: No C=C double bonds. Long chains pack closely → solid at room temperature. Examples: palmitic acid (C16), stearic acid (C18). Found in animal fats (butter, ghee, lard).
- Unsaturated (mono/poly): One or more C=C double bonds. Less dense packing → liquid at room temperature (oils). Examples: oleic acid (C18:1), linoleic acid (C18:2), linolenic acid (C18:3). Found in vegetable oils, fish.
Essential fatty acids: Linoleic acid (omega-6) and α-linolenic acid (omega-3) cannot be synthesised in the body — they must come from diet.
Chemical Tests for Fats
Saponification (soap test): Fat + NaOH → glycerol + soap (sodium salt of fatty acid). The soapy feeling and foam confirm the presence of fat.
Acrolein test: Glycerol (or fats containing it) when heated with anhydrous potassium hydrogen sulphate (KHSO₄) → acrolein (a pungent, irritating aldehyde). Confirmed by the characteristic smell.
Sudan III / Oil Red O test: Lipids stain red/orange with these dyes. Used in histology to identify fat droplets in tissues.
Grease spot test (simple): Press fat on paper and hold to light — a translucent grease spot confirms fat presence (fat doesn’t evaporate unlike water).
Hydrogenation of Oils
Unsaturated vegetable oils can be converted to solid fats (vanaspati) by catalytic hydrogenation: adding H₂ across C=C double bonds in the presence of Ni catalyst at ~200°C. This increases melting point, making the fat solid at room temperature.
Trans fats: Industrial hydrogenation produces some trans isomers (trans-unsaturated fatty acids) as a side product. Trans fats raise LDL (“bad cholesterol”) and lower HDL (“good cholesterol”), increasing heart disease risk — why trans fats are now regulated in food products.
Solved Examples
Example 1 — Reducing vs non-reducing sugar (JEE Main level)
Explain why sucrose does not reduce Fehling’s solution but maltose does.
Solution: Maltose has a free anomeric carbon (C1 of the second glucose unit is in hemiacetal form — it has a free −OH that can open to form an aldehyde in solution). This free aldehyde reduces Cu²⁺ in Fehling’s reagent.
Sucrose is a 1,2’-glycoside — the glycosidic bond connects the C1 of glucose to the C2 of fructose. Both anomeric carbons are involved in the bond. No free aldehyde or ketone can form. Therefore sucrose cannot reduce Fehling’s solution.
Example 2 — Protein test identification (CBSE Class 12)
A student added concentrated HNO₃ to a protein solution and observed yellow colour, which turned orange on adding NaOH. Which amino acids are likely present?
Solution: Yellow colour with concentrated HNO₃ → xanthoproteic test positive → aromatic amino acids present. The candidates are phenylalanine (Phe), tyrosine (Tyr), and tryptophan (Trp). The colour change to orange with NaOH confirms it’s the xanthoproteic reaction (nitration of aromatic ring producing nitrophenyl groups, which ionise in alkali to give orange quinonoid structures).
Example 3 — Saponification number (JEE Main level)
Define saponification number and explain what a high saponification number indicates.
Solution: The saponification number is the number of milligrams of KOH required to saponify (hydrolyse) 1 gram of fat completely.
Since each triglyceride molecule requires 3 moles of KOH for complete hydrolysis, and higher-molecular-weight fats need fewer moles per gram, a high saponification number indicates lower molecular weight fatty acids (short-chain fatty acids). Butter (containing significant short-chain fatty acids) has a higher saponification number than linseed oil (long-chain fatty acids).
Exam-Specific Tips
JEE Main: Expect 1–2 questions per paper on biomolecules. High-probability topics: reducing vs non-reducing sugars (Fehling’s test), protein secondary structure (α-helix vs β-sheet), essential amino acids, saponification. The chapter is entirely conceptual in JEE — no lengthy calculations.
NEET: Biomolecules overlap significantly with Biology (protein structure, DNA). Chemistry biomolecules questions focus on: classification (mono/di/polysaccharide), chemical tests (Biuret, Tollens, Fehling’s), denaturation, and the differences between starch, glycogen, and cellulose. Know the iodine test for starch cold.
Common Mistakes to Avoid
Mistake 1 — Sucrose is a reducing sugar. It is not. Sucrose is non-reducing because both anomeric carbons are in the glycosidic bond. Glucose, fructose, maltose, and lactose ARE reducing sugars. This is probably the #1 error in biomolecule MCQs.
Mistake 2 — Primary structure includes H-bonds. Primary structure is ONLY the amino acid sequence (peptide bonds). H-bonds contribute to secondary structure (α-helix, β-sheet). Disulphide bonds stabilise tertiary structure. Don’t assign structural features to the wrong level.
Mistake 3 — Cellulose is digestible by humans. It is not — humans lack the β-glucosidase enzyme needed to break β(1→4) linkages. Starch (α-linkages) is digestible; cellulose (β-linkages) passes through as dietary fibre.
Mistake 4 — Denaturation destroys primary structure. Denaturation only disrupts secondary, tertiary, and quaternary structure. The peptide bonds (primary structure) remain intact during denaturation. The protein’s amino acid sequence is unchanged; only the folding is disrupted.
Practice Questions
Q1. What is the product when sucrose is hydrolysed? Is the hydrolysate a reducing or non-reducing sugar?
Sucrose + H₂O (acid or sucrase enzyme) → glucose + fructose. Both products are reducing sugars (both have free anomeric carbons). The hydrolysate (invert sugar) gives a positive Fehling’s test even though sucrose itself does not.
Q2. A fat has an iodine value of 90. Is it likely to be solid or liquid at room temperature? What does this indicate about its degree of unsaturation?
Iodine value measures the grams of I₂ absorbed by 100 g of fat (I₂ adds across C=C double bonds). Iodine value 90 means significant unsaturation — the fat is likely liquid (oil) at room temperature. High iodine value = more double bonds = more unsaturated = lower melting point = liquid/oil. Saturated fats have low iodine values (butter: ~30); highly unsaturated oils have high values (linseed oil: ~180).
Q3. Explain why the Biuret test doesn’t give a positive result with single amino acids.
The Biuret test requires at least two peptide bonds (−CO−NH−) for Cu²⁺ coordination. A single amino acid has no peptide bond — it has a free amino group and a free carboxyl group. Dipeptides (one peptide bond) also give a weak or negative response. Tripeptides and above give the characteristic violet-purple colour. Named “Biuret” after the compound H₂N−CO−NH−CO−NH₂ (biuret), which has two amide groups and gives a positive test.
FAQs
What is the difference between starch and glycogen? Both are polysaccharides made of glucose with α(1→4) main-chain linkages. Glycogen has more branching — α(1→6) branch points occur every ~10 glucose units (compared to every ~25 for amylopectin in starch). Glycogen is the animal storage form; starch is the plant storage form.
Why are unsaturated fats liquid at room temperature? The C=C double bonds (especially cis configuration) create “kinks” in the fatty acid chain that prevent close packing of adjacent chains. This reduces intermolecular van der Waals forces and lowers the melting point, making the fat liquid (oil) at room temperature.
Are all 20 amino acids essential? No — only 9 are essential (cannot be synthesised by humans and must come from diet). The other 11 are non-essential (the body can synthesise them), though under certain conditions (illness, rapid growth), some become “conditionally essential.”
What is the difference between a peptide and a protein? A peptide is a short chain of amino acids (typically fewer than 50–100 residues). A protein is a polypeptide chain long enough to fold into a stable 3D structure with a specific function. The distinction is somewhat arbitrary — insulin (51 amino acids) is typically called a protein; a dipeptide is clearly a peptide.
Why does cooking denature egg white but not scramble its primary structure? Heat breaks weak interactions (H-bonds, hydrophobic interactions) that maintain the protein’s folded structure, but does not provide enough energy to break covalent peptide bonds. The amino acid sequence (primary structure) is encoded in DNA and can only be changed by mutation — cooking doesn’t alter DNA.