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Biomolecules
Chemistry · Class 12NCERT pp. 281–3012 sets × 25 questions
Key facts and formulas for Biomolecules, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Key facts for this chapter
| Idea | Rule or fact |
|---|
| Glucose oxidation | Bromine water: CHO → COOH (gluconic acid). Nitric acid: CHO and CH₂OH → saccharic acid. HI heat: n-hexane. |
| D and L | Compare the stereocentre farthest from CHO with glyceraldehyde. D/L is configuration, not the sign of rotation. |
| Anomers and epimers | Anomers differ at C-1 only (α, β). Epimers differ at one other centre: glucose and galactose at C-4. |
| Ring forms | Glucose: pyranose (six-membered). Fructose: furanose (C-5 OH attacks C-2). Pyranose glucose has 5 chiral carbons; open chain has 4. |
| Disaccharides | Sucrose: α-Glc + β-Fru, C1–C2, non-reducing. Maltose: Glc + Glc, C1–C4. Lactose: Gal + Glc, C1–C4. |
| Polysaccharides | Amylose 15–20%, linear, α C1–C4. Amylopectin 80–85%, branched C1–C6. Cellulose: β-glucose, C1–C4. |
| Peptides | Peptide bonds = residues − 1. Hydrolysis needs one water per bond. Biuret needs 2 or more bonds. |
| Amino acids | Zwitter ion has NH₃⁺ and COO⁻. Essential: valine, lysine, tryptophan. Ninhydrin gives purple colour. |
| Vitamins | B2 cheilosis, B6 convulsions, B12 pernicious anaemia, D rickets. B12 is the water soluble vitamin that is stored. |
| Nucleic acids | Chargaff: A = T, G = C, so (A + G)/(T + C) = 1. A–T has 2 H-bonds, G–C has 3. Hydrolysis gives 8 distinct products. |
Three sample questions
Q1. Which of the following fits the formula Cₓ(H₂O)ᵧ but is NOT a carbohydrate?
- Glucose, C₆H₁₂O₆
- Acetic acid, C₂H₄O₂
- Fructose, C₆H₁₂O₆
- Sucrose-derived hexose, C₆H₁₂O₆
Hint and solution
Hint: A carbohydrate must be a polyhydroxy aldehyde or ketone. Which option has no such groups?
- Carbohydrate = polyhydroxy aldehyde/ketone — A carbohydrate is defined by its groups, not only by the formula Cₓ(H₂O)ᵧ.
- Glucose, fructose: C₆H₁₂O₆ = C₆(H₂O)₆ — Glucose and fructose have many OH groups plus CHO or C=O, so they qualify.
- CH₃COOH = C₂(H₂O)₂ — Acetic acid fits the formula on paper but has only COOH and CH₃, with no polyhydroxy aldehyde or ketone.
- Answer: acetic acid — So the formula is only a rough guide, and the exception is acetic acid. Option B.
Answer: B. Acetic acid can be written C₂(H₂O)₂ but is not a polyhydroxy aldehyde or ketone. The formula is only a rough guide.
Q2. Oxidation of glucose with bromine water and with nitric acid gives, respectively,
- saccharic acid and gluconic acid
- gluconic acid and gluconic acid
- hexanoic acid and saccharic acid
- gluconic acid and saccharic acid
Hint and solution
Hint: Bromine water is a mild oxidant that touches only the CHO group. Nitric acid is strong and also oxidises the terminal CH₂OH.
- Br₂/H₂O: CHO → COOH (mild) — Bromine water is a mild oxidant and converts only the aldehyde group to a carboxylic acid.
- Product: gluconic acid (one COOH) — So glucose gives gluconic acid, which has one COOH and a CH₂OH at the end.
- HNO₃: CHO and CH₂OH → 2 COOH — Nitric acid is a strong oxidant and oxidises both ends of the chain.
- Gluconic acid, then saccharic acid — The order asked is bromine water first, then nitric acid. Option D.
Answer: D. Bromine water oxidises only CHO to COOH (one COOH). Nitric acid also oxidises the terminal CH₂OH (two COOH).
Q3. Statement I: Prolonged heating of glucose with HI gives n-hexane.
Statement II: Glucose pentaacetate reacts with hydroxylamine to give an oxime.
- Statement I is correct but Statement II is incorrect
- Both statements are correct
- Statement I is incorrect but Statement II is correct
- Both statements are incorrect
Hint and solution
Hint: One statement shows a straight chain. For the other, ask whether a pentaacetate still has a free CHO group.
- Glucose + HI (heat) → n-hexane — Prolonged heating with HI reduces glucose fully to n-hexane, showing six carbons in a straight chain. Statement I is correct.
- Glucose + (CH₃CO)₂O → pentaacetate — Five OH groups are acetylated. The ring hemiacetal OH is also tied up.
- Pentaacetate + NH₂OH → no oxime — An oxime needs a free CHO. The pentaacetate has no free CHO, so it gives no oxime. Statement II is incorrect.
- I correct, II incorrect — So only the first statement holds. Option A.
Answer: A. n-Hexane proves a straight chain. The pentaacetate does not form an oxime, which suggests the CHO is not free (ring form).
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.