GOC · Hydrocarbons · Functional Groups · Biomolecules & Polymers
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Purification & Characterisation
Purification methods · Lassaigne's test · quantitative estimation · formula determination
1. Purification Methods
Method
Principle
Typical use
Crystallisation
Differential solubility in hot vs cold solvent
Sugar, aspirin, amino acids
Sublimation
Solid → vapour → solid, bypassing the liquid phase
Camphor, naphthalene, iodine, NH₄Cl, benzoic acid
Distillation
Separation by boiling point
Simple (wide bp gap), fractional (close bp), steam (immiscible + thermolabile), vacuum (high bp + thermolabile)
Solvent extraction
Partition between two immiscible solvents
Extracting a solute into ether; many small extractions beat one large
Chromatography
Differential migration over a stationary phase
Column, TLC, paper, gas (GLC)
For TLC/paper chromatography the retardation factor is \(R_f = \dfrac{\text{distance moved by solute}}{\text{distance moved by solvent front}}\), always between 0 and 1. A higher \(R_f\) means the compound is less adsorbed by the stationary phase.
2. Qualitative Analysis — Lassaigne's Sodium Fusion Test
Fusing the compound with sodium converts covalent N, S and halogens into ionic forms (NaCN, Na₂S, NaX) detectable in the water extract (SFE):
Element
Reagent
Observation
Nitrogen
FeSO₄, then conc. H₂SO₄
Prussian blue
Sulphur
Sodium nitroprusside
Violet colour
Chlorine
AgNO₃ (after dil. HNO₃)
White ppt (AgCl), soluble in dil. NH₃
Bromine
AgNO₃
Pale-yellow ppt (AgBr), soluble in conc. NH₃
Iodine
AgNO₃
Yellow ppt (AgI), insoluble in NH₃
3. Quantitative Analysis
Carbon & hydrogen: Liebig combustion over CuO; CO₂ absorbed in KOH, H₂O in anhydrous CaCl₂.
Nitrogen: Dumas method (measure N₂ volume) or Kjeldahl method (back-titrate the liberated NH₃). Kjeldahl fails for N in aromatic rings, nitro, azo and N–N / N–O compounds.
Halogens & sulphur: Carius method — halogen precipitated as AgX, sulphur as BaSO₄.
💡Exam tip: if both N and S are present, fusion gives NaSCN and the nitrogen Prussian-blue test can fail — boil the extract with excess sodium before testing for nitrogen.
⚠Common trap: a lower \(R_f\) means the compound is more strongly adsorbed on the stationary phase (it moved less), not less.
Ring nitrogen (pyridine), nitro, azo and N–N / N–O compounds do not convert to ammonium sulphate under Kjeldahl digestion; use the Dumas method instead.
Four effects govern electron distribution and reactivity in organic molecules:
Inductive Effect (–I / +I)
Through-σ-bonds electron withdrawal (–I: F, Cl, NO₂, CN) or donation (+I: alkyl, O⁻, COO⁻). Decreases rapidly with distance — negligible beyond 3 bonds.
Order –I: F > Cl > Br > I > OH > OR > NH₂ > C₆H₅ > H
Order +I: C(CH₃)₃ > CH(CH₃)₂ > C₂H₅ > CH₃ > H
Resonance (Mesomeric) Effect (–R / +R)
Through-π-system delocalization. –R groups withdraw (NO₂, CN, CHO, COOH, COR), +R groups donate (OH, OR, NH₂, NR₂, halogens via lone pairs).
Conditions: atoms must be coplanar, conjugated system required. Resonance structures differ only in electron positions — not atom positions.
Hyperconjugation (Baker–Nathan)
σ(C–H) electrons of an α-C overlap with adjacent empty/π orbital. Stabilises carbocations, free radicals, alkenes.
Stability ∝ number of α-H: (CH₃)₃C⁺ (9H) > (CH₃)₂CH⁺ (6H) > CH₃CH₂⁺ (3H)
Alkene stability: more substituted = more hyperconjugation = more stable.
Electromeric Effect (E)
Complete, instantaneous π-electron transfer to one atom under reagent attack. Temporary (unlike resonance). +E towards attacking electrophile; –E towards attacking nucleophile.
Key principle: NH₂ basicity depends on lone pair availability on N. Resonance donation into ring (–R) delocalises lone pair → decreases basicity. EWG on ring further decrease basicity; EDG increase it.
CH₃NH₂: no aromatic ring; +I of CH₃ increases e⁻ density on N → most basic (pKₐ of conjugate acid ≈ 10.6)
p-CH₃–C₆H₄–NH₂: CH₃ is +I/+H on ring → slightly more e⁻ density on N than unsubstituted aniline → more basic than aniline
C₆H₅NH₂ (aniline): lone pair delocalised into ring via +R → pKₐ ≈ 4.6
p-O₂N–C₆H₄–NH₂: NO₂ is –R, –I → strongly withdraws e⁻ from N via two resonance paths → least basic (pKₐ ≈ 1.0)
Order (decreasing basicity): CH₃NH₂ > p-toluidine > aniline > p-nitroaniline
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GOC — Key Diagrams
Click any diagram to zoom
Electronic Effects in Benzene Substituents
Activating (o/p-directing) vs. deactivating (m-directing) groups in electrophilic aromatic substitution. Halogens are unique: –I effect dominates (deactivate ring) but +R directs to o/p.
Allylic Carbocation — Resonance Delocalization
Allylic carbocation has positive charge distributed over C1 and C3 (with C2 carrying the double bond), making it more stable than a simple 2° carbocation.
Keto–Enol Tautomerism
Tautomerism is a special type of structural isomerism involving proton shift + π-bond migration. Keto form typically dominates except in β-diketones (acetylacetone ~80% enol due to intramolecular H-bonding).
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GOC — Quick Quiz
Question 1
What is the DBE (degree of unsaturation) of pyridine C₅H₅N?
Allylic carbocation is stabilised by resonance — the +ve charge is delocalized over C1 and C3. This resonance stabilization exceeds the hyperconjugation stabilization of even a 3° carbocation in most cases (allylic ≈ benzylic >> 3°).
Question 3
Arrange in increasing basicity (weakest first): aniline, p-nitroaniline, p-methylaniline, methylamine
p-NO₂ group (–R, –I) heavily withdraws electron density from N → least basic. Aniline: lone pair in resonance with ring → weak base. p-CH₃: +I/+H donates e⁻ slightly → more basic than aniline. CH₃NH₂: no aromatic ring, +I of CH₃ → most basic. Correct: p-NO₂–aniline < aniline < p-CH₃–aniline < CH₃NH₂.
Question 4
Which statement about tautomerism is CORRECT?
Tautomers are structural isomers (not stereoisomers) that interconvert readily — they differ in both proton position AND π-bond position. The keto form and enol form have different connectivity. In resonance structures, atoms don't move — only electrons do. Option D correctly describes keto–enol tautomerism.
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Hydrocarbons
Alkanes · Alkenes · Alkynes · Arenes · Aromaticity · EAS
Alkanes — Nomenclature and Free Radical Halogenation
General formula CₙH₂ₙ₊₂ (acyclic). Sp³ hybridised, tetrahedral. Only σ bonds.
Selectivity (BDE of C–H): 3° > 2° > 1° C–H bonds — 3° radicals most stable. Cl• is less selective (Cl₂ reactive); Br• is highly selective (Br₂ more stable, less reactive).
Anti-Markovnikov product in free radical addition of HBr to alkenes (peroxide/ROOR condition): Br• adds to less substituted C.
Alkenes — Addition Reactions
Reaction
Reagent
Product / Rule
Hydrohalogenation
HX
Markovnikov: H adds to C with more H; X to more substituted C
Hydration
H₂O / H⁺ (acid catalysis)
Markovnikov; forms alcohol
Halogenation
Br₂/CCl₄
Anti addition (trans); bromonium ion intermediate
Ozonolysis (reductive)
O₃ then Zn/H₂O
Cleaves C=C → aldehydes/ketones
Ozonolysis (oxidative)
O₃ then H₂O₂
Aldehydes oxidised to carboxylic acids
Epoxidation
mCPBA or H₂O₂/TiCl₄
Syn addition; forms epoxide
Hydroboration-oxidation
BH₃ then H₂O₂/OH⁻
Anti-Markovnikov; syn addition; alcohol
Alkynes — Key Reactions
Terminal alkynes: Acidic C≡C–H (\(pK_a \sim 25\)). React with NaNH₂, Na metal, AgNO₃ → Ag salt (white ppt), CuCl → Cu salt (red ppt) — diagnostic tests for terminal alkynes.
Hydration: H₂O/H₂SO₄/HgSO₄ (Markovnikov addition) → enol → tautomerises to ketone. Exception: CH≡CH → ethanal (CH₃CHO, anti-Markovnikov product if Wacker).
HX addition: Two additions possible (Markovnikov each time) → geminal dihalide (1,1-dihalide).
Lindlar catalyst: H₂/Pd-BaSO₄ (poisoned) → syn addition → cis alkene. Na/NH₃(l): anti addition → trans alkene (Birch-like).
Arenes — Aromaticity and EAS
Hückel rule: Aromatic if cyclic, planar, fully conjugated, with (4n+2) π electrons.
Benzene: 6π (n=1). Naphthalene: 10π. Pyridine: 6π (N sp², lone pair not in π system). Pyrrole: 6π (N sp³-like, lone pair IN π system). Cyclobutadiene: 4π (anti-aromatic).
EAS mechanism: Electrophile (E⁺) attacks benzene ring → arenium (Wheland) intermediate → deprotonation restores aromaticity.
EAS Reaction
Reagents
Key Point
Nitration
HNO₃ + H₂SO₄ (conc.)
Electrophile: NO₂⁺ (nitronium ion)
Halogenation
Cl₂/Br₂ + Lewis acid (FeCl₃/FeBr₃)
Electrophile: Br⁺ or Cl⁺ (via Br–FeBr₃)
Sulfonation
H₂SO₄ (fuming), SO₃
Reversible; SO₃ is electrophile
Friedel–Crafts alkylation
RCl + AlCl₃
Carbocation R⁺ may rearrange; polyalkylation possible
Friedel–Crafts acylation
RCOCl + AlCl₃
Acylium RCO⁺; no rearrangement; deactivates ring (no polyacylation)
Directing Effects in EAS
Group Type
Direction
Ring Activity
Examples
o/p directors
ortho and para
Activating
–OH, –OR, –NH₂, –NHR, –CH₃, halogens (weak via +M/+I)
o/p directors (halogen)
ortho and para
Deactivating (−I > +M)
–F, –Cl, –Br, –I
m directors
meta
Deactivating
–NO₂, –CN, –CHO, –COOH, –SO₃H, –COR, –NR₃⁺
💡JEE Tip: Markovnikov's rule — H adds to carbon with more H atoms (more H → more stable carbocation forms). Anti-Markovnikov occurs only with: (1) HBr + peroxide/ROOR (free radical), (2) BH₃ hydroboration. Remember: HCl and HI do NOT show anti-Markovnikov even with peroxides.
⚠Common Error: Halogens (–F, –Cl, –Br, –I) are o/p directors yet deactivating — they direct CORRECTLY to o/p via +M resonance, but deactivate via their −I inductive effect. Students often forget this dual character and say halogens are meta directors.
n=1: 4π (cyclobutadiene, cyclopentadienyl cation). Anti-aromatic species are highly reactive or non-existent.
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Hydrocarbons — Solved Examples
Example 1EasyDegree of Unsaturation and Structure
Compound A: C₈H₁₀. Compound B: C₅H₈. Compound C: C₆H₆ (IR shows no C–H stretch above 3000 cm⁻¹, NMR: 1 singlet). Calculate DBE for each and predict structure type.
C₈H₁₀: DBE = (2×8 + 2 − 10)/2 = (18−10)/2 = 4. DBE 4 → one benzene ring (4 DBE = 3 double bonds + 1 ring). C₈H₁₀ is a dimethylbenzene (xylene) — 3 isomers: o-, m-, p-xylene.
C₅H₈: DBE = (10+2−8)/2 = 2. DBE 2 → either 2 double bonds, 1 triple bond, or 1 double bond + 1 ring. Could be cyclopentene (1 ring + 1 C=C) or pentadiene or pent-1-yne.
C₆H₆ with DBE = (12+2−6)/2 = 4. No C–H stretch above 3000 cm⁻¹ = no sp3 C–H (all sp² or sp C–H). 1 singlet NMR = all 6 H equivalent → benzene. 6 equivalent aromatic H, DBE = 4 ✓.
Example 2MediumEAS — Directing Effects and Product Prediction
Predict the major product of nitration of: (a) toluene, (b) chlorobenzene, (c) nitrobenzene. State the reasoning.
Toluene (–CH₃): –CH₃ is o/p director (hyperconjugation and inductive +I) and activating. Major products: o-nitrotoluene + p-nitrotoluene. Para is preferred (less steric hindrance). Major: p-nitrotoluene.
Chlorobenzene (–Cl): –Cl is o/p director (resonance donation +M) but deactivating (−I > +M). Major product: p-nitrochlorobenzene (para favoured over ortho due to steric effect; ring deactivated so harsher conditions needed).
Nitrobenzene (–NO₂): –NO₂ is m director and strongly deactivating (electron withdrawal by both −M and −I). Major product: m-dinitrobenzene. Very slow reaction; requires conc. HNO₃ + H₂SO₄ at elevated temperature.
Example 3HardOzonolysis — Identifying Alkene Structure
An alkene C₅H₁₀ on reductive ozonolysis gives: propanal (CH₃CH₂CHO) and ethanal (CH₃CHO). Identify the alkene and give its IUPAC name.
Reconnect at the C=C bond: CH₃CH₂CH=CHCH₃. This is pent-2-ene (C₅H₁₀ ✓).
Check: DBE = (10+2−10)/2 = 1 ✓ (one double bond). IUPAC name: pent-2-ene. Can be cis or trans depending on substituents.
✓ General method: Each ozonolysis fragment =C gives one R–CHO (or R₂C=O). Reconnect the C=O carbons from each fragment with a C=C double bond to rebuild the original alkene.
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Hydrocarbons — Visual Concepts
Click any diagram to zoom
EAS Mechanism — Benzene Nitration (Arenium Ion Intermediate)
EAS Mechanism (benzene nitration). Step 1 (slow, rate-determining): NO₂⁺ attacks ring → sp³ carbon in arenium (Wheland) intermediate, aromaticity disrupted. Step 2 (fast): base removes H⁺ → aromaticity restored. The slow step (attack) is why activating substituents speed up EAS and deactivating groups slow it.
Markovnikov vs Anti-Markovnikov HBr Addition
Same starting material (propene + HBr) gives opposite regiochemistry based on mechanism. Ionic (no peroxide): Markovnikov → Br on C2 (2-bromopropane). Radical (ROOR peroxide): Anti-Markovnikov → Br on C1 (1-bromopropane). Note: this switch only works for HBr — NOT HCl or HI.
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Hydrocarbons — Quick Quiz
Question 1
Propene reacts with HBr in the presence of organic peroxide (ROOR). The major product is:
Peroxides initiate free radical mechanism. Br• adds to the less substituted carbon (gives more stable secondary radical at C2, but Br• attacks C1 to give 1-bromopropane via termination). Major product = 1-bromopropane (anti-Markovnikov). This peroxide effect applies only to HBr. Answer: B ✓
Question 2
Which of the following is NOT aromatic according to Hückel's rule?
Hückel: (4n+2) π electrons for aromaticity. Cyclobutadiene has 4π electrons (n=1 → 4n=4, NOT 4n+2=6) → anti-aromatic. Anti-aromatic compounds are highly unstable. Cyclobutadiene exists only briefly as a reactive intermediate. Answer: C ✓
Question 3
Nitrobenzene undergoes EAS at the meta position because the –NO₂ group:
–NO₂ withdraws electrons by −M (resonance: O=N=O draws e⁻ from ring via delocalization) and −I (inductive). This withdrawal is strongest at ortho/para → those positions are most depleted. Meta positions have relatively MORE electron density → EAS attack occurs there. Answer: A ✓
Nature of the C–X bond: polar (halogen more electronegative than carbon), so the carbon is electrophilic and open to nucleophilic attack. Bond strength falls C–F > C–Cl > C–Br > C–I, so iodides are the most reactive.
Substitution mechanisms:SN1 — two steps via a carbocation, favoured by 3° halides and polar protic solvents, gives racemisation; SN2 — one step, backside attack, favoured by 1° halides, gives inversion (Walden inversion).
Haloarenes are far less reactive to nucleophilic substitution than haloalkanes (partial double-bond character from resonance + sp² carbon).
Uses & environmental effects:chloroform (CHCl₃) — solvent/former anaesthetic, oxidises in air/light to poisonous phosgene; iodoform (CHI₃) — antiseptic; freons (CFCs) — refrigerants that deplete the ozone layer; DDT — a persistent insecticide that bioaccumulates and is now largely banned.
Acidity: phenol (\(pK_a \approx 10\)) is more acidic than alcohols because the phenoxide ion is resonance-stabilised. Electron-withdrawing groups (–NO₂, especially ortho/para) increase acidity; electron-donating groups decrease it.
Electrophilic substitution: –OH is ortho/para-directing and activating — halogenation (Br₂/H₂O → 2,4,6-tribromophenol), nitration and sulphonation occur readily.
Reimer–Tiemann reaction: phenol + CHCl₃ + NaOH → ortho-hydroxybenzaldehyde (salicylaldehyde), via a dichlorocarbene intermediate.
Kolbe's reaction: sodium phenoxide + CO₂ (under pressure) → sodium salicylate → salicylic acid (used to make aspirin).
Aldehydes and Ketones
Nucleophilic addition to C=O: Nu attacks electrophilic C. Aldehyde more reactive than ketone (less steric hindrance + more electrophilic C).
Tollens' test: Ag(NH₃)₂OH → silver mirror. Positive for aldehydes ONLY (not ketones). Formic acid (HCOOH) and glucosealso positive.
Fehling's test: Cu²⁺ (blue) → Cu₂O (red ppt). Positive for aliphatic aldehydes and reducing sugars. Negative for aromatic aldehydes and ketones.
Iodoform test: I₂/NaOH. Positive for: CH₃CHO, CH₃COR, CH₃CH(OH)R — all have CH₃CO– or CH₃CHOH– group. Yellow CHI₃ ppt (triiodomethane, mp 119°C).
Aldol condensation: Base + aldehyde/ketone having α-H → β-hydroxy carbonyl compound → dehydration → α,β-unsaturated compound.
💡JEE Tip: Iodoform test positive: any compound that gives CH₃CO– or CH₃CH(OH)– on oxidation. Ethanol (CH₃CH₂OH) is positive (oxidises to CH₃CHO → iodoform). Methanol is negative. Propan-2-ol (CH₃CHOHCH₃) is positive; propan-1-ol is negative. This is a favourite JEE distinction.
⚠Common Error: Fehling's test is negative for aromatic aldehydes (like benzaldehyde) even though they are aldehydes. This is because Fehling's reagent only oxidises aliphatic (not aromatic) aldehydes. Tollens' test, however, is positive for ALL aldehydes including benzaldehyde.
Example 1EasyIodoform Test — Which Compounds Test Positive?
Which of the following give a positive iodoform test? (a) Ethanol, (b) Methanol, (c) Propan-2-ol, (d) Propan-1-ol, (e) Acetone, (f) Diethyl ketone.
Positive iodoform test: compound must have CH₃CO– group or be oxidisable to it (i.e., CH₃CH(OH)– group).
Ethanol CH₃CH₂OH: Oxidises to CH₃CHO (acetaldehyde = CH₃CO–H). Positive ✓
Methanol CH₃OH: Oxidises to HCHO (no CH₃ group on C=O). Negative ✗
Propan-2-ol CH₃CH(OH)CH₃: Has CH₃CHOH– group directly. Positive ✓
Propan-1-ol CH₃CH₂CH₂OH: Oxidises to CH₃CH₂CHO (propanal; no CH₃ adjacent to C=O). Negative ✗
Acetone CH₃COCH₃: Has CH₃CO– group. Positive ✓
Diethyl ketone CH₃CH₂COCH₂CH₃: No CH₃ adjacent to C=O (it's CH₂). Negative ✗
Example 2MediumBasicity Order of Amines
Arrange in decreasing order of basicity in aqueous solution: (a) Methylamine (CH₃NH₂), (b) Dimethylamine (CH₃)₂NH, (c) Trimethylamine (CH₃)₃N, (d) Aniline (C₆H₅NH₂), (e) Ammonia (NH₃).
Two factors: inductive effect of alkyl groups (increases basicity) and solvation (NH, NH₂ better than N for H-bonding with water → stabilises conjugate acid).
Aniline: lone pair on N is delocalised into benzene ring → less available → very low basicity (\(pK_b \approx 9.4\)).
Ammonia: no alkyl group → moderate basicity (\(pK_b = 4.74\)).
Methylamine vs Dimethylamine: in water, dimethylamine is MORE basic than methylamine (better inductive donation by 2 CH₃, and still has 1 N–H for solvation). Trimethylamine: 3 CH₃ groups donate more inductively BUT no N–H for solvation → LESS basic than dimethylamine.
✓ Butan-2-ol is a 2° alcohol — accessible via Grignard addition of ethyl group to ethanal. Both fragments (ethanal and ethyl Grignard) come from ethanol.
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Functional Groups — Visual Concepts
Click any diagram to zoom
Reactivity of Carbonyl Derivatives (Nucleophilic Acyl Substitution)
Reactivity of carboxylic acid derivatives in nucleophilic acyl substitution. Reactivity depends on the leaving group ability. Cl⁻ is the best leaving group → acyl chlorides most reactive. NH₂⁻ is the worst leaving group → amides least reactive. This order governs interconversion: more reactive can form less reactive, not vice versa.
Aldol Condensation — Mechanism Summary
Aldol condensation of acetaldehyde. Base removes α-H → enolate attacks another carbonyl → β-hydroxyaldehyde (aldol product). Heating dehydrates to give α,β-unsaturated aldehyde (crotonaldehyde). Key condition: must have α-H (carbon adjacent to C=O). Without α-H → Cannizzaro reaction instead.
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Functional Groups — Quick Quiz
Question 1
Which alcohol does NOT give a positive iodoform test?
Iodoform test positive: compound must have CH₃CO– or CH₃CHOH– group. Propan-1-ol: oxidises to CH₃CH₂CHO (propanal) — no CH₃ on the carbonyl carbon → negative. Ethanol → CH₃CHO ✓; Propan-2-ol has CH₃CHOH– ✓; Butan-2-ol has CH₃CHOH– ✓. Answer: C ✓
Question 2
Formaldehyde (HCHO) does NOT undergo Aldol condensation with itself because:
Aldol condensation requires α-hydrogen (H on the carbon adjacent to C=O). In HCHO (H–CHO), there is no carbon adjacent to the carbonyl — it IS the terminal carbon. No α-H means no enolate can form. HCHO instead undergoes Cannizzaro reaction (disproportionation). Answer: A ✓
Question 3
The Hofmann bromamide reaction of benzamide (C₆H₅CONH₂) gives:
Hofmann bromamide: RCONH₂ + Br₂/KOH → RNH₂. The amide loses the carbonyl carbon as CO₂, leaving the R group attached to NH₂. C₆H₅CONH₂ → C₆H₅NH₂ (aniline). Product is a primary amine with ONE FEWER carbon than the starting amide. Answer: B ✓
Glucose (C₆H₁₂O₆): Aldohexose. Open chain: CHO–(CHOH)₄–CH₂OH. Cyclic: α-D-glucopyranose (OH at C1 axial) and β-D-glucopyranose (OH at C1 equatorial) — anomers. Undergoes mutarotation in solution.
Fructose: Ketohexose. Sweetest sugar. Cyclic form is furanose (5-membered ring).
Disaccharide
Units
Bond
Reducing?
Maltose (malt sugar)
2 glucose
α-1,4 glycosidic
Yes (free anomeric –OH at C1)
Lactose (milk sugar)
Galactose + Glucose
β-1,4 glycosidic
Yes
Sucrose (cane sugar)
Glucose + Fructose
α,β-1,2 glycosidic (both anomeric C involved)
No (no free anomeric –OH)
Reducing sugar: Has free anomeric –OH (hemiacetal/hemiketal) → can reduce Tollens/Fehling → glucose, fructose, maltose, lactose. Non-reducing: sucrose, trehalose.
Starch: Amylose (linear, α-1,4) + Amylopectin (branched, α-1,4 + α-1,6 at branches). Gives blue-black colour with I₂.
Cellulose: β-1,4 glucose units (humans lack β-glucosidase → indigestible). Most abundant organic polymer on Earth.
Proteins and Amino Acids
20 naturally occurring α-amino acids. Each has H₂N–CH(R)–COOH. At physiological pH they exist as zwitterions: ⁺H₃N–CH(R)–COO⁻.
Peptide bond: –CO–NH– formed by condensation of –COOH of one amino acid with –NH₂ of another (water lost). Peptide bond is planar (partial double bond character due to resonance).
Protein structure levels:
1° (primary): amino acid sequence.
2° (secondary): α-helix and β-sheet (H-bonding between C=O and N–H).
3° (tertiary): 3D folding (disulfide bridges, H-bonds, hydrophobic interactions).
4° (quaternary): multiple polypeptide subunits (e.g., haemoglobin = 4 subunits).
Denaturation: Disruption of 2°/3°/4° structure (not peptide bonds). Caused by heat, pH change, organic solvents, heavy metals. Reversible (renaturation) or irreversible.
Enzymes: Biological catalysts (protein). Lock-and-key model vs induced fit. Specific: active site complementary to substrate.
Nucleotide = nitrogenous base + pentose sugar + phosphate. Nucleoside = base + sugar (no phosphate).
Purine bases: Adenine (A), Guanine (G). Pyrimidine bases: Cytosine (C), Thymine (T, in DNA), Uracil (U, in RNA).
Vitamins and Hormones
Vitamin
Type
Deficiency disease
A (retinol)
Fat-soluble
Night blindness, xerophthalmia
B₁ (thiamine)
Water-soluble
Beriberi
B₂ (riboflavin)
Water-soluble
Cheilosis, corneal vascularisation
B₁₂ (cobalamin)
Water-soluble
Pernicious anaemia
C (ascorbic acid)
Water-soluble
Scurvy
D (calciferol)
Fat-soluble
Rickets (children), osteomalacia (adults)
E (tocopherol)
Fat-soluble
Reproductive failure, muscular dystrophy
K (phylloquinone)
Fat-soluble
Impaired blood clotting
Fat-soluble: A, D, E, K (mnemonic: ADEK). Water-soluble: B-complex and C.
Polymers
Polymer
Type
Monomers
Use
Nylon-6,6
Condensation, polyamide
Hexamethylenediamine + adipic acid
Fibres, ropes
Nylon-6
Condensation (ring-opening)
Caprolactam
Fibres
Dacron (Terylene)
Condensation, polyester
Ethylene glycol + terephthalic acid
Fibres, bottles
Bakelite
Condensation, thermosetting
Phenol + formaldehyde
Electrical fittings
PVC
Addition
Vinyl chloride (CH₂=CHCl)
Pipes, insulation
Teflon (PTFE)
Addition
Tetrafluoroethylene CF₂=CF₂
Non-stick coating
Buna-S (SBR)
Addition (copolymer)
Butadiene + styrene
Synthetic rubber tyres
Natural rubber
Addition (natural)
Isoprene (cis-1,4)
Elastic applications
💡JEE Tip: Sucrose is the ONLY common non-reducing disaccharide because BOTH anomeric carbons (C1 of glucose AND C2 of fructose) are involved in the glycosidic bond — no free hemiacetal group left. All monosaccharides (glucose, fructose, galactose) are reducing sugars.
⚠Common Error: Nylon-6 and Nylon-6,6 sound similar but differ: Nylon-6 is made from ONE monomer (caprolactam, ring-opening), while Nylon-6,6 is from TWO monomers (hexamethylenediamine + adipic acid). The numbers refer to the carbon atoms in each monomer unit.
Mutarotation: α-D-glucose ⇌ open chain ⇌ β-D-glucose (in aqueous solution)
Specific rotation changes from +112° (α) or −92° (β) to equilibrium +52.7°. The anomers differ only at C1 (anomeric carbon). Catalysed by acid or base.
Cellulose: Polysaccharide (β-1,4). The reducing end has a free anomeric –OH, but for practical purposes (high MW, insoluble) — treated as non-reducing in JEE context.
Example 2MediumPolymer Classification
Classify each polymer: (a) Nylon-6, (b) Natural rubber, (c) Bakelite, (d) Teflon, (e) Buna-S. State the type of polymerization and give the monomer(s).
Bakelite: Condensation polymer (phenol + formaldehyde). Cross-linked 3D network. Thermosetting (cannot be remelted). Used for electrical fittings.
Teflon (PTFE): Addition polymer. Monomer: tetrafluoroethylene (CF₂=CF₂). Thermoplastic. Chemically inert due to strong C–F bonds. Non-stick surface.
Buna-S (SBR): Addition copolymer. Monomers: butadiene (CH₂=CH–CH=CH₂) + styrene (C₆H₅CH=CH₂). Synthetic rubber. Used in tyres.
Example 3HardDNA Composition — Chargaff's Rules
In a DNA sample, adenine constitutes 30% of the bases. Find the percentages of thymine, guanine, and cytosine. Also find the number of hydrogen bonds if the DNA has 10,000 base pairs.
✓ Total H-bonds = 6,000 + 6,000 = 12,000. T = 30%, G = 20%, C = 20%. Note: A+T-rich DNA has lower melting temperature (fewer H-bonds per bp).
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Biomolecules & Polymers — Visual Concepts
Click any diagram to zoom
Carbohydrate Classification Tree
Carbohydrate classification. Key distinction: sucrose is the only common non-reducing disaccharide (both anomeric carbons used in glycosidic bond). Starch vs cellulose: both are glucose polymers but differ in linkage (α-1,4 vs β-1,4) — this one structural difference makes starch digestible and cellulose indigestible by humans.
DNA Double Helix — Base Pairing
DNA double helix base pairing. A pairs with T via 2 hydrogen bonds; G pairs with C via 3 hydrogen bonds. The blue ribbons are the sugar-phosphate backbones (antiparallel: 5'→3' on one strand pairs with 3'→5' on the other). G-C content determines DNA melting temperature — higher G-C% → higher Tm (more H-bonds to break).
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Biomolecules & Polymers — Quick Quiz
Question 1
Sucrose does NOT reduce Fehling's solution because:
In sucrose, the glycosidic bond (α,β-1,2) involves the anomeric C1 of glucose AND the anomeric C2 of fructose. Both are locked in the bond → no free hemiacetal/hemiketal → cannot open to give free carbonyl → non-reducing. Answer: B ✓
Question 2
Nylon-6,6 is synthesised from:
Nylon-6,6: condensation polymerization of hexamethylenediamine (6C diamine) + adipic acid (6C diacid) → polyamide with –CO–NH– (amide) linkages. The "6,6" refers to 6 carbons in each monomer. Nylon-6 (A) comes from caprolactam alone. Dacron (B) uses ethylene glycol + terephthalic acid (polyester). Answer: C ✓
Question 3
In DNA, if guanine constitutes 22% of the bases, the percentage of adenine is: