State the reagents and conditions for the 4 key benzene ring side-chain and aromatic conversions (OCR/Edexcel/AQA).

1) Benzene → Nitrobenzene: conc. HNO3 + conc. H2SO4 at 50°C; 2) Nitrobenzene → Phenylamine: Sn + conc. HCl under reflux, then NaOH; 3) Benzene → Bromobenzene: Br2 + anhydrous FeBr3 or AlBr3 halogen carrier catalyst; 4) Alkylbenzene (e.g., methylbenzene) → Benzoic acid: heat under reflux with alkaline KMnO4 followed by dilute H2SO4 (oxidises any alkyl side chain with a benzylic H into -COOH).

Write the three stages (Initiation, Propagation, Termination) of the free-radical substitution of methane (CH4) with chlorine (Cl2).

1) Initiation (UV light homolytic fission): Cl2 → 2Cl•. 2) Propagation (chain reaction): Cl• + CH4 → •CH3 + HCl, followed by •CH3 + Cl2 → CH3Cl + Cl• (overall: CH4 + Cl2 → CH3Cl + HCl). 3) Termination (two radicals collide): 2Cl• → Cl2, 2•CH3 → C2H6, or •CH3 + Cl• → CH3Cl. Further substitution produces CH2Cl2, CHCl3, and CCl4.

Compare the conditions and mechanisms for converting a haloalkane (e.g., 2-bromopropane) into an alcohol (Nucleophilic Substitution) vs an alkene (Elimination).

In both reactions the reagent is potassium/sodium hydroxide (KOH/NaOH), where OH⁻ acts in two different roles depending on the solvent: 1) Warm AQUEOUS NaOH: :OH⁻ acts as a NUCLEOPHILE, attacking the δ+ carbon to substitute Br⁻ and form an alcohol. 2) Hot ETHANOLIC (anhydrous) KOH: :OH⁻ acts as a BASE, removing a proton (H⁺) from an adjacent β-carbon to eliminate HBr and form an alkene.

Why is ethanoic anhydride preferred to ethanoyl chloride in the industrial manufacture of Aspirin?

Ethanoyl chloride is more reactive, but ethanoic anhydride is used industrially because: 1) It is cheaper to manufacture; 2) It is less corrosive and less vulnerable to hydrolysis by moisture; 3) Its by-product is weak ethanoic acid (CH3COOH) rather than toxic, corrosive fumes of hydrogen chloride gas (HCl).

Describe the apparatus and experimental steps to purify an organic liquid product (e.g., 2-chloro-2-methylpropane or an ester) using a separating funnel and redistillation.

1) Pour the reaction mixture into a separating funnel, add aqueous sodium hydrogencarbonate (NaHCO3) to neutralise any remaining acid, stopper, invert, and open the tap periodically to vent CO2 gas pressure; 2) Allow the aqueous and organic layers to settle and run off the lower layer; 3) Transfer the organic layer to a dry conical flask and add an anhydrous drying agent (anhydrous MgSO4 or CaCl2) until the liquid turns from cloudy to clear and the solid stops clumping; 4) Filter/decant and redistil, collecting only the fraction boiling at the pure compound's boiling point.

How do you use Percentage Composition, Empirical Formula, Molecular Ion (M⁺), IR, and NMR together to solve a 6–8 mark 'Combined Spectroscopy' question?

Step 1: Divide % mass by Ar of each element and divide by the smallest to get the Empirical Formula; match against the M⁺ molecular ion peak on the mass spectrum to find the Molecular Formula and calculate the Index of Hydrogen Deficiency (IHD = [2C + 2 + N − H − X] / 2; IHD = 1 means 1 double bond or ring, IHD ≥ 4 suggests a benzene ring). Step 2: Use the IR spectrum to confirm C=O (~1700 cm⁻¹) or O–H (broad >3000 cm⁻¹). Step 3: Count ¹³C/¹H NMR environments, use integration ratios and (n + 1) splitting to assemble the fragments, and draw the displayed structure.

In ¹³C and ¹H NMR Spectroscopy, why are Tetramethylsilane (TMS, Si(CH3)4) and deuterated solvents (CDCl3, D2O) used?

TMS is the internal reference standard assigned a chemical shift of δ = 0 ppm because: 1) All 12 H atoms and 4 C atoms are chemically equivalent, giving a single sharp intense peak; 2) Si is less electronegative than C, so its signal is upfield (well away from organic peaks); 3) It is chemically inert, non-toxic, and volatile (b.p. 27°C, easily evaporated off). Deuterated solvents (CDCl3) are used because ²H (deuterium) has an even mass number (no net nuclear spin, I = 1) and does not produce interfering ¹H NMR signals. Shaking with D2O causes -OH and -NH protons to disappear via proton exchange.

What is a racemic mixture (racemate), why is it optically inactive, and why does nucleophilic addition to an unsymmetrical planar carbonyl (RCHO) produce a racemate?

A racemic mixture contains an equimolar (50:50) mixture of two enantiomers; it is optically inactive because the two enantiomers rotate plane-polarised light by equal angles in opposite directions, cancelling out. In nucleophilic addition of HCN (CN⁻) to an unsymmetrical aldehyde (e.g., ethanal), the C=O group is trigonal planar, so the CN⁻ nucleophile attacks with equal probability from above or below the plane, producing a 50:50 racemate.

Draw/describe the structure of a 2-aminocarboxylic acid (α-amino acid) at its isoelectric point (zwitterion), at low pH (acidic), and at high pH (alkaline).

At the isoelectric point (neutral zwitterion): internal proton transfer gives +H3N–CHR–COO⁻ (high melting point crystalline solid due to ionic lattice attractions). At low pH (strongly acidic, excess H⁺): carboxylate group is protonated to give the cation +H3N–CHR–COOH. At high pH (strongly alkaline, excess OH⁻): ammonium group is deprotonated to give the anion H2N–CHR–COO⁻.

State the chemical tests to distinguish between an Aldehyde and a Ketone using Tollens' Reagent and Fehling's Solution.

Aldehydes are easily oxidised to carboxylic acids (or carboxylate ions), whereas ketones cannot be oxidised easily: 1) Tollens' reagent (ammoniacal silver nitrate, [Ag(NH3)2]⁺): warm gently in a water bath → aldehydes reduce Ag⁺ to metallic silver(0), forming a 'silver mirror' on the test tube wall; ketones give no change. 2) Fehling's / Benedict's solution (blue aqueous Cu²⁺ complex): warm → aldehydes reduce blue Cu²⁺ to a brick-red precipitate of copper(I) oxide (Cu2O); ketones remain blue.

How do you oxidise a primary alcohol to an aldehyde vs a carboxylic acid, and what happens with secondary and tertiary alcohols?

Reagent: acidified potassium dichromate(VI) (K2Cr2O7 / H2SO4, colour changes from orange Cr2O7²⁻ to green Cr³⁺). 1) Primary alcohol + distillation: distil off the aldehyde product immediately before further oxidation. 2) Primary alcohol + heating under reflux with excess oxidising agent: oxidises fully to a carboxylic acid. 3) Secondary alcohol + reflux: oxidises to a ketone. 4) Tertiary alcohol: resists oxidation (stays orange).

Write the equations and Electrophilic Substitution mechanism for Friedel–Crafts Acylation of benzene with ethanoyl chloride and AlCl3.

Reagents: acyl chloride (e.g., CH3COCl) + anhydrous aluminium chloride catalyst (AlCl3) under reflux in dry ether. Generation of acylium electrophile: CH3COCl + AlCl3 → CH3C⁺=O + AlCl4⁻. Mechanism: curly arrow from benzene's delocalised π-ring to the positively charged C⁺ of CH3CO⁺, forming the horseshoe intermediate, then curly arrow from the C–H bond back into the ring to form phenylethanone (a ketone) + H⁺, which reacts with AlCl4⁻ → AlCl3 + HCl.

Contrast Structural Isomerism (chain, position, functional group) with Stereoisomerism (E/Z and optical isomerism).

Structural isomers have the same molecular formula but different structural formulae (chain branching, functional group position, or different functional groups such as aldehydes vs ketones). Stereoisomers have the same structural connectivity but a different 3D spatial arrangement of atoms: E/Z isomers arise from restricted rotation around a C=C double bond (using Cahn–Ingold–Prelog atomic number priority rules: Z = 'zusammen' on same side, E = 'entgegen' on opposite sides); Optical isomers (enantiomers) are non-superimposable mirror images around a chiral carbon atom bonded to 4 different groups.

Explain the four pieces of structural information obtained from a high-resolution Proton (¹H) NMR spectrum, including the (n + 1) spin–spin splitting rule.

1) Number of signals = number of distinct non-equivalent proton environments; 2) Chemical shift (δ in ppm) = type of chemical environment; 3) Integration trace (peak area ratio) = relative number of protons in each environment; 4) Spin–spin splitting pattern (n + 1 rule): if a proton environment has n chemically non-equivalent protons on adjacent carbon atoms, its peak splits into (n + 1) sub-peaks (0 → singlet, 1 → doublet, 2 → triplet 1:2:1, 3 → quartet 1:3:3:1). Note: -OH and -NH protons always appear as singlets and do not split adjacent protons.

Compare the two laboratory synthetic routes to prepare a primary aliphatic amine vs an aromatic amine (phenylamine).

1) Primary aliphatic amine (without forming secondary/tertiary/quaternary ammonium salts): either heat a haloalkane with a large EXCESS of ethanolic ammonia, OR (better 2-step route with 1 product) react haloalkane with KCN in aqueous ethanol (nucleophilic substitution to form a nitrile, extending chain by 1 C) followed by catalytic hydrogenation with H2 / Ni catalyst (or LiAlH4 in dry ether) to form RCH2NH2. 2) Aromatic amine (phenylamine): reduce nitrobenzene by heating under reflux with Tin (Sn) and concentrated HCl, followed by excess NaOH to liberate free C6H5NH2.

Compare the rate of hydrolysis of chloroalkanes, bromoalkanes, and iodoalkanes with aqueous silver nitrate in ethanol.

Rate of nucleophilic substitution increases down Group 7: 1-iodobutane > 1-bromobutane > 1-chlorobutane (yellow AgI precipitate appears fastest, then cream AgBr, then white AgCl slowest). Although C–Cl is the most polar bond, bond enthalpy is the dominant factor: the C–I bond is the longest and weakest (lowest bond enthalpy, 238 kJ mol⁻¹ vs 346 kJ mol⁻¹ for C–Cl), so C–I breaks with the lowest activation energy.

Compare the structures of the condensation polymers Terylene (PET), Nylon-6,6, and Kevlar, and explain why condensation polymers are biodegradable whereas poly(alkenes) are not.

Terylene (polyester): formed from benzene-1,4-dicarboxylic acid + ethane-1,2-diol (ester links -COO-). Nylon-6,6 (polyamide): formed from hexanedioic acid (or hexanedioyl dichloride) + hexane-1,6-diamine (amide links -CONH-). Kevlar (polyamide): formed from benzene-1,4-dicarboxylic acid + benzene-1,4-diamine (flat rigid chains packed with strong H-bonds). Polyesters and polyamides have polar C=O bonds susceptible to nucleophilic attack and hydrolysis by water, acids, alkalis, or enzymes; poly(alkenes) have only non-polar, strong C–C and C–H bonds.

In Infrared (IR) Spectroscopy, how do you identify an Alcohol, an Aldehyde/Ketone, and a Carboxylic Acid, and what is the 'fingerprint region'?

IR radiation is absorbed when its frequency matches the natural vibrational frequency of a polar covalent bond: 1) Alcohol O–H: broad absorption at 3200–3550 cm⁻¹ (broad due to hydrogen bonding) and NO C=O peak; 2) Aldehyde/Ketone C=O: sharp, strong peak at ~1680–1750 cm⁻¹ and NO O–H peak; 3) Carboxylic Acid: BOTH a sharp C=O peak at ~1700 cm⁻¹ AND a very broad O–H peak at 2500–3300 cm⁻¹. The fingerprint region (below 1500 cm⁻¹) contains complex C–C/C–O vibrations matched against a database.

How does the anticancer drug Cisplatin ([Pt(NH3)2Cl2]) prevent DNA replication in rapidly dividing tumour cells?

Cisplatin is a square planar platinum(II) complex with two Cl⁻ ligands in the cis (90°) configuration. Inside the cell, water/DNA displaces the two chloride ligands in a ligand substitution reaction, and the platinum ion forms coordinate covalent bonds to nitrogen atoms on two adjacent guanine bases on DNA, kinking the DNA strand so it cannot unwind and replicate.

What is the distinctive ¹H NMR splitting pattern for an ethyl group (–CH2CH3) vs an isopropyl group (–CH(CH3)2)?

An ethyl group (–CH2CH3) always gives a characteristic Quartet (area 2, from the –CH2– split by 3 adjacent H's: 3 + 1 = 4) and a Triplet (area 3, from the –CH3 split by 2 adjacent H's: 2 + 1 = 3). An isopropyl group (–CH(CH3)2) gives a Septet/Multiplet (area 1, split by 6 equivalent H's: 6 + 1 = 7) and a Doublet (area 6, from two equivalent –CH3 groups split by 1 adjacent CH).

Present the three pieces of evidence (bond lengths, enthalpy of hydrogenation, and resistance to bromination) that disprove Kekulé's alternating C=C/C–C model of Benzene.

1) X-ray diffraction shows all six C–C bond lengths in benzene are identical (0.139 nm), intermediate between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm). 2) Thermochemistry: ΔH(hydrogenation) of cyclohexene is −120 kJ mol⁻¹, so Kekulé's cyclohexa-1,3,5-triene would be −360 kJ mol⁻¹, whereas actual benzene is −208 kJ mol⁻¹ (152 kJ mol⁻¹ more stable due to delocalisation energy). 3) Reactivity: Benzene does not decolourise bromine water (undergoes electrophilic substitution rather than addition to preserve the delocalised π-ring).

Arrange phenylamine (C6H5NH2), ammonia (NH3), and ethylamine (CH3CH2NH2) in order of increasing base strength (pKb / Ka of conjugate acid) and explain why.

Weakest base → Strongest base: Phenylamine < Ammonia < Ethylamine. Base strength depends on the availability of the lone pair of electrons on the N atom to accept a proton (H⁺): 1) In phenylamine, the N lone pair delocalises into the aromatic benzene π-ring, making it much less available to accept H⁺; 2) In ethylamine, the ethyl group has a positive inductive effect (+I, electron-pushing), increasing electron density on the N atom and stabilising the ethylammonium ion.

How do you identify the functional groups in a multi-functional organic molecule using a sequence of test-tube reactions?

1) Add bromine water: decolourises orange to colourless → Alkene (C=C) or Phenol. 2) Add Na2CO3 or NaHCO3(aq): effervescence of CO2 → Carboxylic acid (-COOH). 3) Add 2,4-dinitrophenylhydrazine (2,4-DNPH / Brady's reagent): bright orange/yellow precipitate → Carbonyl (Aldehyde or Ketone; recrystallise the 2,4-DNPH derivative and measure its melting point to identify the specific carbonyl). 4) Warm with Tollens' reagent: silver mirror → Aldehyde. 5) Warm with acidified K2Cr2O7: orange to green → Primary/Secondary alcohol or Aldehyde.

Explain the mechanism of Electrophilic Addition of HBr to an unsymmetrical alkene (propene) and why Markovnikov's Rule favours the major product.

1) The high electron density of the C=C π-bond attacks the δ+ H of H–Br, breaking the H–Br bond heterolytically; 2) In propene, protonation forms either a secondary carbocation intermediate (CH3–CH⁺–CH3) or a primary carbocation (CH3–CH2–CH2⁺); 3) Alkyl groups are electron-releasing (+I positive inductive effect), stabilising the positive charge, so the more stable secondary carbocation forms faster and is attacked by the :Br⁻ ion to give 2-bromopropane as the major product.

Describe the reagents and Nucleophilic Addition mechanism for reducing aldehydes/ketones to alcohols and for adding HCN to form hydroxynitriles.

Reduction: aqueous sodium tetrahydridoborate(III) (NaBH4), which generates the hydride nucleophile (:H⁻) that attacks the δ+ carbonyl carbon, followed by protonation from water to form a primary alcohol (from aldehyde) or secondary alcohol (from ketone). Hydroxynitrile formation: KCN followed by dilute acid (HCl/H2SO4)—never HCN gas alone for safety; :CN⁻ attacks the δ+ C=O carbon, and the intermediate alkoxide ion O⁻ is protonated by H⁺/HCN, extending the carbon chain by 1 carbon.

In organic synthesis, which two key reactions increase the length of the carbon chain by forming a new C–C bond?

1) Reaction with Cyanide (:CN⁻): Nucleophilic substitution of a haloalkane with KCN in aqueous ethanol (RX + CN⁻ → RCN + X⁻, adds 1 C) OR Nucleophilic addition of HCN (KCN/H⁺) to an aldehyde/ketone to form a hydroxynitrile. The nitrile (-CN) can then be reduced to a primary amine (-CH2NH2) using H2/Ni, or hydrolysed with dilute HCl under reflux to a carboxylic acid (-COOH). 2) Friedel–Crafts Acylation/Alkylation of benzene with RCOCl / RX and anhydrous AlCl3.

How is High-Resolution Mass Spectrometry used to distinguish between compounds with the same nominal Mr (e.g., C2H4O2, C3H8O, and C3H4O)?

Low-resolution mass spectrometry gives nominal whole-number m/z values (e.g., propanal C3H6O, butane C4H10, and ethanoic acid all have nominal Mr = 58 or 60). High-resolution mass spectrometry measures m/z to 4 or 5 decimal places using exact isotopic masses (¹²C = 12.0000, ¹H = 1.0078, ¹⁶O = 15.9949, ¹⁴N = 14.0031), uniquely identifying the exact molecular formula from the M⁺ peak.

Describe the Nucleophilic Addition–Elimination mechanism of Acyl Chlorides (RCOCl) and Acid Anhydrides with water, alcohols, ammonia, and primary amines.

The lone pair on the nucleophile (:OH2, :OHR', :NH3, :NH2R') attacks the strongly δ+ carbonyl carbon of the acyl chloride; the π-bond breaks to form a tetrahedral intermediate, then the lone pair on oxygen reforms the C=O double bond and expels the Cl⁻ leaving group, followed by loss of H⁺. Products with RCOCl: 1) + H2O → carboxylic acid + misty fumes of HCl; 2) + R'OH → ester + HCl; 3) + 2NH3 → primary amide + NH4Cl; 4) + 2R'NH2 → N-substituted secondary amide + R'NH3⁺Cl⁻.

Why does phenol (C6H5OH) react much more readily with bromine water than benzene does (OCR/Edexcel)?

In phenol, a lone pair of electrons from the p-orbital of the oxygen atom in the -OH group is partially delocalised into the aromatic π-ring. This increases the electron density of the benzene ring (activating it in the 2, 4, and 6 positions), enabling phenol to polarise Br2 molecules directly without a halogen carrier catalyst, decolourising bromine water and forming a white precipitate of 2,4,6-tribromophenol.

Write the equations and Electrophilic Substitution mechanism for the Nitration of Benzene.

Reagents: concentrated nitric acid (HNO3) + concentrated sulfuric acid (H2SO4 catalyst) at 50°C. Generation of the nitronium ion electrophile: HNO3 + 2H2SO4 → NO2⁺ + 2HSO4⁻ + H3O⁺ (or HNO3 + H2SO4 → NO2⁺ + HSO4⁻ + H2O). Mechanism: curly arrow from benzene's delocalised π-electron ring to N of NO2⁺ forming a horseshoe Wheland intermediate (covering 5 carbons with a + charge), then curly arrow from C–H bond back into the ring to restore aromaticity and release H⁺ (which regenerates the H2SO4 catalyst: H⁺ + HSO4⁻ → H2SO4).

How do Thin-Layer Chromatography (TLC), Column Chromatography, and Gas Chromatography–Mass Spectrometry (GC–MS) separate mixtures?

All chromatography relies on the balance between adsorption onto the stationary phase (often silica SiO2 or alumina Al2O3) and solubility in the mobile phase. In Gas Chromatography (GC), the stationary phase is a high-boiling liquid lining a long coiled capillary tube in a temperature-controlled oven, and the mobile phase is an inert carrier gas (He or N2). Each component has a characteristic retention time, and the coupled Mass Spectrometer (GC–MS) identifies each component via its molecular ion peak and fragmentation pattern.

Describe the experimental steps to purify an impure organic solid (e.g., Aspirin) by Recrystallisation and test its purity.

1) Dissolve the crude solid in the MINIMUM volume of hot solvent (so the solution is saturated on cooling and yield is maximised); 2) Filter the hot solution rapidly through fluted filter paper to remove insoluble impurities; 3) Allow the filtrate to cool slowly in an ice bath so pure crystals reform while soluble impurities remain dissolved in the cold solvent; 4) Filter under reduced pressure using a Büchner funnel and vacuum pump, wash with ice-cold solvent, and dry in a warm oven; 5) Measure its melting point: a pure solid has a sharp melting point matching the data book value.

A-Level Chemistry II (Organic & Analysis): Reaction Mechanisms, Carbonyls, Aromatic Chemistry, Amines, Polymers & NMR/IR Spectroscopy

High-yield A-Level Organic Chemistry and Spectroscopy across AQA, OCR, and Edexcel: E/Z & optical isomerism, free-radical substitution, electrophilic/nucleophilic mechanisms, Friedel-Crafts acylation, condensation polymers, and 1H/13C NMR & IR interpretation.

Level: intermediate
Languages: en → en
Cards: 32
Markdown: A-Level Chemistry II (Organic & Analysis): Reaction Mechanisms, Carbonyls, Aromatic Chemistry, Amines, Polymers & NMR/IR Spectroscopy (.md)

Flashcards (32)

#	Front (Term / Question)	Back (Definition / Answer)
1	State the reagents and conditions for the 4 key benzene ring side-chain and aromatic conversions (OCR/Edexcel/AQA).	1) Benzene → Nitrobenzene: conc. HNO3 + conc. H2SO4 at 50°C; 2) Nitrobenzene → Phenylamine: Sn + conc. HCl under reflux, then NaOH; 3) Benzene → Bromobenzene: Br2 + anhydrous FeBr3 or AlBr3 halogen carrier catalyst; 4) Alkylbenzene (e.g., methylbenzene) → Benzoic acid: heat under reflux with alkaline KMnO4 followed by dilute H2SO4 (oxidises any alkyl side chain with a benzylic H into -COOH).
2	Write the three stages (Initiation, Propagation, Termination) of the free-radical substitution of methane (CH4) with chlorine (Cl2).	1) Initiation (UV light homolytic fission): Cl2 → 2Cl•. 2) Propagation (chain reaction): Cl• + CH4 → •CH3 + HCl, followed by •CH3 + Cl2 → CH3Cl + Cl• (overall: CH4 + Cl2 → CH3Cl + HCl). 3) Termination (two radicals collide): 2Cl• → Cl2, 2•CH3 → C2H6, or •CH3 + Cl• → CH3Cl. Further substitution produces CH2Cl2, CHCl3, and CCl4.
3	Compare the conditions and mechanisms for converting a haloalkane (e.g., 2-bromopropane) into an alcohol (Nucleophilic Substitution) vs an alkene (Elimination).	In both reactions the reagent is potassium/sodium hydroxide (KOH/NaOH), where OH⁻ acts in two different roles depending on the solvent: 1) Warm AQUEOUS NaOH: :OH⁻ acts as a NUCLEOPHILE, attacking the δ+ carbon to substitute Br⁻ and form an alcohol. 2) Hot ETHANOLIC (anhydrous) KOH: :OH⁻ acts as a BASE, removing a proton (H⁺) from an adjacent β-carbon to eliminate HBr and form an alkene.
4	Why is ethanoic anhydride preferred to ethanoyl chloride in the industrial manufacture of Aspirin?	Ethanoyl chloride is more reactive, but ethanoic anhydride is used industrially because: 1) It is cheaper to manufacture; 2) It is less corrosive and less vulnerable to hydrolysis by moisture; 3) Its by-product is weak ethanoic acid (CH3COOH) rather than toxic, corrosive fumes of hydrogen chloride gas (HCl).
5	Describe the apparatus and experimental steps to purify an organic liquid product (e.g., 2-chloro-2-methylpropane or an ester) using a separating funnel and redistillation.	1) Pour the reaction mixture into a separating funnel, add aqueous sodium hydrogencarbonate (NaHCO3) to neutralise any remaining acid, stopper, invert, and open the tap periodically to vent CO2 gas pressure; 2) Allow the aqueous and organic layers to settle and run off the lower layer; 3) Transfer the organic layer to a dry conical flask and add an anhydrous drying agent (anhydrous MgSO4 or CaCl2) until the liquid turns from cloudy to clear and the solid stops clumping; 4) Filter/decant and redistil, collecting only the fraction boiling at the pure compound's boiling point.
6	How do you use Percentage Composition, Empirical Formula, Molecular Ion (M⁺), IR, and NMR together to solve a 6–8 mark 'Combined Spectroscopy' question?	Step 1: Divide % mass by Ar of each element and divide by the smallest to get the Empirical Formula; match against the M⁺ molecular ion peak on the mass spectrum to find the Molecular Formula and calculate the Index of Hydrogen Deficiency (IHD = [2C + 2 + N − H − X] / 2; IHD = 1 means 1 double bond or ring, IHD ≥ 4 suggests a benzene ring). Step 2: Use the IR spectrum to confirm C=O (~1700 cm⁻¹) or O–H (broad >3000 cm⁻¹). Step 3: Count ¹³C/¹H NMR environments, use integration ratios and (n + 1) splitting to assemble the fragments, and draw the displayed structure.
7	In ¹³C and ¹H NMR Spectroscopy, why are Tetramethylsilane (TMS, Si(CH3)4) and deuterated solvents (CDCl3, D2O) used?	TMS is the internal reference standard assigned a chemical shift of δ = 0 ppm because: 1) All 12 H atoms and 4 C atoms are chemically equivalent, giving a single sharp intense peak; 2) Si is less electronegative than C, so its signal is upfield (well away from organic peaks); 3) It is chemically inert, non-toxic, and volatile (b.p. 27°C, easily evaporated off). Deuterated solvents (CDCl3) are used because ²H (deuterium) has an even mass number (no net nuclear spin, I = 1) and does not produce interfering ¹H NMR signals. Shaking with D2O causes -OH and -NH protons to disappear via proton exchange.
8	What is a racemic mixture (racemate), why is it optically inactive, and why does nucleophilic addition to an unsymmetrical planar carbonyl (RCHO) produce a racemate?	A racemic mixture contains an equimolar (50:50) mixture of two enantiomers; it is optically inactive because the two enantiomers rotate plane-polarised light by equal angles in opposite directions, cancelling out. In nucleophilic addition of HCN (CN⁻) to an unsymmetrical aldehyde (e.g., ethanal), the C=O group is trigonal planar, so the CN⁻ nucleophile attacks with equal probability from above or below the plane, producing a 50:50 racemate.
9	Draw/describe the structure of a 2-aminocarboxylic acid (α-amino acid) at its isoelectric point (zwitterion), at low pH (acidic), and at high pH (alkaline).	At the isoelectric point (neutral zwitterion): internal proton transfer gives +H3N–CHR–COO⁻ (high melting point crystalline solid due to ionic lattice attractions). At low pH (strongly acidic, excess H⁺): carboxylate group is protonated to give the cation +H3N–CHR–COOH. At high pH (strongly alkaline, excess OH⁻): ammonium group is deprotonated to give the anion H2N–CHR–COO⁻.
10	State the chemical tests to distinguish between an Aldehyde and a Ketone using Tollens' Reagent and Fehling's Solution.	Aldehydes are easily oxidised to carboxylic acids (or carboxylate ions), whereas ketones cannot be oxidised easily: 1) Tollens' reagent (ammoniacal silver nitrate, [Ag(NH3)2]⁺): warm gently in a water bath → aldehydes reduce Ag⁺ to metallic silver(0), forming a 'silver mirror' on the test tube wall; ketones give no change. 2) Fehling's / Benedict's solution (blue aqueous Cu²⁺ complex): warm → aldehydes reduce blue Cu²⁺ to a brick-red precipitate of copper(I) oxide (Cu2O); ketones remain blue.
11	How do you oxidise a primary alcohol to an aldehyde vs a carboxylic acid, and what happens with secondary and tertiary alcohols?	Reagent: acidified potassium dichromate(VI) (K2Cr2O7 / H2SO4, colour changes from orange Cr2O7²⁻ to green Cr³⁺). 1) Primary alcohol + distillation: distil off the aldehyde product immediately before further oxidation. 2) Primary alcohol + heating under reflux with excess oxidising agent: oxidises fully to a carboxylic acid. 3) Secondary alcohol + reflux: oxidises to a ketone. 4) Tertiary alcohol: resists oxidation (stays orange).
12	Write the equations and Electrophilic Substitution mechanism for Friedel–Crafts Acylation of benzene with ethanoyl chloride and AlCl3.	Reagents: acyl chloride (e.g., CH3COCl) + anhydrous aluminium chloride catalyst (AlCl3) under reflux in dry ether. Generation of acylium electrophile: CH3COCl + AlCl3 → CH3C⁺=O + AlCl4⁻. Mechanism: curly arrow from benzene's delocalised π-ring to the positively charged C⁺ of CH3CO⁺, forming the horseshoe intermediate, then curly arrow from the C–H bond back into the ring to form phenylethanone (a ketone) + H⁺, which reacts with AlCl4⁻ → AlCl3 + HCl.
13	Contrast Structural Isomerism (chain, position, functional group) with Stereoisomerism (E/Z and optical isomerism).	Structural isomers have the same molecular formula but different structural formulae (chain branching, functional group position, or different functional groups such as aldehydes vs ketones). Stereoisomers have the same structural connectivity but a different 3D spatial arrangement of atoms: E/Z isomers arise from restricted rotation around a C=C double bond (using Cahn–Ingold–Prelog atomic number priority rules: Z = 'zusammen' on same side, E = 'entgegen' on opposite sides); Optical isomers (enantiomers) are non-superimposable mirror images around a chiral carbon atom bonded to 4 different groups.
14	Explain the four pieces of structural information obtained from a high-resolution Proton (¹H) NMR spectrum, including the (n + 1) spin–spin splitting rule.	1) Number of signals = number of distinct non-equivalent proton environments; 2) Chemical shift (δ in ppm) = type of chemical environment; 3) Integration trace (peak area ratio) = relative number of protons in each environment; 4) Spin–spin splitting pattern (n + 1 rule): if a proton environment has n chemically non-equivalent protons on adjacent carbon atoms, its peak splits into (n + 1) sub-peaks (0 → singlet, 1 → doublet, 2 → triplet 1:2:1, 3 → quartet 1:3:3:1). Note: -OH and -NH protons always appear as singlets and do not split adjacent protons.
15	Compare the two laboratory synthetic routes to prepare a primary aliphatic amine vs an aromatic amine (phenylamine).	1) Primary aliphatic amine (without forming secondary/tertiary/quaternary ammonium salts): either heat a haloalkane with a large EXCESS of ethanolic ammonia, OR (better 2-step route with 1 product) react haloalkane with KCN in aqueous ethanol (nucleophilic substitution to form a nitrile, extending chain by 1 C) followed by catalytic hydrogenation with H2 / Ni catalyst (or LiAlH4 in dry ether) to form RCH2NH2. 2) Aromatic amine (phenylamine): reduce nitrobenzene by heating under reflux with Tin (Sn) and concentrated HCl, followed by excess NaOH to liberate free C6H5NH2.
16	Compare the rate of hydrolysis of chloroalkanes, bromoalkanes, and iodoalkanes with aqueous silver nitrate in ethanol.	Rate of nucleophilic substitution increases down Group 7: 1-iodobutane > 1-bromobutane > 1-chlorobutane (yellow AgI precipitate appears fastest, then cream AgBr, then white AgCl slowest). Although C–Cl is the most polar bond, bond enthalpy is the dominant factor: the C–I bond is the longest and weakest (lowest bond enthalpy, 238 kJ mol⁻¹ vs 346 kJ mol⁻¹ for C–Cl), so C–I breaks with the lowest activation energy.
17	Compare the structures of the condensation polymers Terylene (PET), Nylon-6,6, and Kevlar, and explain why condensation polymers are biodegradable whereas poly(alkenes) are not.	Terylene (polyester): formed from benzene-1,4-dicarboxylic acid + ethane-1,2-diol (ester links -COO-). Nylon-6,6 (polyamide): formed from hexanedioic acid (or hexanedioyl dichloride) + hexane-1,6-diamine (amide links -CONH-). Kevlar (polyamide): formed from benzene-1,4-dicarboxylic acid + benzene-1,4-diamine (flat rigid chains packed with strong H-bonds). Polyesters and polyamides have polar C=O bonds susceptible to nucleophilic attack and hydrolysis by water, acids, alkalis, or enzymes; poly(alkenes) have only non-polar, strong C–C and C–H bonds.
18	In Infrared (IR) Spectroscopy, how do you identify an Alcohol, an Aldehyde/Ketone, and a Carboxylic Acid, and what is the 'fingerprint region'?	IR radiation is absorbed when its frequency matches the natural vibrational frequency of a polar covalent bond: 1) Alcohol O–H: broad absorption at 3200–3550 cm⁻¹ (broad due to hydrogen bonding) and NO C=O peak; 2) Aldehyde/Ketone C=O: sharp, strong peak at ~1680–1750 cm⁻¹ and NO O–H peak; 3) Carboxylic Acid: BOTH a sharp C=O peak at ~1700 cm⁻¹ AND a very broad O–H peak at 2500–3300 cm⁻¹. The fingerprint region (below 1500 cm⁻¹) contains complex C–C/C–O vibrations matched against a database.
19	How does the anticancer drug Cisplatin ([Pt(NH3)2Cl2]) prevent DNA replication in rapidly dividing tumour cells?	Cisplatin is a square planar platinum(II) complex with two Cl⁻ ligands in the cis (90°) configuration. Inside the cell, water/DNA displaces the two chloride ligands in a ligand substitution reaction, and the platinum ion forms coordinate covalent bonds to nitrogen atoms on two adjacent guanine bases on DNA, kinking the DNA strand so it cannot unwind and replicate.
20	What is the distinctive ¹H NMR splitting pattern for an ethyl group (–CH2CH3) vs an isopropyl group (–CH(CH3)2)?	An ethyl group (–CH2CH3) always gives a characteristic Quartet (area 2, from the –CH2– split by 3 adjacent H's: 3 + 1 = 4) and a Triplet (area 3, from the –CH3 split by 2 adjacent H's: 2 + 1 = 3). An isopropyl group (–CH(CH3)2) gives a Septet/Multiplet (area 1, split by 6 equivalent H's: 6 + 1 = 7) and a Doublet (area 6, from two equivalent –CH3 groups split by 1 adjacent CH).
21	Present the three pieces of evidence (bond lengths, enthalpy of hydrogenation, and resistance to bromination) that disprove Kekulé's alternating C=C/C–C model of Benzene.	1) X-ray diffraction shows all six C–C bond lengths in benzene are identical (0.139 nm), intermediate between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm). 2) Thermochemistry: ΔH(hydrogenation) of cyclohexene is −120 kJ mol⁻¹, so Kekulé's cyclohexa-1,3,5-triene would be −360 kJ mol⁻¹, whereas actual benzene is −208 kJ mol⁻¹ (152 kJ mol⁻¹ more stable due to delocalisation energy). 3) Reactivity: Benzene does not decolourise bromine water (undergoes electrophilic substitution rather than addition to preserve the delocalised π-ring).
22	Arrange phenylamine (C6H5NH2), ammonia (NH3), and ethylamine (CH3CH2NH2) in order of increasing base strength (pKb / Ka of conjugate acid) and explain why.	Weakest base → Strongest base: Phenylamine < Ammonia < Ethylamine. Base strength depends on the availability of the lone pair of electrons on the N atom to accept a proton (H⁺): 1) In phenylamine, the N lone pair delocalises into the aromatic benzene π-ring, making it much less available to accept H⁺; 2) In ethylamine, the ethyl group has a positive inductive effect (+I, electron-pushing), increasing electron density on the N atom and stabilising the ethylammonium ion.
23	How do you identify the functional groups in a multi-functional organic molecule using a sequence of test-tube reactions?	1) Add bromine water: decolourises orange to colourless → Alkene (C=C) or Phenol. 2) Add Na2CO3 or NaHCO3(aq): effervescence of CO2 → Carboxylic acid (-COOH). 3) Add 2,4-dinitrophenylhydrazine (2,4-DNPH / Brady's reagent): bright orange/yellow precipitate → Carbonyl (Aldehyde or Ketone; recrystallise the 2,4-DNPH derivative and measure its melting point to identify the specific carbonyl). 4) Warm with Tollens' reagent: silver mirror → Aldehyde. 5) Warm with acidified K2Cr2O7: orange to green → Primary/Secondary alcohol or Aldehyde.
24	Explain the mechanism of Electrophilic Addition of HBr to an unsymmetrical alkene (propene) and why Markovnikov's Rule favours the major product.	1) The high electron density of the C=C π-bond attacks the δ+ H of H–Br, breaking the H–Br bond heterolytically; 2) In propene, protonation forms either a secondary carbocation intermediate (CH3–CH⁺–CH3) or a primary carbocation (CH3–CH2–CH2⁺); 3) Alkyl groups are electron-releasing (+I positive inductive effect), stabilising the positive charge, so the more stable secondary carbocation forms faster and is attacked by the :Br⁻ ion to give 2-bromopropane as the major product.
25	Describe the reagents and Nucleophilic Addition mechanism for reducing aldehydes/ketones to alcohols and for adding HCN to form hydroxynitriles.	Reduction: aqueous sodium tetrahydridoborate(III) (NaBH4), which generates the hydride nucleophile (:H⁻) that attacks the δ+ carbonyl carbon, followed by protonation from water to form a primary alcohol (from aldehyde) or secondary alcohol (from ketone). Hydroxynitrile formation: KCN followed by dilute acid (HCl/H2SO4)—never HCN gas alone for safety; :CN⁻ attacks the δ+ C=O carbon, and the intermediate alkoxide ion O⁻ is protonated by H⁺/HCN, extending the carbon chain by 1 carbon.
26	In organic synthesis, which two key reactions increase the length of the carbon chain by forming a new C–C bond?	1) Reaction with Cyanide (:CN⁻): Nucleophilic substitution of a haloalkane with KCN in aqueous ethanol (RX + CN⁻ → RCN + X⁻, adds 1 C) OR Nucleophilic addition of HCN (KCN/H⁺) to an aldehyde/ketone to form a hydroxynitrile. The nitrile (-CN) can then be reduced to a primary amine (-CH2NH2) using H2/Ni, or hydrolysed with dilute HCl under reflux to a carboxylic acid (-COOH). 2) Friedel–Crafts Acylation/Alkylation of benzene with RCOCl / RX and anhydrous AlCl3.
27	How is High-Resolution Mass Spectrometry used to distinguish between compounds with the same nominal Mr (e.g., C2H4O2, C3H8O, and C3H4O)?	Low-resolution mass spectrometry gives nominal whole-number m/z values (e.g., propanal C3H6O, butane C4H10, and ethanoic acid all have nominal Mr = 58 or 60). High-resolution mass spectrometry measures m/z to 4 or 5 decimal places using exact isotopic masses (¹²C = 12.0000, ¹H = 1.0078, ¹⁶O = 15.9949, ¹⁴N = 14.0031), uniquely identifying the exact molecular formula from the M⁺ peak.
28	Describe the Nucleophilic Addition–Elimination mechanism of Acyl Chlorides (RCOCl) and Acid Anhydrides with water, alcohols, ammonia, and primary amines.	The lone pair on the nucleophile (:OH2, :OHR', :NH3, :NH2R') attacks the strongly δ+ carbonyl carbon of the acyl chloride; the π-bond breaks to form a tetrahedral intermediate, then the lone pair on oxygen reforms the C=O double bond and expels the Cl⁻ leaving group, followed by loss of H⁺. Products with RCOCl: 1) + H2O → carboxylic acid + misty fumes of HCl; 2) + R'OH → ester + HCl; 3) + 2NH3 → primary amide + NH4Cl; 4) + 2R'NH2 → N-substituted secondary amide + R'NH3⁺Cl⁻.
29	Why does phenol (C6H5OH) react much more readily with bromine water than benzene does (OCR/Edexcel)?	In phenol, a lone pair of electrons from the p-orbital of the oxygen atom in the -OH group is partially delocalised into the aromatic π-ring. This increases the electron density of the benzene ring (activating it in the 2, 4, and 6 positions), enabling phenol to polarise Br2 molecules directly without a halogen carrier catalyst, decolourising bromine water and forming a white precipitate of 2,4,6-tribromophenol.
30	Write the equations and Electrophilic Substitution mechanism for the Nitration of Benzene.	Reagents: concentrated nitric acid (HNO3) + concentrated sulfuric acid (H2SO4 catalyst) at 50°C. Generation of the nitronium ion electrophile: HNO3 + 2H2SO4 → NO2⁺ + 2HSO4⁻ + H3O⁺ (or HNO3 + H2SO4 → NO2⁺ + HSO4⁻ + H2O). Mechanism: curly arrow from benzene's delocalised π-electron ring to N of NO2⁺ forming a horseshoe Wheland intermediate (covering 5 carbons with a + charge), then curly arrow from C–H bond back into the ring to restore aromaticity and release H⁺ (which regenerates the H2SO4 catalyst: H⁺ + HSO4⁻ → H2SO4).
31	How do Thin-Layer Chromatography (TLC), Column Chromatography, and Gas Chromatography–Mass Spectrometry (GC–MS) separate mixtures?	All chromatography relies on the balance between adsorption onto the stationary phase (often silica SiO2 or alumina Al2O3) and solubility in the mobile phase. In Gas Chromatography (GC), the stationary phase is a high-boiling liquid lining a long coiled capillary tube in a temperature-controlled oven, and the mobile phase is an inert carrier gas (He or N2). Each component has a characteristic retention time, and the coupled Mass Spectrometer (GC–MS) identifies each component via its molecular ion peak and fragmentation pattern.
32	Describe the experimental steps to purify an impure organic solid (e.g., Aspirin) by Recrystallisation and test its purity.	1) Dissolve the crude solid in the MINIMUM volume of hot solvent (so the solution is saturated on cooling and yield is maximised); 2) Filter the hot solution rapidly through fluted filter paper to remove insoluble impurities; 3) Allow the filtrate to cool slowly in an ice bath so pure crystals reform while soluble impurities remain dissolved in the cold solvent; 4) Filter under reduced pressure using a Büchner funnel and vacuum pump, wash with ice-cold solvent, and dry in a warm oven; 5) Measure its melting point: a pure solid has a sharp melting point matching the data book value.