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Unit 4

Unit 4: Organic 2 and Physical 2 Flashcards

Active recall practice for key definitions, equations, mechanisms, and experimental concepts across optical isomerism, aldehydes and ketones, carboxylic acids, aromatic chemistry, amines, polymers, amino acids, organic synthesis, NMR spectroscopy and chromatography.

Unit 4: Organic 2 & Physical 2 Master Deck: read all 26 questions and answers
  1. What is meant by the overall order of a reaction?
    The sum of the powers to which the reactant concentrations are raised in the rate equation: Overall Order = m + n in Rate = k[A]m[B]n.
  2. How do you derive the units of the rate constant k?
    Rearrange the rate equation for k: k = Rate / ([A]m[B]n). Substitute mol dm⁻³ s⁻¹ for rate and mol dm⁻³ for each concentration term, then cancel out units.
  3. How is activation energy (Ea) calculated from an Arrhenius plot of ln k against 1/T?
    The gradient of the straight line equals -Ea / R. Therefore, Ea = -gradient × R (in J mol⁻¹; divide by 1000 for kJ mol⁻¹).
  4. Why does a small temperature increase cause a significant increase in reaction rate?
    A much greater fraction of molecules possess kinetic energy greater than or equal to the activation energy (E ≥ Ea), leading to a significantly higher frequency of successful collisions per second.
  5. What is the formula for calculating the partial pressure of a gas in an equilibrium mixture?
    p(A) = x(A) × Ptotal
    where x(A) is the mole fraction: n(A) / ntotal.
  6. Which single experimental factor alters the numerical value of Kp?
    Temperature only. Changes in pressure, concentration, or the addition of a catalyst have zero effect on the numerical value of Kp.
  7. What is a chiral centre in organic molecules?
    A carbon atom bonded to four different atoms or groups of atoms, producing non-superimposable mirror image stereoisomers.
  8. Why is an equimolar (racemic) mixture of enantiomers optically inactive?
    It contains equal amounts of both (+) and (-) enantiomers. The clockwise rotation of plane-polarised light by one enantiomer is exactly cancelled out by the equal anticlockwise rotation of the other.
  9. Why does nucleophilic addition to an unsymmetrical planar carbonyl form a racemic mixture?
    The carbonyl group (C=O) is planar. The incoming nucleophile has an equal probability (50:50) of attacking from above or below the plane, forming equal amounts of both enantiomers.
  10. What reagents and nucleophile are used to reduce aldehydes and ketones to alcohols?
    Reagent: Sodium tetrahydridoborate(III) (NaBH₄) in aqueous or ethanolic solution.
    Nucleophile: Hydride ion (:H⁻).
  11. What visual observation occurs when Tollens' reagent is warmed with an aldehyde versus a ketone?
    Aldehyde: Silver mirror forms on the inner wall of the test tube ([Ag(NH₃)₂]⁺ + e⁻ → Ag(s) + 2NH₃).
    Ketone: No reaction (solution remains colourless).
  12. What visual observation occurs when Fehling's solution is warmed with an aldehyde versus a ketone?
    Aldehyde: Blue solution turns into a brick-red precipitate of copper(I) oxide (Cu₂O).
    Ketone: No reaction (solution remains blue).
  13. What is the difference between acid and alkaline hydrolysis of an ester?
    Acid hydrolysis: Reversible, heated under reflux with dilute acid (HCl/H₂SO₄) to yield carboxylic acid + alcohol.
    Alkaline hydrolysis: Irreversible (saponification), heated under reflux with dilute NaOH to yield carboxylate salt + alcohol.
  14. What reaction mechanism type occurs when acyl chlorides react with nucleophiles?
    Nucleophilic addition-elimination. The nucleophile attacks the carbonyl carbon, and a chloride ion (Cl⁻) is subsequently eliminated.
  15. Why are acid anhydrides preferred over acyl chlorides in industrial manufacturing (e.g. aspirin)?
    Acid anhydrides are less corrosive, react less violently, do not produce hazardous, toxic hydrogen chloride (HCl) gas fumes, and are cheaper to manufacture.
  16. What thermochemical evidence demonstrates that benzene is more stable than the theoretical Kekule cyclohexa-1,3,5-triene?
    Hydrogenation of cyclohexene is -120 kJ mol⁻¹ (predicting -360 kJ mol⁻¹ for three double bonds). The experimental enthalpy of hydrogenation of benzene is only -208 kJ mol⁻¹, proving benzene is 152 kJ mol⁻¹ more stable due to delocalised pi electron ring resonance.
  17. What equations show how the electrophile NO₂⁺ is generated for benzene nitration?
    HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻
    Concentrated sulfuric acid acts as a stronger acid and catalyst, protonating nitric acid.
  18. What are the structural criteria for the arenium intermediate carbocation in electrophilic aromatic substitution?
    A horseshoe-shaped delocalised pi system. The horseshoe covers the five carbons other than the one bonded to both H and the electrophile (for example NO₂), open at that carbon, with the positive charge inside the ring.
  19. Why does methylbenzene react faster in electrophilic substitution than benzene?
    The methyl group releases electron density into the aromatic pi cloud via a positive inductive effect (+I), increasing electron density and making the ring more attractive to electrophiles.
  20. Arrange phenylamine, ammonia, and ethylamine in order of increasing base strength.
    Phenylamine < Ammonia < Ethylamine
    Phenylamine is weakest (lone pair delocalises into ring). Ammonia is intermediate. Ethylamine is strongest (+I effect of ethyl group pushes electron density onto nitrogen).
  21. What reagents and conditions are used to reduce nitrobenzene to phenylamine?
    Heat under reflux with tin (Sn) and concentrated hydrochloric acid (HCl), followed by neutralisation with aqueous sodium hydroxide (NaOH):
    ArNO₂ + 6[H] → ArNH₂ + 2H₂O.
  22. What monomers form the polyamide Nylon-6,6 and the polyester Terylene (PET)?
    Nylon-6,6: Hexane-1,6-diamine + hexanedioic acid (forming amide links, -CO-NH-).
    Terylene: Benzene-1,4-dicarboxylic acid + ethane-1,2-diol (forming ester links, -COO-).
  23. Why are condensation polymers biodegradable whereas addition polymers are not?
    Condensation polymers contain polar bonds (C-N or C-O) in their chain backbone that are susceptible to nucleophilic attack and hydrolysis by acids, alkalis, or enzymes. Addition polymers have unreactive, non-polar C-C backbones.
  24. What is an amino acid zwitterion and how does it react in acidic versus basic solution?
    A dipolar ion with both -NH₃⁺ and -COO⁻ groups.
    In acid: -COO⁻ accepts H⁺ to become -COOH (overall +1 cation).
    In base: -NH₃⁺ loses H⁺ to become -NH₂ (overall -1 anion).
  25. What is the (n+1) splitting rule in high-resolution ¹H NMR spectroscopy?
    If a proton has n equivalent protons on adjacent carbon atoms, its resonance signal is split into (n+1) peaks: 0 adjacent → Singlet, 1 → Doublet, 2 → Triplet, 3 → Quartet, 6 → Septet.
  26. What are the roles of hot filtration and cold solvent washing in recrystallisation?
    Hot filtration: Removes insoluble impurities while keeping the product dissolved in hot solvent.
    Ice-cold wash: Removes soluble impurities on crystal surfaces without dissolving the purified solid.
Spec 3.3.8: Aldehydes & Ketones (Carbonyls): read all 6 questions and answers
  1. Why are aldehydes and ketones susceptible to nucleophilic addition?
    The carbonyl group is strongly polarised (C(δ+)=O(δ−)) due to the higher electronegativity of oxygen. The electron-deficient carbonyl carbon is readily attacked by nucleophiles.
  2. What reagent reduces aldehydes and ketones to alcohols, and what is the nucleophile?
    Reagent: Sodium tetrahydridoborate(III) (NaBH₄) in aqueous or ethanolic solution.
    Nucleophile: Hydride ion (:H⁻) with a lone pair.
  3. What visual observation occurs when Tollens' reagent is warmed with an aldehyde versus a ketone?
    Aldehyde: Silver mirror forms on the glass wall (reduction of [Ag(NH₃)₂]⁺ to metallic Ag).
    Ketone: No reaction / solution remains colourless.
  4. What visual observation occurs when Fehling's solution is warmed with an aldehyde versus a ketone?
    Aldehyde: Blue solution forms a brick-red precipitate of copper(I) oxide (Cu₂O).
    Ketone: No reaction / solution remains blue.
  5. What reagents and conditions are used to add cyanide to aldehydes or ketones?
    Potassium cyanide (KCN) followed by dilute acid (or acidified KCN), yielding a hydroxynitrile. Using KCN provides a higher concentration of free :CN⁻ nucleophile.
  6. Why is the reaction of carbonyl compounds with KCN valuable in synthetic pathways?
    It increases the carbon chain length by one carbon atom and produces a bifunctional molecule containing both an -OH group and a -C≡N group (which can be reduced to an amine).
Spec 3.3.11: Amines & Nitrogen Compounds: read all 6 questions and answers
  1. Rank phenylamine, ammonia, and ethylamine in order of increasing base strength (weakest to strongest).
    Phenylamine < Ammonia < Ethylamine
    Phenylamine: lone pair delocalises into aromatic ring (least available). Ammonia: neutral baseline. Ethylamine: ethyl group pushes electron density onto nitrogen via positive inductive (+I) effect.
  2. Why does reacting a halogenoalkane with excess ammonia favour primary amine formation?
    A large excess of ammonia ensures that incoming haloalkane molecules are far more likely to collide with ammonia rather than the primary amine product, minimising multiple alkylations (secondary and tertiary amines).
  3. How can a pure primary aliphatic amine be synthesised without producing a mixture of amines?
    React a halogenoalkane with potassium cyanide (KCN) in ethanol to form a nitrile, then reduce the nitrile with H₂ and a nickel (Ni) catalyst (or LiAlH₄).
  4. What reagents and conditions are used to reduce nitrobenzene to phenylamine?
    Heat under reflux with tin (Sn) and concentrated hydrochloric acid (HCl), followed by treatment with aqueous sodium hydroxide (NaOH):
    C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O.
  5. What are quaternary ammonium salts and what is their primary commercial application?
    Compounds where four alkyl groups are bonded to nitrogen ([R₄N]⁺ X⁻). They are used as cationic surfactants in fabric softeners and hair conditioners.
  6. What type of reaction occurs when an amine reacts with an acyl chloride?
    Nucleophilic addition-elimination. The amine nitrogen lone pair attacks the carbonyl carbon, and a molecule of HCl is eliminated to form an N-substituted amide.
Spec 3.3.13: Amino Acids and Proteins: read all 5 questions and answers
  1. What is a zwitterion and how does an alpha-amino acid form one in aqueous solution?
    A zwitterion is an internal dipolar ion carrying both positive and negative formal charges with an overall net charge of zero. In an amino acid, the basic -NH₂ group accepts a proton from the acidic -COOH group, forming ⁺H₃N-CH(R)-COO⁻.
  2. What ionic structure does an alpha-amino acid adopt in strongly acidic conditions (low pH)?
    In strongly acidic conditions (excess H⁺), the carboxylate group accepts a proton to become -COOH while the amino group remains protonated as -NH₃⁺. The amino acid exists as a cation with a net +1 charge (⁺H₃N-CH(R)-COOH).
  3. What ionic structure does an alpha-amino acid adopt in strongly alkaline conditions (high pH)?
    In strongly alkaline conditions (excess OH⁻), the -NH₃⁺ group loses a proton to reform -NH₂ while the carboxyl group remains deprotonated as -COO⁻. The amino acid exists as an anion with a net -1 charge (H₂N-CH(R)-COO⁻).
  4. How is a peptide bond formed between two amino acids, and what is the primary structure of a protein?
    A peptide bond (-CONH-) forms through a condensation reaction between the -COOH group of one amino acid and the -NH₂ group of another, eliminating H₂O. Primary structure is the specific linear sequence of amino acids in a polypeptide chain.
  5. What interactions stabilise the secondary and tertiary structures of proteins?
    Secondary structure (alpha-helices and beta-pleated sheets) is stabilised by hydrogen bonding between backbone C=O and N-H groups. Tertiary structure (3D folding) is held by disulfide bridges (-S-S-), ionic bonds between charged R-groups, hydrogen bonds, and van der Waals interactions between R-groups.
Spec 3.3.10: Aromatic Chemistry (Arenes): read all 6 questions and answers
  1. What thermochemical evidence demonstrates that benzene is more stable than theoretical cyclohexa-1,3,5-triene?
    Hydrogenation of cyclohexene is -120 kJ mol⁻¹ (predicting -360 kJ mol⁻¹ for three double bonds). The experimental value for benzene is only -208 kJ mol⁻¹. Benzene is 152 kJ mol⁻¹ more stable due to delocalised pi electron ring resonance.
  2. Why does benzene undergo electrophilic substitution instead of electrophilic addition?
    Electrophilic addition would permanently destroy the stable aromatic pi electron ring (losing 152 kJ mol⁻¹ of stabilisation). Substitution regenerates and preserves the stable delocalised aromatic system.
  3. What chemical equation shows the generation of the nitronium ion (NO₂⁺) electrophile?
    HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻
    Concentrated sulfuric acid acts as a catalyst and protonates nitric acid.
  4. What are the essential drawing rules for the intermediate arenium carbocation in benzene substitution?
    A horseshoe-shaped delocalised pi system spanning 5 carbons, open at the sp³ carbon that carries both the incoming electrophile and hydrogen, with the positive charge located inside the inner ring.
  5. What catalyst is required for Friedel-Crafts acylation, and what equation shows electrophile formation?
    Anhydrous aluminium chloride (AlCl₃).
    Equation: CH₃COCl + AlCl₃ ⇌ CH₃CO⁺ + AlCl₄⁻
    AlCl₃ acts as a Lewis acid catalyst.
  6. Why does methylbenzene react significantly faster than benzene in electrophilic substitution?
    The methyl group pushes electron density into the aromatic pi ring via a positive inductive effect (+I), increasing pi electron density and activating the ring towards electrophilic attack.
Spec 3.3.9: Carboxylic Acids & Derivatives: read all 6 questions and answers
  1. Why do carboxylic acids have higher boiling points than alcohols of similar relative molecular mass?
    Carboxylic acids form strong intermolecular hydrogen bonds, frequently existing as stable hydrogen-bonded dimers in solid and liquid states.
  2. What reagents and conditions are required for laboratory esterification?
    Heat a carboxylic acid and an alcohol under reflux in the presence of a concentrated sulfuric acid (conc. H₂SO₄) catalyst. The reaction is reversible.
  3. Compare the reversibility and products of acid versus alkaline hydrolysis of esters.
    Acid hydrolysis: Reversible, heated with dilute acid to form carboxylic acid + alcohol.
    Alkaline hydrolysis (saponification): Irreversible, heated with aqueous NaOH to form carboxylate salt + alcohol.
  4. What type of mechanism takes place when acyl chlorides react with water, alcohols, or amines?
    Nucleophilic addition-elimination. The nucleophile attacks the carbonyl carbon, and a chloride ion (Cl⁻) is eliminated.
  5. What are the observations when ethanoyl chloride reacts with cold water?
    A vigorous exothermic reaction producing steamy, acidic white fumes of hydrogen chloride (HCl) gas and ethanoic acid.
  6. Why are acid anhydrides preferred over acyl chlorides in industrial manufacturing (such as aspirin synthesis)?
    Acid anhydrides are less expensive, react less violently, do not produce hazardous, corrosive HCl fumes (producing a harmless carboxylic acid byproduct), and are less sensitive to moisture.
Spec 3.3.7: Optical Isomerism & Chirality: read all 6 questions and answers
  1. What is a chiral centre in organic molecules?
    A carbon atom bonded to four different atoms or groups of atoms, creating non-superimposable mirror images.
  2. Define the term enantiomers.
    Optical isomers that are non-superimposable mirror images of each other.
  3. How do enantiomers differ in their physical and chemical properties?
    They have identical physical properties (melting point, boiling point, density) and identical chemical reactivity with achiral reagents. They differ only in the direction they rotate plane-polarised light and in reactions with other chiral molecules.
  4. How does a polarimeter distinguish between two enantiomers?
    Plane-polarised light is passed through the sample. One enantiomer rotates the plane of polarisation clockwise (+ or d-enantiomer), while the other rotates it by the same angle anticlockwise (- or l-enantiomer).
  5. What is a racemic mixture (racemate) and why is it optically inactive?
    A 50:50 (equimolar) mixture of both enantiomers. It is optically inactive because the clockwise rotation by one enantiomer is exactly cancelled out by the equal anticlockwise rotation of the other.
  6. Why does nucleophilic addition of HCN to an unsymmetrical planar aldehyde or ketone produce a racemate?
    The carbonyl group (C=O) has trigonal planar geometry. The nucleophile (:CN⁻) has an equal probability (50:50) of attacking from above or below the plane, yielding equal quantities of both enantiomers.
Spec 3.3.15: Organic Analysis and NMR: read all 4 questions and answers
  1. What does the number of peaks in a Carbon-13 (¹³C) NMR spectrum tell you about a molecule?
    The number of separate peaks in a ¹³C NMR spectrum equals the number of non-equivalent (chemically distinct) carbon environments in the molecule. Symmetrical molecules exhibit fewer peaks than total carbon atoms.
  2. Why is tetramethylsilane (TMS) used as the standard reference in NMR spectroscopy?
    TMS, Si(CH₃)₄, gives a single intense, sharp reference peak at exactly δ = 0.00 ppm because all 12 protons (and 4 carbons) are equivalent. It is chemically inert, non-toxic, volatile (easily removed), and absorbs far upfield from most organic signals.
  3. What three key pieces of information are obtained from a high-resolution Proton (¹H) NMR spectrum?
    1. Chemical shift (δ in ppm): indicates the chemical and electronic environment of protons. 2. Integration (peak area): gives the relative ratio of protons in each environment. 3. Splitting pattern (multiplet): reveals the number of non-equivalent protons on adjacent carbons via the (n+1) rule.
  4. State the (n+1) splitting rule in ¹H NMR and describe the patterns for n = 0, 1, 2, and 3 adjacent protons.
    A signal splits into (n+1) sub-peaks, where n is the number of non-equivalent protons attached to adjacent carbon atoms. n=0: singlet (1); n=1: doublet (1:1); n=2: triplet (1:2:1); n=3: quartet (1:3:3:1).
Spec 3.3.12: Condensation Polymers: read all 6 questions and answers
  1. What is the defining distinction between condensation and addition polymerisation?
    Condensation polymerisation joins monomers with two functional groups together with the elimination of a small molecule (such as H₂O or HCl) at each bond. Addition polymerisation involves opening C=C double bonds with no byproducts.
  2. What monomers form Nylon-6,6 and what is the linkage?
    Hexane-1,6-diamine and hexanedioic acid react with the elimination of water to form a polyamide containing amide linkages (-CO-NH-).
  3. What monomers form Terylene (PET) and what is the linkage?
    Benzene-1,4-dicarboxylic acid and ethane-1,2-diol react with the elimination of water to form a polyester containing ester linkages (-COO-).
  4. Why do polyamides have higher tensile strength and melting points than polyalkenes?
    Polyamides form extensive intermolecular hydrogen bonds between adjacent chains (between C=O oxygen lone pairs and polar N-H hydrogens), which require substantial energy to overcome.
  5. Why are condensation polymers biodegradable whereas polyalkenes are non-biodegradable?
    Condensation polymers have polar bonds (C-N or C-O) in their chain backbones that can be attacked by nucleophiles and hydrolysed by acids, alkalis, or environmental enzymes. Polyalkenes have non-polar, unreactive C-C backbones.
  6. What are the products of the alkaline hydrolysis of Nylon-6,6?
    Heating with aqueous sodium hydroxide (NaOH) hydrolyses the amide bonds to yield hexane-1,6-diamine and the sodium salt of hexanedioic acid (sodium hexanedioate).
Spec 3.3.14: Organic Synthesis: read all 10 questions and answers
  1. Why do chemists try to design processes that do not need a solvent?
    Solvents can be toxic and flammable. They must be removed and disposed of, and they add cost and waste.
  2. Why do chemists try to use non-hazardous starting materials?
    They are safer for workers and the environment, and the safety measures needed are cheaper.
  3. Why do chemists prefer a synthesis with fewer steps?
    Each step loses some product, so fewer steps give a higher overall yield. Fewer steps also use less energy, time and materials, and make less waste.
  4. How do you find the overall yield of a three-step synthesis where each step gives 80%?
    Multiply the step yields: 0.8 × 0.8 × 0.8 = 0.512, so the overall yield is 51%.
  5. Which type of reaction has 100% atom economy?
    Addition reactions. Substitution and elimination reactions have a lower atom economy, because they also make a second product.
  6. How can you add one carbon atom to the chain of a halogenoalkane?
    Reflux with KCN in ethanol/water. This makes a nitrile with one more carbon atom.
  7. Give the reagent that reduces a nitrile to a primary amine.
    H₂ with a nickel catalyst (or LiAlH₄).
  8. How do you make phenylamine from benzene?
    1. Conc. HNO₃ and conc. H₂SO₄ below 55 °C gives nitrobenzene. 2. Sn and conc. HCl, then NaOH, gives phenylamine.
  9. How do you convert propene into propanone in two steps?
    1. Conc. H₂SO₄, then water, gives propan-2-ol. 2. Acidified K₂Cr₂O₇ under reflux oxidises it to propanone.
  10. A primary alcohol is oxidised with acidified K₂Cr₂O₇. How do you get the aldehyde rather than the carboxylic acid?
    For the aldehyde, distil it off as it forms. For the carboxylic acid, use excess oxidising agent and heat under reflux.
Spec 3.3.16: Chromatography: read all 10 questions and answers
  1. What does separation in chromatography depend on?
    The balance between solubility in the moving (mobile) phase and retention by the stationary phase.
  2. In thin-layer chromatography (TLC), what are the stationary and mobile phases?
    Stationary: a solid coated on a plate. Mobile: a solvent that moves up the plate.
  3. In column chromatography, what are the stationary and mobile phases?
    Stationary: a solid packed in a column. Mobile: a solvent that moves down the column.
  4. In gas chromatography (GC), what are the stationary and mobile phases?
    Stationary: a solid, or a solid coated by a liquid, packed in a column. Mobile: a gas passed through the column under pressure at high temperature.
  5. How do you calculate an Rf value?
    Rf = distance moved by the spot ÷ distance moved by the solvent front. Measure both from the start line.
  6. Why is the TLC start line drawn in pencil, above the solvent level?
    Pencil does not dissolve in the solvent. If the spots were below the solvent level, they would dissolve into the solvent in the beaker.
  7. How can you see colourless amino acid spots on a TLC plate?
    Spray with ninhydrin, which turns the spots purple, or use ultraviolet light.
  8. What is the retention time in GC?
    The time from injection of the sample to detection of a component. Compare it with standards run under the same conditions.
  9. Why is gas chromatography linked to mass spectrometry (GC-MS)?
    GC separates the mixture. MS identifies each separated component from its mass spectrum.
  10. How does TLC show that a product is pure?
    A pure substance gives only one spot.
Spec 3.3.11, 3.3.12, 3.3.13 and 3.3.15: Amines, Polymers, Proteins and NMR: read all 11 questions and answers
  1. A halogenoalkane reacts with ammonia. Show the series of products as substitution continues.
    RNH₂ → R₂NH → R₃N → R₄N⁺X⁻. Each step adds one alkyl group. The last product is a quaternary ammonium salt.
  2. What are aromatic amines, made by reducing nitro compounds, used to make?
    Dyes.
  3. What forces act between the chains of a polyamide such as nylon 6,6 or Kevlar?
    Hydrogen bonds between N-H on one chain and C=O on a neighbouring chain.
  4. What forces act between the chains of a polyester such as Terylene?
    Permanent dipole-dipole forces.
  5. Why are polyalkenes not biodegradable?
    They have only non-polar C-C and C-H bonds, so there is nothing for water or nucleophiles to attack.
  6. Give one advantage and one disadvantage of disposing of polymers by incineration.
    Advantage: it reduces the volume of waste and the energy can be recovered. Disadvantage: it makes CO₂, and toxic gases such as HCl (from PVC) must be removed.
  7. Give the structure of glycine in acid and in alkali.
    In acid: H₃N⁺CH₂COOH. In alkali: H₂NCH₂COO⁻.
  8. What holds the secondary structure of a protein (α-helix and β-pleated sheet)?
    Hydrogen bonds between N-H and C=O groups.
  9. How does cisplatin stop cancer cells dividing?
    One Cl⁻ ligand is replaced by water. The Pt then bonds to a nitrogen atom on a guanine base in DNA, usually forming two bonds. This distorts the DNA so it cannot replicate.
  10. Why is TMS, Si(CH₃)₄, used as the standard in NMR?
    All 12 H atoms are equivalent, so it gives a single peak, set at δ = 0. It is inert and volatile, and its peak lies away from most other peaks.
  11. Why are CDCl₃ or CCl₄ used as solvents in ¹H NMR?
    They contain no ¹H atoms, so they give no peak in the spectrum.
Spec 3.1.11, 3.3.8, 3.3.9 and 3.3.10: Rates, Carbonyls and Aromatic Chemistry: read all 18 questions and answers
  1. Sketch the rate-concentration graph for orders 0, 1 and 2.
    Order 0: horizontal line (rate does not change with [A]). Order 1: straight line through the origin. Order 2: upward curve through the origin (rate ∝ [A]²).
  2. How do you find the rate at a given time from a concentration-time graph?
    Draw a tangent to the curve at that time. The rate is the gradient of the tangent (ignore the minus sign for a reactant). The initial rate is the gradient of the tangent at t = 0.
  3. Ea = 50.0 kJ mol⁻¹, A = 1.00 × 10¹⁰ s⁻¹, R = 8.31 J K⁻¹ mol⁻¹. Find k at 298 K.
    Ea/RT = 50 000 / (8.31 × 298) = 20.19. k = 1.00 × 10¹⁰ × e^(-20.19) = 17.0 s⁻¹.
  4. Outline the mechanism for the reduction of ethanal by NaBH₄.
    1. Lone pair on H⁻ attacks the δ+ carbonyl C; the C=O π electrons move to O, giving an alkoxide ion. 2. The O⁻ takes H⁺ from water, giving ethanol. CH₃CHO + 2[H] → CH₃CH₂OH.
  5. Why does ethanal + KCN, then dilute acid, give a racemic mixture?
    The C=O group is planar, so CN⁻ attacks from either side with equal chance. Equal amounts of the two enantiomers form, so the product is optically inactive.
  6. Give the hazards of using KCN to make a hydroxynitrile.
    KCN is highly toxic. With acid it can release very toxic HCN gas. Use a fume cupboard and gloves. KCN is used instead of HCN because HCN is a very toxic gas.
  7. Give four uses of esters.
    Solvents (for example ethyl ethanoate in glues and nail-varnish remover), plasticisers, perfumes and food flavourings.
  8. What forms when a fat is heated with NaOH(aq)?
    Glycerol (propane-1,2,3-triol) and soap, the sodium salts of long-chain carboxylic acids (RCOO⁻Na⁺).
  9. How is biodiesel made, and what is it?
    Vegetable oil + 3CH₃OH → 3 methyl esters + glycerol, with a catalyst such as KOH. Biodiesel is a mixture of methyl esters of long-chain carboxylic acids.
  10. Give the three steps of the addition-elimination mechanism for an acyl chloride and a nucleophile.
    1. Lone pair on the nucleophile attacks the δ+ carbon; C=O π electrons move to O. 2. C=O reforms and Cl⁻ leaves. 3. H⁺ is lost from the nucleophile (often taken by Cl⁻ to give HCl).
  11. Sulfonation of benzene: reagent, electrophile, product and uses?
    Fuming sulfuric acid (conc. H₂SO₄ containing SO₃), warm. Electrophile SO₃. C₆H₆ + SO₃ → C₆H₅SO₃H (benzenesulfonic acid). Used to make surfactants (detergents) and sulfonamide drugs.
  12. Friedel-Crafts alkylation: how is the electrophile formed and the catalyst regenerated?
    RCl + AlCl₃ → R⁺ + AlCl₄⁻. Then H⁺ + AlCl₄⁻ → AlCl₃ + HCl. Example: C₆H₆ + CH₃CH₂Cl → C₆H₅CH₂CH₃ (ethylbenzene) + HCl.
  13. Describe the intermediate in electrophilic substitution of benzene.
    A partial ring (horseshoe) with a + charge. It covers the five carbons not bonded to both H and the electrophile. An arrow from the C-H bond back into the ring restores the delocalised ring as H⁺ is lost.
  14. Why does methylbenzene react faster than benzene, and where does substitution happen?
    The CH₃ group releases electrons into the ring, so the ring is more electron-rich. Substitution is mainly at positions 2 and 4 (for example nitration gives 1-methyl-2-nitrobenzene and 1-methyl-4-nitrobenzene).
  15. What forms when benzene reacts with Cl₂ in UV light?
    Addition to the ring: C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (1,2,3,4,5,6-hexachlorocyclohexane). The ring loses its delocalisation.
  16. Methylbenzene + Cl₂: compare UV light (boiling) with AlCl₃ in the dark.
    UV light, no catalyst: substitution in the side chain, C₆H₅CH₃ + Cl₂ → C₆H₅CH₂Cl + HCl; excess Cl₂ gives C₆H₅CHCl₂ and C₆H₅CCl₃. AlCl₃ in the dark: substitution in the ring, giving 1-chloro-2-methylbenzene and 1-chloro-4-methylbenzene.
  17. Why does (chloromethyl)benzene react with warm NaOH(aq) but chlorobenzene does not?
    Side-chain Cl behaves like a halogenoalkane: substitution gives phenylmethanol. In chlorobenzene a Cl lone pair overlaps with the ring, so the C-Cl bond is stronger, and the electron-rich ring repels nucleophiles.
  18. How can you show that (chloromethyl)benzene and chlorobenzene differ in reactivity?
    Warm each with NaOH(aq), acidify with dilute HNO₃, then add AgNO₃(aq). (Chloromethyl)benzene gives a white AgCl precipitate. Chlorobenzene gives no precipitate.
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