chem p2

Explain why alkanes are generally unreactive towards nucleophiles and electrophiles.

They are non‑polar molecules; C‑C and C‑H bonds are strong and non‑polar, so there is no polar site for attack by charged or electron‑rich species.

1. Describe the mechanism for the reaction of ethane with chlorine under UV light. Name the three stages involved.

Free‑radical substitution. Stages: Initiation (Cl₂ → 2Cl•), Propagation (C₂H₆ + Cl• → •C₂H₅ + HCl; •C₂H₅ + Cl₂ → C₂H₅Cl + Cl•), Termination (e.g. •C₂H₅ + Cl• → C₂H₅Cl).

Write equations for the complete and incomplete combustion of propane. State one problem associated with incomplete combustion.

Complete: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O. Incomplete: C₃H₈ + 3½O₂ → 3CO + 4H₂O (or C + H₂O). Problem: Carbon monoxide is toxic / soot causes pollution.

State the two structural requirements for E/Z isomerism. Draw the structure of Z‑but‑2‑ene. (3 marks)

C=C double bond (restricted rotation around the bond); two different groups attached to each carbon atom of the double bond. (Diagram: CH₃ groups on the same side).

Describe the mechanism for the reaction of but‑2‑ene with hydrogen bromide. Explain why only one organic product forms here, whereas propene forms two products. (5 marks)

A Electrophilic addition. HBr polarises; H⁺ is the electrophile and attacks the C=C bond forming a carbocation. But‑2‑ene forms identical secondary carbocations whichever end is attacked → only one product. Propene forms primary and secondary carbocations → major/minor products.

Describe a chemical test to distinguish between cyclohexane and cyclohexene. Include the observation and an equation. (2 marks)

A Add bromine water. Cyclohexene: orange solution turns colourless (addition: C₆H₁₀ + Br₂ → C₆H₁₀Br₂). Cyclohexane: no visible change (reacts only under UV light).

Explain why addition polymers such as poly(ethene) are chemically inert and non‑biodegradable. (2 marks)

A Made of strong, non‑polar C‑C and C‑H bonds; no reactive functional groups present for chemicals or enzymes to attack.

8. Classify 2‑methylbutan‑2‑ol and butan‑1‑ol as primary, secondary or tertiary alcohols. Explain your classification. (2 marks)

A 2‑methylbutan‑2‑ol = tertiary (carbon bonded to –OH is attached to 3 other carbons). Butan‑1‑ol = primary (carbon bonded to –OH is attached to 1 other carbon).

Describe the oxidation of butan‑1‑ol using acidified potassium dichromate(VI) under distillation and under reflux. Give the organic products and colour changes observed. (4 marks)

Distillation → butanal (aldehyde); Reflux → butanoic acid (carboxylic acid). Orange dichromate(VI) ions are reduced to green chromium(III) ions in both cases.

Give reagents, conditions and an equation for the dehydration of propan‑2‑ol to form propene. (3 marks)

A Reagent: concentrated sulfuric acid OR aluminium oxide catalyst. Conditions: 170°C (acid) OR 300°C (Al₂O₃). Equation: CH₃CH(OH)CH₃ → CH₃CH=CH₂ + H₂O.

How would you convert ethanol to iodoethane in the laboratory? Give reagents and conditions. (2 marks)

A Heat with red phosphorus and iodine (PI₃ formed in situ), or react with concentrated hydrogen iodide.

12. Explain why haloalkanes react with nucleophiles. Define the term nucleophilic substitution. (3 marks)

A The C‑Halogen bond is polar (Cδ⁺, Halδ⁻) → the carbon is electron‑poor and is attacked by electron‑rich nucleophiles. Nucleophilic substitution: a nucleophile replaces the halogen atom in the molecule.

Arrange chloroethane, bromoethane and iodoethane in order of increasing rate of hydrolysis. Explain the order you give. (3 marks)

Chloroethane < bromoethane < iodoethane. C‑I bond is weakest (lowest bond enthalpy) so breaks most easily; C‑Cl bond is strongest.

Compare Sₙ1 and Sₙ2 mechanisms. State which mechanism is favoured by primary and tertiary haloalkanes, and explain why. (4 marks

Sₙ2: 1 step, transition state formed, favoured by primary haloalkanes — less steric hindrance allows nucleophile to attack easily. Sₙ1: 2 steps, carbocation intermediate formed, favoured by tertiary haloalkanes — alkyl groups stabilise the positive charge.

Describe how you would convert 2‑bromopropane into propene. State reagents and conditions. (2 marks)

A Elimination reaction: heat with sodium or potassium hydroxide dissolved in ethanol (alcoholic NaOH/KOH).

• Draw the structures of ethanal and propanone. Explain why aldehydes are easily oxidised but ketones are resistant to oxidation. (3 marks)

Ethanal: CH₃CHO; Propanone: CH₃COCH₃. Aldehydes have a hydrogen atom attached to the carbonyl carbon — easily removed. Ketones have two alkyl groups attached — no hydrogen to lose, and C‑C bonds are strong.

Describe chemical tests using Tollens’ reagent and Fehling’s solution to distinguish between aldehydes and ketones. Include reagents, conditions and observations. (4 marks)

Tollens’: [Ag(NH₃)₂]⁺ complex, heat gently → silver mirror forms with aldehyde, no change with ketone. Fehling’s: Cu²⁺ ions in alkaline solution, heat → blue solution turns to brick‑red precipitate with aldehyde, remains blue with ketone.

Write an equation and name the product formed when butanone is reduced using NaBH₄. (2 marks)

CH₃COCH₂CH₃ + 2[H] → CH₃CH(OH)CH₂CH₃ → product = butan‑2‑ol.

Outline the mechanism for the reaction of propanal with NaBH₄. Name the mechanism type. (4 marks)

A Nucleophilic addition. :H⁻ ion attacks δ+ carbonyl carbon; π bond breaks, oxygen gains negative charge; O⁻ bonds to H⁺ ion from solvent to form alcohol group.

20. Explain why ethanoic acid is a stronger acid than ethanol. Refer to bond strength and ion stability in your answer. (3 marks)

A The adjacent C=O group withdraws electrons, weakening the O‑H bond; the carboxylate ion formed is stabilised by delocalisation of the negative charge over two oxygen atoms. The ethoxide ion has no such delocalisation and is unstable.

Write an equation and state the observation for the reaction of methanoic acid with sodium carbonate. (2 marks)

A 2HCOOH + Na₂CO₃ → 2HCOONa + H₂O + CO₂. Observation: Fizzing / gas evolved.

Describe the preparation of methyl propanoate in the laboratory. Give reagents, conditions, catalyst and an equation. (4 marks)

Esterification. Reagents: propanoic acid + methanol. Catalyst: concentrated sulfuric acid. Conditions: heat under reflux. Equation: CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O.

Compare and contrast the acidic and alkaline hydrolysis of ethyl methanoate. State reversibility and products formed. (4 marks)

Acidic: reversible reaction → methanoic acid + ethanol. Alkaline: irreversible reaction → sodium methanoate + ethanol.

• Describe the structure and bonding in benzene. Explain why benzene undergoes substitution reactions rather than addition reactions. (4 marks)

Planar hexagonal ring; all C‑C bonds equal length; delocalised π‑electron cloud above and below the ring. Addition would break delocalisation and lose large stability gain; substitution preserves delocalisation.

Outline the nitration of benzene. Include reagents, conditions, formation of the electrophile and type of mechanism. (5 marks)

Electrophilic substitution. Reagents: concentrated HNO₃ + concentrated H₂SO₄. Conditions: 50°C. Electrophile NO₂⁺ formed: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. NO₂⁺ attacks ring; H⁺ lost to reform delocalisation.

Give reagents, catalyst and conditions for Friedel‑Crafts acylation of benzene. Why is this reaction useful in synthesis? (3 marks)

Reagents: benzene + acyl chloride. Catalyst: aluminium chloride (AlCl₃). Conditions: heat. Useful: introduces carbonyl group onto ring; forms ketone; no multiple substitution occurs.

Explain why phenol is more reactive towards electrophiles than benzene. Refer to electron density in your answer. (3 marks)

–OH group donates lone pair of electrons into benzene ring system; increases electron density of the ring (especially at 2,4,6 positions; attracts electrophiles more strongly — no catalyst needed for bromination.

Describe the reaction of phenol with bromine water. Give observations and product formed. (3 marks)

Orange bromine water decolourises; white precipitate of 2,4,6‑tribromophenol forms.

Explain the directing effects of –OH and –NO₂ groups on a benzene ring. Classify each group as activating or deactivating. (4 marks)

A –OH: 2,4‑directing, activating — increases ring electron density. –NO₂: 3‑directing, deactivating — withdraws electrons, decreases ring electron density.

Classify CH₃CH₂NH₂, (CH₃CH₂)₂NH, (CH₃CH₂)₃N as primary, secondary or tertiary. Explain why ethylamine is a stronger base than ammonia. (4 marks)

A Primary, secondary, tertiary. Ethyl group has +I (electron‑releasing) effect → increases electron density on nitrogen lone pair → better ability to accept H⁺ ions.

How would you prepare a pure sample of ethylamine from bromoethane? Why must ammonia be used in excess? (3 marks)

Heat bromoethane with concentrated ammonia solution in a sealed tube. Excess NH₃ prevents further substitution to form diethylamine or triethylamine.

Write an equation for the reaction of propanoyl chloride with ammonia. Name the organic product. (2 marks)

CH₃CH₂COCl + 2NH₃ → CH₃CH₂CONH₂ + NH₄Cl. Product: propanamide.

Explain why acyl chlorides react readily with nucleophiles whereas carboxylic acids are much less reactive. (3 marks)

Chlorine is strongly electron‑withdrawing → makes carbonyl carbon very δ+; Cl⁻ is an excellent leaving group. In acids, –OH group is a poor leaving group and resonance reduces δ+ charge.

How do you convert propanenitrile into propylamine? Give reagents and conditions. (2 marks)

A Reduce with LiAlH₄ in dry ether, or react with hydrogen gas and nickel catalyst under heat.

Distinguish between addition and condensation polymerisation. Give one example of each type. (4 marks)

Addition: monomers contain C=C double bonds; only polymer is formed — e.g. poly(ethene). Condensation: two monomers each with two functional groups; small molecule (H₂O or HCl) lost per linkage — e.g. polyester, polyamide.

Draw the repeating unit of the polyester formed from ethane‑1,2‑diol and propanedioic acid. (2 marks)

A –[‑O‑CH₂CH₂‑O‑CO‑CH₂‑CO‑]–

Explain why polyamides generally have higher melting points and greater strength than polyesters. (2 marks)

A N‑H groups in polyamides enable hydrogen bonding between polymer chains — stronger intermolecular forces require more energy to separate chains.

• Define structural isomerism. Name the three types and give one example for each. (3 marks)

• A Same molecular formula, different structural formula. Chain: butane / 2‑methylpropane. Position: propan‑1‑ol / propan‑2‑ol. Functional group: ethanol / methoxymethane.

• State the structural requirement for optical isomerism. Identify the chiral carbon atom in 2‑chlorobutane. (2 marks)

• A Presence of a chiral carbon atom (carbon attached to four different groups). Carbon‑2 in 2‑chlorobutane is chiral.

• In mass spectrometry, define the molecular ion peak. Explain how high‑resolution mass spectrometry helps identify a molecular formula. (2 marks)

• A Peak with highest m/z value = relative molecular mass. High‑res gives exact mass value → calculate molecular formula from precise atomic masses.

• Use IR absorption values to explain how you would distinguish between an alcohol, an aldehyde and a carboxylic acid. (4 marks)

• A Alcohol: broad O‑H absorption 3200–3600 cm⁻¹. Aldehyde: C=O 1720–1740 cm⁻¹ + characteristic C‑H absorption ~2720 cm⁻¹. Carboxylic acid: very broad O‑H 2500–3300 cm⁻¹ + C=O 1700–1725 cm⁻¹.

• Predict the number of peaks in the ¹³C NMR spectrum of butan‑1‑ol and butan‑2‑ol. Explain your answer. (3 marks)

• A Butan‑1‑ol: 4 peaks (all carbon environments different). Butan‑2‑ol: 4 peaks (different environments due to position of –OH group).

• For propanal (CH₃CH₂CHO), explain what information you get from integration values and splitting patterns in ¹H NMR. (3 marks)

• A Integration ratio 3:2:1 → relative number of hydrogens in each environment. Splitting: CH₃ triplet, CH₂ quartet, CHO triplet → shows number of adjacent hydrogens.

• State the n+1 rule in NMR and apply it to the molecule CH₃CH₂Cl. (2 marks)

• A A peak is split into n+1 peaks where n = number of adjacent non‑equivalent hydrogens. CH₃ appears as triplet; CH₂ appears as quartet.

• What is meant by the fingerprint region in IR spectroscopy? How is it used in analysis? (2 marks)

• A Region below 1500 cm⁻¹ — unique absorption pattern for every compound. Match spectrum against database/library to confirm identity.

• Explain how thin‑layer chromatography (TLC) separates components and how the Rf value is calculated. (3 marks)

Separation based on different solubility in mobile phase and adsorption to stationary phase. Rf = distance moved by spot ÷ distance moved by solvent front.

• Explain how gas chromatography (GC) separates mixtures. What does the retention time indicate? (3 marks)

Components separated by different retention on stationary phase vs volatility in carrier gas. Retention time = time taken to pass through column — characteristic for identification.

• Compound X has formula C₄H₈O. IR shows strong absorption at 1715 cm⁻¹. ¹H NMR shows two singlets. Deduce structure and explain reasoning. (4 marks)

1715 cm⁻¹ confirms C=O (ketone). Two singlets = no adjacent hydrogens → structure is 2‑methylpropanal.

Compound Y has formula C₃H₆O₂. IR shows absorption at 1740 cm⁻¹, no broad OH absorption. Suggest compound class and possible structure. (3 marks)

A Absorption at 1740 cm⁻¹ (ester C=O position), no OH → Ester. Structure: methyl ethanoate or ethyl methanoate.

• How could you use IR spectroscopy and mass spectrometry to confirm synthesis of ethanoyl chloride? (3 marks)

IR: C=O absorption ~1790 cm⁻¹ (higher wavenumber than acids/esters). MS: molecular ion peaks at m/z = 78 and 80 due to Cl isotopes.

• Describe two chemical tests and one spectroscopic method to distinguish between propanal and propanone. (4 marks)

Chemical: Tollens’ reagent / Fehling’s solution — only propanal reacts. IR: propanal has C‑H absorption ~2720 cm⁻¹, propanone does not.

• Explain how you would identify an unknown alcohol as primary, secondary or tertiary using oxidation reactions and IR spectroscopy. (4 marks)

Oxidise with acidified K₂Cr₂O₇: primary → aldehyde/acid; secondary → ketone; tertiary → no reaction. IR: check product for C=O absorption and exact wavenumber to confirm type.

• What information does the integration trace give in a ¹H NMR spectrum? (1 mark)

Relative number of hydrogen atoms in each chemical environment.

• Explain why tetramethylsilane (TMS) is used as the reference standard in NMR spectroscopy. (2 marks)

Produces a single peak well away from most organic signals; inert, volatile and easily removed from sample.

Compound Z has formula C₂H₅OCl. IR shows absorption at 1800 cm⁻¹. Deduce functional group and structure. (3 marks)

High wavenumber 1800 cm⁻¹ = acyl chloride group. Structure: ethanoyl chloride.

Describe how combining data from IR, MS and NMR allows you to determine the full structure of an unknown organic compound. (4 marks)

IR → identifies functional groups present. MS → gives molecular mass and molecular formula. NMR → shows number/position of C and H environments and connectivity. Combine all data to build and confirm final structure.