Organic Chemistry

Propan-1-ol and propan-2-ol can be distinguished by warming each with acidified potassium dichromate(VI) and then testing the organic product. Which result identifies propan-2-ol?

• AIt is not oxidised at all, so the dichromate stays orange
• BIt is oxidised to a carboxylic acid that turns blue litmus red
• CIt is oxidised to a ketone (propanone) which gives no reaction with Tollens' reagent✓
• DIt is oxidised to an aldehyde that gives a silver mirror with Tollens' reagent

Answer: Propan-2-ol is a secondary alcohol, oxidised by acidified dichromate(VI) to the ketone propanone. Ketones are not oxidised further and give a negative Tollens' test. Propan-1-ol (primary) would be oxidised via an aldehyde to propanoic acid, and the aldehyde gives a positive Tollens' (silver mirror).

The IUPAC name of (CH₃)₃CCH₂OH is:

• A2,2-dimethylpropan-1-ol✓
• B2-methylbutan-2-ol
• C3,3-dimethylpropan-1-ol
• D2,2-dimethylbutan-1-ol

Answer: (CH₃)₃CCH₂OH has a central carbon bearing three methyl groups and one CH₂OH. The longest chain containing the OH is propan-1-ol (C1 = CH₂OH, C2 = the quaternary carbon) with two methyl substituents on C2: 2,2-dimethylpropan-1-ol.

Which of the following compounds exhibits E/Z (cis–trans) isomerism?

• Abut-1-ene
• Bbut-2-ene✓
• C2-methylpropene
• Dpropene

Answer: E/Z isomerism requires each carbon of the C=C double bond to carry two different groups. In but-2-ene (CH₃CH=CHCH₃) each double-bond carbon bears a CH₃ and an H, so E and Z forms exist. But-1-ene and propene have a CH₂ end (two H's), and 2-methylpropene has two CH₃ on one carbon, so none of these show E/Z isomerism.

Which compound contains a chiral centre and can therefore exist as a pair of optical isomers (enantiomers)?

• Apropan-1-ol
• Bpropan-2-ol
• Cbutan-2-ol✓
• D2-methylpropan-2-ol

Answer: A chiral centre is a carbon bonded to four different groups. In butan-2-ol, C2 carries H, OH, CH₃ and C₂H₅ — four different groups — so it is chiral. Propan-2-ol's central carbon has two identical CH₃ groups, and the others lack four different groups.

Bromine reacts with ethene at room temperature. What is the mechanism, and what is the nature of the species that attacks the alkene first?

• AElectrophilic addition; bromine acts as an electrophile✓
• BFree-radical substitution; a bromine radical
• CNucleophilic substitution; a bromide ion nucleophile
• DElectrophilic substitution; a Br⁺ electrophile replaces an H

Answer: Alkenes have an electron-rich C=C π bond. As Br₂ approaches, the π electrons polarise it (induced dipole), and the δ+ bromine acts as an electrophile, adding across the double bond. The mechanism is electrophilic addition, giving 1,2-dibromoethane.

When methane reacts with chlorine in the presence of ultraviolet light, the propagation steps of the mechanism are:

• ACl₂ → 2Cl• then Cl• + CH₄ → HCl + CH₃•
• BCH₃• + Cl• → CH₃Cl then 2Cl• → Cl₂
• CCH₄ + Cl₂ → CH₃Cl + HCl in a single step
• DCl• + CH₄ → HCl + CH₃• then CH₃• + Cl₂ → CH₃Cl + Cl•✓

Answer: In free-radical substitution the propagation steps regenerate a radical so the chain continues: Cl• + CH₄ → HCl + CH₃• and CH₃• + Cl₂ → CH₃Cl + Cl•. Option A's first step is initiation (homolysis); option B shows termination steps.

The rate of alkaline hydrolysis of halogenoalkanes by warm aqueous NaOH increases in the order C–F < C–Cl < C–Br < C–I. What is the main reason for this trend?

• AThe C–I bond is the strongest, so it reacts fastest
• BIodine is the most electronegative halogen
• CThe C–I bond is the weakest, so it breaks most easily✓
• DFluoroalkanes have the lowest boiling points

Answer: Bond enthalpy decreases C–F > C–Cl > C–Br > C–I. The reaction rate is governed mainly by how easily the carbon–halogen bond breaks, so the weakest bond (C–I) is hydrolysed fastest, despite iodine being the least electronegative halogen.

1-bromopropane is warmed with sodium hydroxide. Under which conditions is nucleophilic substitution (forming propan-1-ol) favoured over elimination (forming propene)?

• AAqueous NaOH (NaOH dissolved in water), warmed✓
• BConcentrated NaOH dissolved in ethanol, heated under reflux
• CAnhydrous conditions with no solvent
• DDilute sulfuric acid as the reagent

Answer: Warm aqueous NaOH favours substitution: the hydroxide acts as a nucleophile to give the alcohol. Hot NaOH in ethanol favours elimination, with hydroxide acting as a base to give the alkene. Reagent and solvent therefore determine the product.

A student has two unlabelled bottles, one containing an aldehyde (propanal) and one a ketone (propanone). Which reagent distinguishes them, giving a positive result with the aldehyde only?

• AFehling's solution, which gives a brick-red precipitate of Cu₂O with the aldehyde✓
• BBromine water, which decolourises with the aldehyde
• CUniversal indicator, which turns red with the aldehyde
• DAnhydrous copper(II) sulfate, which turns blue with the ketone

Answer: Aldehydes are readily oxidised, so they reduce Fehling's solution, giving a brick-red precipitate of copper(I) oxide, Cu₂O. Ketones cannot be oxidised easily and give no reaction. (Tollens' reagent — a silver mirror — works equally well.)

Which reagent and observation correctly identify the presence of a carboxylic acid such as ethanoic acid?

• AAcidified KMnO₄, which turns from purple to colourless
• B2,4-dinitrophenylhydrazine, giving an orange precipitate
• CTollens' reagent, giving a silver mirror
• DSodium carbonate, which gives effervescence (CO₂ released)✓

Answer: Carboxylic acids are weak acids that react with carbonates (or hydrogencarbonates) to release carbon dioxide gas, seen as effervescence: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. 2,4-DNPH is a test for carbonyls; Tollens' is for aldehydes.

When benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid at about 50 °C, nitrobenzene is formed. Which species is the electrophile that attacks the benzene ring?

• ANO₂⁻ (the nitrite ion)
• BHNO₃ molecule
• CNO₂⁺ (the nitronium ion)✓
• DNO• radical

Answer: Concentrated H₂SO₄ protonates HNO₃, which then loses water to generate the nitronium ion: HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺. The electrophile NO₂⁺ attacks the delocalised π system in an electrophilic substitution, giving nitrobenzene.