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 straight through to a carboxylic acid (propanoic acid) that turns litmus red
• CIt is oxidised to a ketone (propanone) which gives no reaction with Tollens' reagent✓
• DIt is oxidised to an aldehyde, which then 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⁺ ion 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 taking place in one concerted step with no radicals
• 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•. Cl₂ → 2Cl• is initiation (homolysis), and radicals combining to give CH₃Cl or Cl₂ are 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 of the four, so it reacts fastest
• BIodine is the most electronegative halogen; it pulls the electrons hardest
• CThe C–I bond is the weakest, so it breaks most easily✓
• DFluoroalkanes have the lowest boiling points, so they escape from the mixture

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 is quickly decolourised by the aldehyde but not by the ketone
• CUniversal indicator paper, which turns red (pH 3) with the aldehyde but not the ketone
• DAnhydrous copper(II) sulfate, which changes from white to blue with the ketone only

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 a bright orange precipitate
• CTollens' reagent, which deposits a silver mirror on warming
• 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 from HNO₂)
• Bthe undissociated HNO₃ molecule itself
• CNO₂⁺ (the nitronium ion)✓
• Dthe NO• nitrogen monoxide 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.

What is the general formula of an alkane?

• ACₙH₂ₙ
• BCₙH₂ₙ₋₂
• CCₙH₂ₙO
• DCₙH₂ₙ₊₂✓

Answer: Alkanes are saturated, so each carbon carries the maximum hydrogen. Alkenes with one double bond are CₙH₂ₙ, and each further degree of unsaturation removes two more hydrogens.

What type of reaction occurs between an alkane and chlorine in UV light?

• AElectrophilic substitution
• BFree radical substitution✓
• CNucleophilic substitution
• DElectrophilic addition

Answer: UV light homolytically splits Cl₂ into radicals, which then abstract hydrogen from the alkane in a chain of initiation, propagation and termination steps. The lack of selectivity produces a mixture of products.

Why do alkenes undergo electrophilic addition?

• AThe C=C double bond is an area of low electron density
• BAlkenes carry a permanent negative charge overall
• CAlkenes contain a highly polar carbon-carbon bond
• DThe C=C double bond is an area of high electron density✓

Answer: The pi bond's electrons sit above and below the carbons and are readily attacked by electron-deficient species. The pi bond breaks and two new sigma bonds form, so the product is saturated.

What is Markovnikov's rule used to predict?

• AWhich carbon of an unsymmetrical alkane loses its hydrogen atom
• BWhich carbon of an unsymmetrical alkene gains the hydrogen✓
• CThe rate at which an electrophilic addition reaction takes place
• DThe stereochemistry of the product that forms

Answer: The hydrogen adds to the carbon already carrying more hydrogens, because that route passes through the more stable carbocation. Alkyl groups are electron-releasing and stabilise the positive charge.

Why does a tertiary halogenoalkane hydrolyse faster than a primary one?

• AIt forms a more stable carbocation intermediate✓
• BIt forms a less stable carbocation intermediate
• CThe carbon-halogen bond is far more polar in it
• DIt has less steric hindrance around the carbon

Answer: Tertiary halogenoalkanes react by an SN1 route through a carbocation, which three alkyl groups stabilise. Primary ones must go by SN2, where the nucleophile attacks the crowded carbon directly.

What are structural isomers?

• ACompounds with the same molecular formula but different arrangement✓
• BCompounds with the same arrangement but different molecular formula
• CCompounds with the same empirical and molecular formula
• DCompounds differing only in the spatial arrangement of groups

Answer: Structural isomers differ in which atoms are joined to which — chain, position or functional group isomerism. Differing only in spatial arrangement describes stereoisomerism, such as E-Z isomers.

Why does E-Z isomerism occur in some alkenes?

• ARestricted rotation about the C=C double bond✓
• BFree rotation about the C=C double bond
• CThe presence of a chiral carbon atom in the chain
• DThe presence of a lone pair on one of the carbons

Answer: The pi bond stops the carbons rotating relative to one another, so groups are locked on one side or the other. It only occurs when each carbon of the double bond carries two different groups.

What is the product of oxidising a primary alcohol under reflux with excess oxidising agent?

• AAn aldehyde and nothing further
• BA ketone and nothing further
• CAn ester of that alcohol
• DA carboxylic acid✓

Answer: Distilling immediately traps the aldehyde, but heating under reflux with excess acidified dichromate oxidises it further to the carboxylic acid. Secondary alcohols give ketones and go no further.

What is an elimination reaction of a halogenoalkane?

• ALoss of hydrogen halide to form an alkene✓
• BLoss of hydrogen halide to form an alcohol
• CGain of a hydroxide ion to form an alcohol
• DGain of a hydrogen halide to form an alkane

Answer: Warm ethanolic potassium hydroxide favours elimination, where OH⁻ acts as a base and removes a proton. Aqueous conditions favour substitution instead — the solvent decides which route dominates.