Organic Chemistry MCQs – 40 Practice Questions with Answers and Explanations

Answer: B – Alkanes are saturated hydrocarbons with the general formula C_nH_2n+2. C_nH_2n is the general formula of the alkenes.

Answer: C – Alkenes have the general formula C_nH_2n, so with n = 6 the formula is C₆H₁₂. C₆H₁₄ would be the alkane hexane.

Answer: A – Alcohols contain the hydroxyl functional group, −OH, giving them the general formula C_nH_2n+1OH. −COOH is the carboxylic acid group and C=C is found in alkenes.

Answer: D – There are exactly 2: butane (a straight chain) and 2-methylpropane (a branched chain). Structural isomers have the same molecular formula but different structural formulas.

Answer: B – There are 3 isomers: pentane, 2-methylbutane and 2,2-dimethylpropane. The number of isomers grows quickly as the carbon chain gets longer.

Answer: C – The longest chain has 3 carbons (propane) with a methyl group on the middle carbon, so the name is 2-methylpropane. It is the branched isomer of butane.

Answer: A – The empirical formula is the simplest whole-number ratio of atoms. Dividing 6:12:6 by 6 gives 1:2:1, so the empirical formula is CH₂O.

Answer: D – CH₃COOH is ethanoic acid; it contains the −COOH group. C₂H₅OH is ethanol, CH₃CHO is ethanal and CH₃COCH₃ is propanone.

Answer: C – Each member of a homologous series differs from the next by a CH₂ group. Members share the same functional group and general formula, and show similar chemical properties.

Answer: B – Butane and 2-methylpropane are both C₄H₁₀ but have different structures. The other pairs have different molecular formulas, so they cannot be isomers.

Answer: A – In a plentiful supply of air, hydrocarbons burn completely: CH₄ + 2O₂ → CO₂ + 2H₂O.

Answer: D – When the air supply is limited, carbon monoxide (and sometimes soot/carbon) forms instead of carbon dioxide. Carbon monoxide is toxic because it binds strongly to haemoglobin and reduces oxygen transport in the blood.

Answer: C – UV light provides the energy to split Cl₂ into chlorine free radicals, which starts the chain reaction: CH₄ + Cl₂ → CH₃Cl + HCl.

Answer: B – Ethene adds bromine across its C=C double bond to form colourless dibromoethane, so the orange colour disappears. This is the standard test for an alkene (unsaturation); alkanes do not react.

Answer: D – Hydrogenation of alkenes uses hydrogen gas with a nickel catalyst (about 150°C): C₂H₄ + H₂ → C₂H₆. The same type of reaction hardens vegetable oils into margarine.

Answer: A – Markovnikov’s rule: the hydrogen adds to the carbon of the double bond that already has more hydrogens, so bromine ends up on the middle carbon, giving 2-bromopropane as the major product. This is because the more stable secondary carbocation forms in the mechanism.

Answer: C – C₂H₄ + H₂O → C₂H₅OH using steam at about 300°C, 60-70 atm pressure and a concentrated phosphoric acid catalyst.

Answer: B – In addition polymerisation the C=C bonds open up and thousands of monomer units join into long chains with no other product. Chloroethene gives poly(chloroethene), commonly called PVC.

Answer: D – Cracking splits a long-chain alkane into a shorter alkane and an alkene. Only option D balances (10 C and 22 H on both sides) and gives an alkane plus an alkene (octane + ethene).

Answer: A – The pi bond in C=C is weaker and more exposed than a sigma bond, so it breaks readily and other atoms add across it. Alkanes have only strong single bonds, so they react mainly by free-radical substitution.

Answer: C – Yeast enzymes convert glucose to ethanol and carbon dioxide: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂. The mixture is kept warm (about 37°C) and air is excluded so the yeast respires anaerobically.

Answer: B – Industry hydrates ethene (from cracking crude oil) with steam over a phosphoric acid catalyst. Fermentation is used for alcoholic drinks and in countries with plentiful sugar crops, but it is slower and gives a dilute product.

Answer: D – The orange dichromate(VI) is reduced to green chromium(III) as the ethanol is oxidised. With excess oxidising agent and reflux, a primary alcohol is oxidised fully to the carboxylic acid, ethanoic acid.

Answer: A – A primary alcohol is oxidised first to an aldehyde (propanal) and then, with excess oxidant and reflux, to the carboxylic acid, propanoic acid. Propanal is obtained only if it is distilled off as it forms; propanone comes from the secondary alcohol propan-2-ol.

Answer: C – Carboxylic acids react with reactive metals like other acids: 2CH₃COOH + Mg → (CH₃COO)₂Mg + H₂. The reaction is slower than with a strong acid of the same concentration because ethanoic acid is a weak acid.

Answer: B – Acids react with carbonates to give a salt, water and carbon dioxide: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂. The CO₂ turns limewater milky.

Answer: D – A carboxylic acid reacts with an alcohol to form an ester plus water (esterification): CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. The concentrated sulfuric acid is a catalyst; ethyl ethanoate has a sweet, fruity smell.

Answer: A – In an ester name, the first part comes from the alcohol (methanol → methyl) and the second part from the acid (propanoic acid → propanoate), giving methyl propanoate.

Answer: B – Dehydration removes H₂O from the alcohol to form the alkene: C₂H₅OH → C₂H₄ + H₂O. Concentrated sulfuric (or phosphoric) acid, or passing the vapour over hot aluminium oxide, catalyses the elimination.

Answer: C – Secondary alcohols are oxidised to ketones, so propan-2-ol gives propanone. Ketones cannot be oxidised further by dichromate, unlike primary alcohols, which go on to carboxylic acids; tertiary alcohols are not oxidised at all.

Answer: D – The hydroxide ion is a nucleophile (an electron-pair donor) that replaces the bromine: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻. Aqueous conditions favour substitution; hot ethanolic KOH would give elimination instead.

Answer: A – The CN⁻ nucleophile replaces Br and adds one carbon to the chain: CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻. The product is propanenitrile (3 carbons). This is a key chain-lengthening step, as the nitrile can then be hydrolysed to a carboxylic acid or reduced to an amine.

Answer: B – Hot, concentrated ethanolic KOH acts as a base and removes HBr from the halogenoalkane (elimination), forming the alkene ethene. The solvent matters: aqueous KOH gives substitution to ethanol, ethanolic KOH gives elimination.

Answer: C – The high electron density of the pi bond attracts the electrophile; the pi bond breaks, one carbon bonds to Br and the other becomes a positively charged carbocation. Br⁻ then attacks the carbocation to complete the addition, giving dibromoethane.

Answer: A – Tollens’ reagent is a mild oxidising agent that oxidises aldehydes to carboxylic acids, depositing a silver mirror. Ketones are not oxidised, so propanone gives no reaction – this test distinguishes aldehydes from ketones.

Answer: D – 2,4-DNP reacts with the C=O group of both aldehydes and ketones to give an orange/yellow crystalline precipitate. It does not distinguish between aldehydes and ketones – Tollens’ or Fehling’s reagent is used for that – but the melting point of the crystals can identify the specific carbonyl compound.

Answer: B – Fehling’s solution contains blue Cu²⁺ ions, which an aldehyde reduces to brick-red copper(I) oxide, Cu₂O, while the aldehyde is oxidised to a carboxylic acid. Ketones give no reaction, so Fehling’s also distinguishes aldehydes from ketones.

Answer: C – The iodoform test is positive only for compounds containing a CH₃CO− group (methyl ketones, and ethanal – the only aldehyde) or a CH₃CH(OH)− group, which is oxidised to CH₃CO− during the test. Propanone contains CH₃CO−; pentan-3-one does not.

Answer: A – A broad peak at 3200-3600 cm⁻¹ is the O−H stretch of an alcohol. The absence of a strong C=O peak (1680-1750 cm⁻¹) rules out esters, ketones and carboxylic acids.

Answer: D – The molecular ion peak gives M_r = 58. Propanone, C₃H₆O, has M_r = (3 × 12) + 6 + 16 = 58 and contains C=O but no O−H. Butane also has M_r 58 but no C=O; propan-1-ol and ethanoic acid both have M_r 60.