Cameroon GCE A-Level Chemistry: Organic Chemistry: Hydrocarbons and Functional Groups

Organic Chemistry: Hydrocarbons and Functional Groups

The exam tests naming, isomerism, characteristic reactions of each functional group and distinguishing tests, plus simple structure deduction from data.

Nomenclature and isomerism

Find the longest chain containing the principal functional group, number to give the lowest locants, and use suffix priority: carboxylic acid > ester > amide > nitrile > aldehyde > ketone > alcohol > amine > halogenoalkane. Isomerism: chain, positional, functional group, E-Z and optical. E-Z requires two different groups on each double-bond carbon; optical isomerism requires a carbon with four different groups.

Reactions by group

Alkanes undergo free-radical substitution with Cl2 or Br2 under ultraviolet light. Alkenes undergo electrophilic addition with HBr, Br2 and steam (H3PO4 catalyst); bromine water is decolourised. Alcohols are oxidised: primary to aldehyde by distillation with acidified dichromate, primary to carboxylic acid under reflux, secondary to ketone, tertiary not oxidised. Alcohols dehydrate to alkenes with hot concentrated H2SO4 or Al2O3, and esterify with carboxylic acids using concentrated H2SO4. Aldehydes reduce Tollens' reagent to a silver mirror and Fehling's solution to a red precipitate; ketones do not. Both give orange crystals with 2,4-dinitrophenylhydrazine.

Worked example

A compound X of formula C4H8O gives an orange precipitate with 2,4-dinitrophenylhydrazine but no silver mirror with Tollens' reagent and is not oxidised by acidified dichromate. The precipitate confirms a carbonyl. The negative Tollens test separates ketone from aldehyde, and lack of oxidation rules out an alcohol. So X is a ketone: butanone, CH3COCH2CH3. Aldehydes such as butanal would give a silver mirror, and non-carbonyl isomers would not give the orange precipitate.

Traps

Calling a ketone an aldehyde; giving a formula that violates valency; allowing E-Z isomerism when one carbon carries two identical groups; writing an ester without the bridging oxygen; and forgetting that a primary alcohol can give aldehyde or acid depending on distillation versus reflux conditions.

Key ideas

Suffix priority determines the parent name and numbering.
Tollens and Fehling distinguish aldehydes from ketones; 2,4-DNP confirms a carbonyl.
Primary alcohols give aldehyde on distillation, acid on reflux.
Bromine water decolourisation tests for C=C unsaturation.
Chirality requires four different groups on one carbon.

Organic Reaction Mechanisms and Synthesis

The exam tests curly-arrow mechanisms for substitution, addition and elimination, reagent and condition selection, and the design of short synthetic routes with correct reagents, conditions and products.

Mechanism families

Arrows show movement of an electron pair, starting from a lone pair, a bond or a negative centre and pointing to an electron-deficient atom or bond. Learn four families: free-radical substitution (initiation, propagation, termination), electrophilic addition to alkenes, nucleophilic substitution of halogenoalkanes, and nucleophilic addition to carbonyls. Benzene reacts by electrophilic substitution, retaining aromaticity; the nitrating agent is NO2+, generated from concentrated HNO3 and concentrated H2SO4 at 50-60 degrees C.

Reagents and conditions

Aqueous NaOH or KOH with a halogenoalkane gives an alcohol by substitution: primary via SN2 (rate depends on both substrate and hydroxide), tertiary via SN1 (rate depends on substrate only, with a carbocation intermediate). Hot ethanolic KOH gives elimination to an alkene. HBr addition to propene follows Markovnikov's rule via the more stable secondary carbocation. NaBH4 reduces aldehydes and ketones to alcohols but not carboxylic acids; LiAlH4 in dry ether is stronger.

Worked example

Describe the mechanism for bromoethane with aqueous hydroxide. OH- attacks the delta-positive carbon from the side opposite the C-Br bond while the C-Br bond breaks, passing through a five-coordinate transition state; the products are ethanol and Br-. This is SN2: the rate equation is rate = k[CH3CH2Br][OH-], and a primary halogenoalkane favours this pathway because the alternative primary carbocation is unstable. With 2-bromo-2-methylpropane the mechanism is SN1: Br- leaves first to form a tertiary carbocation, then OH- attacks.

Synthesis routes

Ethene to ethanol: steam with H3PO4. Ethanol to ethanal: distil with acidified K2Cr2O7. Ethanal to ethanoic acid: reflux with excess acidified dichromate. Ethanoic acid to ethyl ethanoate: ethanol with concentrated H2SO4 under reflux. Each step needs the reagent and condition, not just the product.

Traps

Arrows from the wrong atom or drawn as single-headed; omitting lone pairs, dipoles and charges; writing NaOH without aqueous or ethanolic, which changes substitution into elimination; confusing reflux with distillation; and forgetting that the Markovnikov product arises from the more stable carbocation.

Key ideas

Curly arrows start at an electron pair and end at an electron-deficient site.
Aqueous hydroxide gives substitution; ethanolic hot KOH gives elimination.
Primary halogenoalkanes favour SN2; tertiary favour SN1.
Markovnikov addition goes via the more stable carbocation.
Synthetic answers must include reagents and conditions.

Laboratory Practical Skills and Data Analysis

The exam tests apparatus choice, correct recording of measurements, titration and calorimetry calculations with uncertainties, graph work, and describing qualitative tests with the observation, not just the conclusion.

Measurement and technique

Use a pipette for 25.0 cm3 aliquots, a burette for titres and a volumetric flask for making standard solutions. Read the bottom of the meniscus at eye level and record burette readings to two decimal places, to the nearest 0.05 cm3. Rinse the burette with the titrant and the pipette with the solution being measured. Place a white tile under the flask to judge the end-point sharply, and add the last drops slowly with swirling. Choose an indicator to match the titration: phenolphthalein for weak acid with strong base, methyl orange for weak base with strong acid.

Processing data

Average only concordant titres, within about 0.10 cm3 of each other. Percentage error = (absolute uncertainty divided by the measured value) x 100. For a burette, each reading has an uncertainty of about 0.05 cm3, so a titre from two readings has about 0.10 cm3. For rate experiments, take the gradient of a tangent; for enthalpy work, extrapolate the temperature curve back to the time of mixing to correct for heat loss.

Worked example

25.0 cm3 portions of 0.100 mol dm-3 NaOH were titrated with hydrochloric acid of unknown concentration; the mean titre was 24.35 cm3. Moles of NaOH = 0.100 x 25.0/1000 = 2.50 x 10-3 mol. The equation is 1:1, so moles HCl = 2.50 x 10-3 mol. Concentration = 2.50 x 10-3 / (24.35/1000) = 0.1027 mol dm-3, reported as 0.103 mol dm-3. The titre uncertainty is 0.10 cm3, so percentage error = 0.10/24.35 x 100 = 0.41%, which justifies quoting three significant figures.

Traps

Using the 250 cm3 flask volume instead of the pipetted 25.0 cm3; adding uncertainties in the wrong direction; averaging non-concordant titres; quoting more significant figures than the apparatus supports; omitting extrapolation in calorimetry; and writing conclusions without observations. Say limewater turns milky, a white precipitate forms, a silver mirror appears, or damp red litmus is bleached.

Key ideas

Burette readings go to two decimal places, to the nearest 0.05 cm3.
Percentage error = (uncertainty/measurement) x 100.
Average only concordant titres.
Extrapolate temperature curves to the mixing time to correct heat loss.
Qualitative answers need the observation, not just the name of the test.