JAMB Chemistry 2017
Questions & Answers

A candidate might select 3, 4, or 6 by failing to account for all possible positional isomers across the various carbon skeletons of pentane. The alkyl bromide formula C₅H₁₁Br derives from pentane, which features three base skeletons: n-pentane, isopentane, and neopentane. Positioning the bromine atom at every possible location on these skeletons yields 3 isomers from n-pentane, 4 from isopentane, and 1 from neopentane, culminating in a total of 8 structural isomers. Common mistake: Forgetting to evaluate all structural variations of the parent alkane skeleton.

An individual might select Ethane, Bromoethane, or Ethanol by misidentifying the halogen addition mechanism as a substitution reaction or a hydration process. Ethene (C₂H₄) undergoes an addition reaction when exposed to bromine (Br₂), resulting in the saturation of the double bond and the formation of 1,2-dibromoethane (C₂H₄Br₂). Common mistake: Confusing addition reactions of alkenes with substitution reactions of alkanes.

One might choose 8.0 g, 12.0 g, or 20.0 g by failing to set up the proper stoichiometric mole ratio between the hydrocarbon and the halogen reactant. The molar mass of ethene (C₂H₄) is 28 g/mol, which means 2.8 g represents 0.1 moles. According to the balanced equation C₂H₄ + Br₂ → C₂H₄Br₂, the reaction proceeds in a 1:1 mole ratio; thus, multiplying 0.1 moles by the molar mass of Br₂ (160 g/mol) results in 16.0 g of bromine required. Common mistake: Miscalculating the molar mass of bromine gas (Br₂).

A student might select Diethyl ether, Ethyl sulphate, or Ethane by stopping halfway through the reaction sequence or confusing the final product of hydrolysis. Ethene initially reacts with concentrated sulfuric acid to generate ethyl hydrogen sulphate, which then undergoes hydrolysis and warming to yield ethanol as the final alcohol product. Common mistake: Identifying the intermediate ethyl hydrogen sulphate as the final product rather than continuing to the hydrated alcohol.

An examinee might choose ease of manufacture, high lather volume, or biodegradability without identifying the specific chemical drawback of traditional soaps in hard water. Detergents create soluble salts when interacting with calcium (Ca²⁺) and magnesium (Mg²⁺) ions present in hard water, thereby preventing the unwanted precipitation of soap scum. Common mistake: Confusing general cleaning efficiency with the specific ability to avoid hard water scum.

A student might mistakenly select Dehydration, Neutralization, or Substitution by confusing the elimination of hydrogen halide with water loss or ionic replacement. The reaction removes a hydrogen atom and a bromine atom from CH₃CH₂Br to construct an ethene double bond, which classifies the chemical transformation as dehydrohalogenation. Common mistake: Confusing dehydrohalogenation with simple dehydration reactions that remove water.

One might pick Propan-2-ethanol, Ethanol, or Methanol by failing to match all the physical characteristics such as high boiling point and hygroscopic viscosity. Propan-1-ol (CH₃CH₂CH₂OH) exhibits a higher boiling point and greater viscosity than shorter-chain alcohols like methanol or ethanol, while remaining completely miscible with water and strongly hygroscopic. Common mistake: Confusing the physical properties of primary propyl alcohols with lower-chain alcohols.

An examinee might select Ethane, Polyethene, or Ethanol by incorrectly assuming ethyne undergoes addition polymerization like alkenes rather than cyclic trimerization. Ethyne (C₂H₂) undergoes a specific polymerization pathway under controlled conditions to form benzene (C₆H₆) via the balanced relation 3C₂H₂ → C₆H₆. Common mistake: Assuming ethyne forms linear polyethene chains instead of aromatic cyclic compounds.

A student might select Ethene, But-1-ene, or Propane by overlooking the structural symmetry requirements necessary for restricted rotation around double bonds. But-2-ene (CH₃CH=CHCH₃) possesses a double-bonded carbon framework where each carbon atom carries two different substituent groups (a methyl group and a hydrogen atom), enabling cis and trans geometric isomerism. Common mistake: Assuming any alkene with a double bond automatically exhibits geometric isomerism.

One might choose Acetyl chloride, Chloroform, or Trichloroethanoic acid by mistaking alpha-hydrogen substitution for complete halogenation or acyl chloride formation. Chlorinating ethanoic acid (CH₃COOH) in the presence of sunlight selectively substitutes a hydrogen atom on the alpha carbon, producing chloroethanoic acid (CH₂ClCOOH). Common mistake: Over-halogenating the acid or confusing it with acid chloride derivatives.

A candidate might select Pent-1-ene, Pent-3-ene, or 2-methylbut-1-ene by numbering the carbon chain from the wrong end or misidentifying the longest continuous path. The longest continuous carbon chain contains 5 carbon atoms, and numbering from the end closest to the unsaturation places the double bond at carbon 2, yielding the correct IUPAC designation pent-2-ene. Common mistake: Numbering the carbon chain from the end that gives the double bond a higher locant number.

Students might choose other options by confusing halogenation and alkyl coupling reagents, but the correct choice is A because CH₃CH₂Br, derived from ethene and HBr, undergoes a Wurtz reaction with sodium to build propane (CH₃CH₂CH₃). Common mistake: confusing the roles of sodium and potassium hydroxide in alkane synthesis.

Students could erroneously guess multiple isomers if they confuse cyclic structures with branched-chain alkenes, but the correct choice is A because cyclopentane is an unsubstituted cyclic alkane that possesses only a single unique structural form (C₅H₁₀). Common mistake: assuming all hydrocarbons with five carbons must have structural isomers.

Students might incorrectly select other options by misnumbering the carbon chain from the wrong end, but the correct choice is A because the longest continuous chain contains four carbons (butane) with the chlorine substituent positioned on carbon-2, yielding 2-chlorobutane. Common mistake: failing to number the chain from the end nearest the substituent.

Students may mistakenly select options containing ethanol or free carboxylic acids by confusing esterification with saponification, but the correct choice is A because the alkaline hydrolysis of a glyceride with NaOH produces glycerol alongside sodium salts of fatty acids, known as sodium alkanoate. Common mistake: confusing the products of hydrolysis with the original reactants of ester formation.

Students might mistakenly pick substitution products like ethyl sulphate or ether by altering reaction temperatures, but the correct choice is A because treating ethanol with concentrated sulfuric acid at 180°C drives an intramolecular elimination reaction to yield ethene (C₂H₄). Common mistake: confusing the 180°C elimination temperature with the lower 140°C temperature that favors ether formation.

Students could mistakenly select secondary or tertiary alcohols by misinterpreting hydroxyl group positions, but the correct choice is C because propan-1-ol features its -OH group attached to a terminal carbon bonded to only one other carbon atom, defining it as a primary alcohol. Common mistake: misidentifying the degree of substitution on the carbon bearing the hydroxyl group.

Students might mistakenly choose saturated or polyunsaturated formulas by applying the wrong general formula, but the correct choice is B because a four-carbon hydrocarbon featuring one double bond belongs to the alkene family with the general formula CₙH₂ₙ, giving C₄H₈ when n equals 4. Common mistake: using the alkane formula (CₙH₂ₙ₊₂) instead of the alkene formula for unsaturated hydrocarbons.

Students might mistakenly select substitution or neutralization reagents like sodium hydroxide, but the correct choice is A because bromine water serves as a standard test for unsaturation, readily decolorizing as bromine adds across multiple carbon-carbon bonds. Common mistake: assuming aqueous sodium hydroxide reacts with double bonds.

Students might mistakenly stop at the aldehyde stage or select a ketone by misjudging oxidation depths, but the correct choice is C because excess acidified potassium dichromate(VI) fully oxidizes the primary alcohol propan-1-ol past propanal straight to propanoic acid. Common mistake: failing to account for complete oxidation of a primary alcohol with excess oxidizing agent.