JAMB Past Questions

JAMB Chemistry 2018
Questions & Answers

40 questions · Correct answers highlighted · 40 with explanations

40 Total Questions
2018 Exam Year
40 With Explanations
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1
Question 1 of 40
JAMB · Chemistry · 2018

The fraction of crude oil used as jet fuel is

A. refinery gas
B. diesel oil
C. kerosene
D. gasoline
Explanation

Distractor check: Options A, B, and D mention other crude oil distillation products like refinery gas, diesel, or gasoline, which have different boiling ranges and engine applications. Reasoning to the answer: Jet fuel consists primarily of kerosene, a petroleum fraction recovered during fractional distillation within a boiling range of roughly 150–275°C that provides the necessary stability and energy for commercial and military aircraft engines. Common mistake: Confusing jet fuel with heavier diesel fractions or lighter gasoline.

2
Question 2 of 40
JAMB · Chemistry · 2018

What is the IUPAC name of the compound CH₃-CH₂-CH(CH₃)-C=C(CH₃)-CH₂-CH₃?

A. 3,5-dimethylhept-3-ene
B. 2,4-dimethylhex-3-ene
C. 3,5-dimethylhept-2-ene
D. 2,4-dimethylhex-2-ene
Explanation

A student might incorrectly choose B or D by miscounting the carbon chain as a hex-3-ene or hex-2-ene derivative rather than identifying the longest 7-carbon parent chain. To determine the correct IUPAC name, first identify the principal chain of seven carbons forming a heptane skeleton, then locate the double bond so it receives the lowest possible locant, placing it at position 3 (hept-3-ene). Afterward, assign the methyl substituents at carbons 3 and 5 in ascending order. Common mistake: failing to number the longest continuous carbon chain from the end that gives the double bond the lowest locant.

3
Question 3 of 40
JAMB · Chemistry · 2018

It is not desirable to use lead tetraethyl as an anti-knock agent because

A. it decreases the octane rating of petrol
B. it is expensive
C. it causes pollution from exhaust fumes
D. it lowers the flash point of petrol
Explanation

Option A or D might look plausible if a student confuses an anti-knock agent's role with altering fuel volatility or octane numbers. Historically used to suppress engine knocking, lead tetraethyl releases harmful lead particles through tailpipe exhausts that cause severe environmental contamination and neurological risks, making exhaust pollution the primary reason for its discontinuation. Common mistake: confusing fuel performance metrics with environmental toxicity outcomes.

4
Question 4 of 40
JAMB · Chemistry · 2018

The carbon atoms in ethene (C₂H₄) are

A. sp³ hybridized
B. sp² hybridized
C. sp hybridized
D. sp³d hybridized
Explanation

A student might incorrectly guess A or C by confusing the hybridization of an alkene with single-bonded alkanes or triple-bonded alkynes. In ethene (C₂H₄), each carbon atom forms a double bond consisting of one sigma and one pi bond, alongside two single sigma bonds, which necessitates three hybrid orbitals. This structural requirement yields sp² hybridization and establishes a trigonal planar geometry with characteristic bond angles of approximately 120°. Common mistake: counting individual bonds rather than regions of electron density to determine hybridization state.

5
Question 5 of 40
JAMB · Chemistry · 2018

Catalytic hydrogenation of benzene (C₆H₆) produces

A. an aromatic hydrocarbon
B. margarine
C. cyclohexane
D. DDT
Explanation

Option A or B might be selected if a student mistakes the benzene ring for a generic aromatic or lipid-like substance like margarine. The catalytic hydrogenation of benzene (C₆H₆) involves introducing hydrogen gas (H₂) over a nickel catalyst, which breaks the aromatic stability and transforms the cyclic structure into a saturated cycloalkane. Consequently, the reaction C₆H₆ + 3H₂ yields cyclohexane, representing a six-carbon ring containing only single bonds. Common mistake: forgetting that adding hydrogen to an aromatic ring eliminates unsaturation and yields a cycloalkane.

6
Question 6 of 40
JAMB · Chemistry · 2018

The functional group present in an organic compound that reacts with a saturated solution of NaHCO₃ to produce CO₂ gas is

A. hydroxyl group
B. carboxyl group
C. carbonyl group
D. amino group
Explanation

A student could mistakenly select option A or C by confusing the reactions of alcohols or carbonyls with standard acid-base tests. Carboxylic acids feature the carboxyl group (-COOH), which contains an acidic proton capable of reacting with a sodium bicarbonate (NaHCO₃) solution to release carbon dioxide gas, water, and a salt via the reaction RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂. Common mistake: assuming all oxygen-containing organic functional groups react with bicarbonate to release carbon dioxide.

7
Question 7 of 40
JAMB · Chemistry · 2018

The characteristic reaction of carbonyl compounds (aldehydes and ketones) is

A. substitution
B. elimination
C. addition
D. saponification
Explanation

Option A or B might be chosen by a student who confuses carbonyl reactivity with alkane or alkyl halide reaction pathways. Aldehydes and ketones contain a polar carbon-oxygen double bond where the carbon atom is electrophilic, making them prone to nucleophilic attack. Because of this electron deficiency, they engage in nucleophilic addition reactions with reagents like HCN or NaHSO₃ across the π-bond to form cyanohydrins or bisulfite addition products. Common mistake: assuming C=O double bonds undergo substitution instead of addition.

8
Question 8 of 40
JAMB · Chemistry · 2018

Palm wine turns sour with time because

A. the sugar content is converted into alcohol
B. the carbohydrate content forms a sour taste
C. it is adulterated by tappers
D. microbial activity produces organic acids
Explanation

A student might pick option A if they think the sugar content turning into alcohol makes the wine taste sour, overlooking the subsequent chemical conversion. Palm wine becomes sour over time because bacteria such as Acetobacter drive a microbial oxidation process that converts ethanol into acetic acid through the reaction C₂H₅OH + O₂ → CH₃COOH + H₂O. Common mistake: stopping at alcohol fermentation and missing the secondary microbial oxidation step that produces organic acids.

9
Question 9 of 40
JAMB · Chemistry · 2018

Which of the following represents saponification?

A. Reaction of a carboxylic acid with NaOH
B. Reaction of an ester with NaOH
C. Reaction of an alcohol with NaOH
D. Reaction of an alkane with NaOH
Explanation

Option A might seem correct if a student confuses general carboxylic acid neutralization with the alkaline hydrolysis of an ester. Saponification is specifically the base-promoted cleavage of an ester using a base such as sodium hydroxide (NaOH), which yields an alcohol and a carboxylate salt (soap) following the general equation RCOOR' + NaOH → RCOONa + R'OH. Common mistake: confusing standard acid-base neutralization with ester hydrolysis.

10
Question 10 of 40
JAMB · Chemistry · 2018

The confirmatory test for alkanoic acids in organic qualitative analysis is

A. turning wet blue litmus paper red
B. reaction with alkanols to form esters
C. reaction with NaOH to form a gas
D. reaction with Na₂CO₃ to liberate CO₂ gas
Explanation

Options A or B might be mistakenly chosen because they are general reactions of acids or functional group conversions, but they lack the specificity required for a confirmatory test. Alkanoic acids react with sodium carbonate (Na₂CO₃) to liberate carbon dioxide gas, which produces effervescence and turns lime water milky according to the equation 2RCOOH + Na₂CO₃ → 2RCOONa + H₂O + CO₂. Common mistake: confusing a general pH indicator color change with a definitive functional group confirmatory test.

11
Question 11 of 40
JAMB · Chemistry · 2018

Which of the following is a secondary alcohol?

A. CH₃CH₂OH
B. CH₃CH(OH)CH₃
C. CH₃CH₂CH₂OH
D. CH₃C(OH)(CH₃)₂
Explanation

A student might select option A or C by choosing a straight-chain primary alcohol instead of evaluating the carbon attachment points. A secondary alcohol is defined by a hydroxyl (-OH) group bonded to a carbon atom that is itself attached to two other carbon groups, which is precisely the arrangement found in propan-2-ol (CH₃CH(OH)CH₃). Common mistake: counting the total number of carbons in the chain rather than checking the degree of substitution on the hydroxyl-bearing carbon.

12
Question 12 of 40
JAMB · Chemistry · 2018

Which of the following compounds reacts with both sodium metal and silver nitrate solution?

A. CH₃C≡CCH₃
B. CH₃C≡CH
C. CH₃CH₂C≡CH
D. CH₃CH₂C≡CCH₃
Explanation

Option A or D might be mistakenly picked by a student who assumes any alkyne will react with sodium metal and silver nitrate. Only terminal alkynes possess an acidic hydrogen atom bonded to a triple-bonded carbon (≡C-H), allowing CH₃C≡CH (propyne) to react with sodium to form a sodium acetylide and with AgNO₃ to yield a white silver acetylide precipitate. Common mistake: ignoring the position of the triple bond when predicting acidity and metal substitution reactions.

13
Question 13 of 40
JAMB · Chemistry · 2018

Ethanol and dimethyl ether have the same molecular formula (C₂H₅OH and CH₃OCH₃). This phenomenon is known as

A. isomerism
B. isotopy
C. allotropy
D. homology
Explanation

Options B, C, or D might be selected if a student confuses structural formulas with allotropes or elemental isotopes. Ethanol (C₂H₅OH) and dimethyl ether (CH₃OCH₃) share an identical molecular formula of C₂H₆O but possess entirely different atom arrangements and functional groups, placing them in the category of structural isomers. Common mistake: confusing isomerism in organic compounds with allotropy or isotopic variations.

14
Question 14 of 40
JAMB · Chemistry · 2018

The reaction of ethene with bromine water is an example of

A. substitution reaction
B. addition reaction
C. elimination reaction
D. redox reaction
Explanation

Option A might be chosen if a student confuses the breaking of a pi bond with the replacement of a hydrogen atom. Ethene (C₂H₄) reacts with bromine water (Br₂/H₂O) via electrophilic addition, wherein the carbon-carbon double bond opens up and bromine atoms attach to adjacent carbons to yield 1,2-dibromoethane (C₂H₄Br₂). Common mistake: assuming multiple bonds undergo substitution rather than addition.

15
Question 15 of 40
JAMB · Chemistry · 2018

What gas is produced when ethanol is heated with concentrated H₂SO₄ at 170°C?

A. Carbon dioxide
B. Ethene
C. Ethane
D. Hydrogen
Explanation

A student might mistakenly choose option A or C by confusing dehydration products with combustion byproducts or alkane reductions. When ethanol (C₂H₅OH) is subjected to heating at 170°C with concentrated sulfuric acid (H₂SO₄), it undergoes an elimination reaction where a molecule of water is removed to generate ethene (C₂H₄) via the pathway C₂H₅OH → C₂H₄ + H₂O. Common mistake: overlooking the role of concentrated sulfuric acid as a dehydrating agent at high temperatures.

16
Question 16 of 40
JAMB · Chemistry · 2018

The compound formed when ethanoic acid reacts with ethanol in the presence of a catalyst is

A. ethyl ethanoate
B. methyl ethanoate
C. ethyl methanoate
D. methyl methanoate
Explanation

Option B or C might be chosen if a student mixes up the alkyl groups derived from the starting acid and alcohol reactants. Ethanoic acid (CH₃COOH) combines with ethanol (C₂H₅OH) via an acid-catalyzed esterification reaction to produce ethyl ethanoate (CH₃COOC₂H₅) and water, deriving its name from the ethyl group of the alcohol and the ethanoate portion of the acid. Common mistake: reversing the order of the acid and alcohol components when naming the resulting ester.

17
Question 17 of 40
JAMB · Chemistry · 2018

The oxidation of a primary alcohol with acidified potassium permanganate produces

A. an aldehyde
B. a ketone
C. a carboxylic acid
D. an ester
Explanation

Option A might be mistakenly selected by a student who thinks strong oxidizing agents stop at the intermediate aldehyde stage. Primary alcohols treated with an excess of a strong oxidizing agent such as acidified potassium permanganate (KMnO₄) undergo complete oxidation past the aldehyde level to yield a carboxylic acid, converting ethanol directly to ethanoic acid via the sequence C₂H₅OH → CH₃CHO → CH₃COOH. Common mistake: stopping the oxidation pathway at the aldehyde stage when using strong oxidizing agents.

18
Question 18 of 40
JAMB · Chemistry · 2018

Which of the following is a test for unsaturation in organic compounds?

A. Addition of bromine water, which decolorizes
B. Reaction with sodium metal
C. Formation of a white precipitate with AgNO₃
D. Production of CO₂ with NaHCO₃
Explanation

Options B, C, or D might be selected if a student confuses unsaturation tests with functional group tests for acids or terminal alkynes. Unsaturation in alkenes is identified by treating the sample with bromine water (Br₂/H₂O), which causes the characteristic red-brown color of bromine to vanish as it adds across the multiple bond to form a dibromo compound like 1,2-dibromoethane. Common mistake: confusing tests for double bonds with tests for specific functional groups like carboxylic acids.

19
Question 19 of 40
JAMB · Chemistry · 2018

The reaction of ethanamide (CH₃CONH₂) with bromine and NaOH produces

A. methylamine
B. ethanamine
C. ethylamine
D. methanol
Explanation

Option B or C might be mistakenly chosen by a student who forgets that the Hofmann rearrangement alters the carbon chain length. The reaction of ethanamide (CH₃CONH₂) with bromine and sodium hydroxide (NaOH) removes a carbonyl carbon as sodium carbonate, yielding methylamine (CH₃NH₂), which contains one fewer carbon atom than the starting amide. Common mistake: keeping the original carbon chain length when predicting the products of an amide degradation reaction.

20
Question 20 of 40
JAMB · Chemistry · 2018

Glucose can be distinguished from fructose by

A. Benedict’s test
B. Tollen’s test
C. Fehling’s test
D. Seliwanoff’s test
Explanation

Options A, B, or C might be selected if a student uses general reducing sugar tests that react positively with both monosaccharide types. Seliwanoff’s test successfully differentiates aldoses like glucose from ketoses like fructose because fructose is a ketose that reacts rapidly with resorcinol in hydrochloric acid to yield a red-colored complex. Common mistake: using general tests like Benedict's or Tollen's to distinguish between different classes of monosaccharides.

21
Question 21 of 40
JAMB · Chemistry · 2018

Which of the following is a natural polymer?

A. Polythene
B. Nylon
C. Starch
D. PVC
Explanation

Option A, B, or D might be picked by a student who confuses naturally occurring macromolecules with synthetically manufactured plastics. Starch is a natural polymer composed of glucose monomer units linked together, serving as an energy storage carbohydrate in plants, whereas polythene, nylon, and PVC are entirely synthetic man-made materials. Common mistake: confusing bio-based macromolecules with petroleum-derived synthetic polymers.

22
Question 22 of 40
JAMB · Chemistry · 2018

The compound formed when propene reacts with hydrogen chloride is

A. 1-chloropropane
B. 2-chloropropane
C. 1,2-dichloropropane
D. 3-chloropropane
Explanation

Option A or C might be chosen by a student who overlooks regioselectivity or adds two halogen atoms instead of one. Propene (CH₃-CH=CH₂) reacts with hydrogen chloride through electrophilic addition in accordance with Markovnikov’s rule, where the hydrogen ion attaches to the carbon with more hydrogen atoms and the chloride ion bonds to the substituted carbon to yield 2-chloropropane (CH₃-CHCl-CH₃). Common mistake: ignoring Markovnikov's rule and placing the substituent on the terminal carbon.

23
Question 23 of 40
JAMB · Chemistry · 2018

What is the product of the reaction between ethyne and excess bromine?

A. 1,2-dibromoethene
B. 1,1,2,2-tetrabromoethane
C. 1,2-dibromoethane
D. 1,1-dibromoethene
Explanation

Option A or C might be selected by a student who assumes only one molecule of halogen adds across a triple bond. Ethyne (HC≡CH) reacts with an excess of bromine (Br₂) to break both pi bonds of the triple bond, allowing four bromine atoms to attach and produce 1,1,2,2-tetrabromoethane (CHBr₂-CHBr₂) as the final saturated product. Common mistake: stopping the addition reaction at the alkene stage when excess halogen is present.

24
Question 24 of 40
JAMB · Chemistry · 2018

The reaction of benzene with concentrated HNO₃ and H₂SO₄ at 55°C produces

A. nitrobenzene
B. phenol
C. toluene
D. benzoic acid
Explanation

Option B, C, or D might be chosen if a student confuses nitration conditions with alkylation or oxidation pathways for benzene. Benzene (C₆H₆) reacts with a mixture of concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄) at 55°C via electrophilic aromatic substitution, where the generated nitronium ion (NO₂⁺) substitutes a ring hydrogen to form nitrobenzene (C₆H₅NO₂). Common mistake: confusing nitrating acid mixtures with reagents used for alkylation or ring oxidation.

25
Question 25 of 40
JAMB · Chemistry · 2018

The molecular formula of a hydrocarbon that undergoes polymerization to form polyethene is

A. C₂H₆
B. C₂H₄
C. C₃H₈
D. C₃H₆
Explanation

Option A or C might be chosen by a student who confuses saturated alkanes with unsaturated alkene monomers. Polyethene is manufactured through addition polymerization using ethene (C₂H₄) as the monomer, where multiple ethene units open their carbon-carbon double bonds to link together into a long continuous chain represented by n(C₂H₄) → [-(CH₂-CH₂)-]n. Common mistake: selecting a saturated alkane formula as the monomer for addition polymerization.

26
Question 26 of 40
JAMB · Chemistry · 2018

The compound CH₃CH₂CHO can be reduced to

A. propan-1-ol
B. propan-2-ol
C. propane
D. ethanoic acid
Explanation

Option B or D might be mistakenly picked by a student who confuses aldehyde reduction products with secondary alcohols or carboxylic acids. The compound CH₃CH₂CHO (propanal) is an aldehyde, and its reduction using agents like NaBH₄ or H₂/Ni converts the carbonyl group into a hydroxyl group, yielding propan-1-ol (CH₃CH₂CH₂OH) as a primary alcohol product. Common mistake: reducing an aldehyde to a secondary alcohol or oxidizing it instead of reducing it.

27
Question 27 of 40
JAMB · Chemistry · 2018

What is the IUPAC name of CH₃COCH₂CH₃?

A. Butan-2-one
B. Propan-2-one
C. Butanal
D. Propan-1-one
Explanation

Students might choose Propan-2-one because they miscount the total carbon atoms in the chain, or they might pick Butanal due to confusing a ketone for an aldehyde. The correct name is derived by identifying the four-carbon chain containing the carbonyl function. The main hydrocarbon structure is butane, and positioning the ketone (C=O) at the second carbon location requires the suffix ‘-one’, leading to butan-2-one. Common mistake: Counting the chain incorrectly or misidentifying the functional group class.

28
Question 28 of 40
JAMB · Chemistry · 2018

Which of the following is a condensation polymer?

A. Polyethene
B. Polyvinyl chloride
C. Nylon
D. Polystyrene
Explanation

An examinee might select Polyethene, Polyvinyl chloride, or Polystyrene assuming all polymers are made through identical mechanisms, without differentiating addition from condensation pathways. Nylon forms via a condensation polymerization route wherein a diamine like hexamethylenediamine reacts with a dicarboxylic acid such as adipic acid. During this process, amide bonds are created and water molecules are simultaneously released: H₂N-(CH₂)₆-NH₂ + HOOC-(CH₂)₄-COOH → nylon 6,6 + H₂O. Common mistake: Confusing addition polymers with condensation polymers that eliminate small molecules.

29
Question 29 of 40
JAMB · Chemistry · 2018

The oxidation number of sulfur in H₂SO₃ is

A. +2
B. +4
C. +6
D. -2
Explanation

One might mistakenly calculate +2 or -2 by misapplying the standard oxidation states for hydrogen and oxygen or by failing to balance the overall neutral charge of the molecule. To find the correct oxidation state for sulfur in sulfurous acid (H₂SO₃), set up the algebraic sum: two hydrogen atoms at +1 each, plus the sulfur atom, plus three oxygen atoms at -2 each, equating to zero. This establishes 2 + S - 6 = 0, which solves to S = +4, perfectly corresponding to sulfur's state in sulfite ions. Common mistake: Forgetting to multiply the oxidation number of oxygen by three.

30
Question 30 of 40
JAMB · Chemistry · 2018

What is the mass of sodium hydroxide required to prepare 500 cm³ of a 0.2 mol/dm³ solution? [Molar mass of NaOH = 40 g/mol]

A. 2 g
B. 4 g
C. 8 g
D. 16 g
Explanation

A student might incorrectly select 8 g or 16 g by miscalculating either the required number of moles or misinterpreting the solution volume conversion from cubic centimeters to cubic decimeters. The required moles are found by multiplying the concentration by the volume in liters, giving 0.2 mol/dm³ × 0.5 dm³ = 0.1 moles. Multiplying these moles by the given molar mass of 40 g/mol yields a mass of 0.1 × 40 = 4 g. Common mistake: Failing to convert 500 cm³ into 0.5 dm³ before calculating moles.

31
Question 31 of 40
JAMB · Chemistry · 2018

Which of the following gases will have the highest rate of diffusion at the same temperature and pressure? [Molar masses: He = 4 g/mol, N₂ = 28 g/mol, O₂ = 32 g/mol, CO₂ = 44 g/mol]

A. He
B. N₂
C. O₂
D. CO₂
Explanation

A candidate might mistakenly choose N₂, O₂, or CO₂ by confusing Graham's law and incorrectly assuming heavier molecules move faster. Graham's Law dictates that a gas's diffusion rate varies inversely with the square root of its molecular weight. Because helium (He) features the smallest molar mass at 4 g/mol compared to the heavier options, it travels at the highest speed under identical pressure and temperature conditions. Common mistake: Believing that heavier gases possess greater diffusion rates.

32
Question 32 of 40
JAMB · Chemistry · 2018

The compound used as an anaesthetic in surgery is

A. chloroform
B. methanol
C. ethanol
D. ethanoic acid
Explanation

An individual might mistakenly pick ethanol, methanol, or ethanoic acid, perhaps recalling organic solvents or general laboratory reagents without knowing historical medical applications. Chloroform (CHCl₃) functioned historically as a surgical general anaesthetic because it depresses the central nervous system to induce unconsciousness, although modern medical practices phased it out due to high toxicity. Common mistake: Confusing common laboratory alcohols or acids with historical surgical agents.

33
Question 33 of 40
JAMB · Chemistry · 2018

The pH of a 0.01 mol/dm³ solution of NaOH is

A. 10
B. 11
C. 12
D. 13
Explanation

One might select 10 or 11 by incorrectly calculating the pOH or subtracting it from the wrong base value instead of working through the pH and pOH relationship. Since sodium hydroxide is a strong base that completely dissociates, the hydroxide ion concentration is 0.01 M, leading to a pOH of -log(0.01) = 2. Subtracting this pOH value from 14 gives pH = 14 - 2 = 12, which describes a strongly basic environment. Common mistake: Forgetting that pH and pOH add up to 14.

34
Question 34 of 40
JAMB · Chemistry · 2018

Which of the following elements forms an amphoteric oxide?

A. Sodium
B. Calcium
C. Zinc
D. Potassium
Explanation

A student could mistakenly choose sodium, calcium, or potassium because they are reactive metals, failing to recall that their oxides are purely basic rather than amphoteric. Zinc produces zinc oxide (ZnO), an amphoteric substance capable of reacting with both acids, such as ZnO + 2HCl → ZnCl₂ + H₂O, and bases, such as ZnO + 2NaOH + H₂O → Na₂Zn(OH)₄. This dual nature sets zinc apart from the other listed metals whose oxides exhibit strictly basic behavior. Common mistake: Assuming all metal oxides are basic.

35
Question 35 of 40
JAMB · Chemistry · 2018

What is the product when propan-2-ol is oxidized with acidified potassium dichromate?

A. Propanal
B. Propanone
C. Propanoic acid
D. Ethanoic acid
Explanation

A test-taker might select Propanal or Propanoic acid by confusing secondary alcohol oxidation pathways with those of primary alcohols or mistaking the product class. Treating propan-2-one, a secondary alcohol designated as CH₃CH(OH)CH₃, with acidified potassium dichromate (K₂Cr₂O₇) yields a ketone via the transformation CH₃CH(OH)CH₃ → CH₃COCH₃ + H₂O. Secondary alcohols specifically produce ketones rather than aldehydes or carboxylic acids under these reaction parameters. Common mistake: Treating secondary alcohols as if they oxidize to form aldehydes.

36
Question 36 of 40
JAMB · Chemistry · 2018

The reaction between an alkene and ozone followed by hydrolysis is used to produce

A. alcohols
B. aldehydes or ketones
C. carboxylic acids
D. esters
Explanation

An examinee might select alcohols or carboxylic acids by confusing ozonolysis products with hydration or oxidation reactions. The ozonolysis process breaks alkene double bonds by first reacting with ozone (O₃) to create an ozonide intermediate, which is subsequently hydrolyzed (using zinc and water) to yield aldehydes or ketones. For instance, CH₃CH=CH₂ converts into CH₃CHO and HCHO depending on the specific starting material's structure. Common mistake: Assuming ozonolysis directly produces carboxylic acids without further oxidation.

37
Question 37 of 40
JAMB · Chemistry · 2018

Which of the following is NOT a characteristic of a homologous series?

A. Members have the same general formula
B. Members have similar chemical properties
C. Members differ by a CH₂ unit
D. Members have the same physical properties
Explanation

A student might incorrectly pick any of the first three options, believing that homologous series members share identical physical properties rather than systematic variations. A homologous series shares a common general formula, steps upward by a fixed CH₂ increment, and maintains similar chemical characteristics owing to identical functional groups. However, physical traits like boiling points change systematically with molecular size rather than staying uniform. Common mistake: Confusing chemical similarity with identical physical properties across a homologous series.

38
Question 38 of 40
JAMB · Chemistry · 2018

The Haber process is used industrially to produce

A. nitric acid
B. ammonia
C. sulfuric acid
D. hydrochloric acid
Explanation

One might mistakenly select nitric acid, sulfuric acid, or hydrochloric acid by confusing different industrial synthesis pathways like the Contact or Ostwald processes with the nitrogen-fixing route. The Haber process is specifically designed to manufacture ammonia (NH₃) from elemental nitrogen and hydrogen via the reversible reaction N₂ + 3H₂ ⇌ 2NH₃, utilizing an iron catalyst under 200 atm of pressure and 450°C. Common mistake: Mixing up the industrial catalysts and products of the Haber process with acid-producing methods.

39
Question 39 of 40
JAMB · Chemistry · 2018

Which of the following salts will produce a basic solution when dissolved in water?

A. NaCl
B. NH₄Cl
C. Na₂CO₃
D. KNO₃
Explanation

An examinee might pick NaCl, NH₄Cl, or KNO₃ by overlooking salt hydrolysis principles and assuming all dissolved salts yield neutral solutions. Sodium carbonate (Na₂CO₃) dissolves to release carbonate ions (CO₃²⁻), which undergo hydrolysis according to CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻, thereby generating hydroxide ions that create an alkaline environment. In contrast, substances like sodium chloride and potassium nitrate remain neutral, whereas ammonium chloride yields acidity. Common mistake: Assuming all salts dissolve to form neutral solutions regardless of their parent ions.

40
Question 40 of 40
JAMB · Chemistry · 2018

The primary source of energy for Earth’s climate system is

A. volcanic activity
B. solar radiation
C. ocean currents
D. earth’s rotation
Explanation

A student might select volcanic activity, ocean currents, or earth's rotation by confusing internal geological forces or fluid dynamics with the external driver of weather. Solar radiation acts as the primary energy engine for Earth's climate engine, powering evaporation cycles, temperature gradients, and atmospheric movement. The sun's incoming energy warms the planet, directly dictating meteorological patterns and broader climatic phenomena. Common mistake: Attributing primary climate energy to internal planetary heat instead of extraterrestrial solar input.

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