A-Level Chemistry Practice Test 17

60 original multiple-choice questions with answers & explanations — covering the full A-Level Chemistry syllabus (AQA, Edexcel & OCR), AS & A2.

Note: These are original practice questions written in A-Level multiple-choice style — not reproduced exam questions. Use them to test your readiness for the real A-Level Chemistry exams.

📱 How to use: Try each question first, then click “Show Answer” to check. Covers the full A-Level course: atomic structure, amount of substance, bonding, energetics & thermodynamics, kinetics, equilibria, redox & electrode potentials, acids & buffers, inorganic chemistry, and organic chemistry with spectroscopy.

⚛ Atomic Structure & Periodicity

Q1A sample of magnesium contains 79.0% Mg-24, 10.0% Mg-25 and 11.0% Mg-26. What is the relative atomic mass of magnesium in this sample?

(A) 24.3
(B) 24.0
(C) 24.8
(D) 25.0
Show Answer

Correct: (A)

Ar = (24 × 79.0 + 25 × 10.0 + 26 × 11.0) / 100 = (1896 + 250 + 286) / 100 = 24.3.

Q2The first four ionisation energies of an element are 590, 1145, 4912 and 6474 kJ mol^-1. In which group of the Periodic Table is this element?

(A) Group 1
(B) Group 3
(C) Group 2
(D) Group 13
Show Answer

Correct: (C)

There is a large jump between the 2nd and 3rd ionisation energies, so the third electron is removed from an inner shell; the element has 2 outer electrons and is in Group 2.

Q3What is the electronic configuration of an Fe^3+ ion? (Fe: atomic number 26)

(A) [Ar] 4s^2 3d^3
(B) [Ar] 3d^5
(C) [Ar] 4s^1 3d^4
(D) [Ar] 3d^3 4s^2
Show Answer

Correct: (B)

Fe is [Ar] 4s^2 3d^6. The two 4s electrons are removed first, then one 3d electron, giving [Ar] 3d^5.

Q4In a time-of-flight mass spectrometer, an ion with m/z = 32 takes 4.2 × 10^-6 s to reach the detector. How long does an ion with m/z = 128 take, if all ions are given the same kinetic energy?

(A) 1.05 × 10^-6 s
(B) 2.1 × 10^-6 s
(C) 16.8 × 10^-6 s
(D) 8.4 × 10^-6 s
Show Answer

Correct: (D)

Flight time is proportional to sqrt(m/z). sqrt(128/32) = 2, so the time is 2 × 4.2 × 10^-6 = 8.4 × 10^-6 s.

Q5Why does the first ionisation energy decrease down Group 2?

(A) The outer electrons are further from the nucleus and more shielded, so they are less strongly attracted
(B) The nuclear charge decreases down the group
(C) The outer electrons are in shells closer to the nucleus
(D) There are fewer shielding electrons down the group
Show Answer

Correct: (A)

Although nuclear charge increases, the extra shielding and greater distance outweigh this, so the outer electron is removed more easily.

⚖ Amount of Substance

Q6A 0.500 g sample of a gas occupies 154 cm^3 at 100 kPa and 298 K. What is the molar mass of the gas? (R = 8.31 J K^-1 mol^-1)

(A) 40 g mol^-1
(B) 64 g mol^-1
(C) 80 g mol^-1
(D) 160 g mol^-1
Show Answer

Correct: (C)

n = pV/RT = (100000 × 154 × 10^-6) / (8.31 × 298) = 0.00622 mol. M = 0.500 / 0.00622 = 80 g mol^-1.

Q7A 25.0 cm^3 sample of sodium hydroxide solution is neutralised by 22.4 cm^3 of 0.150 mol dm^-3 sulfuric acid. What is the concentration of the sodium hydroxide?

(A) 0.134 mol dm^-3
(B) 0.269 mol dm^-3
(C) 0.336 mol dm^-3
(D) 0.538 mol dm^-3
Show Answer

Correct: (B)

Moles H2SO4 = 0.150 × 22.4/1000 = 0.00336. One mole of H2SO4 reacts with 2 mol of NaOH, so moles NaOH = 0.00672. Concentration = 0.00672 / 0.0250 = 0.269 mol dm^-3.

Q8Aspirin (Mr = 180) is made from salicylic acid (Mr = 138). When 5.00 g of salicylic acid is used, 4.89 g of aspirin is obtained. What is the percentage yield?

(A) 50.0%
(B) 66.7%
(C) 90.0%
(D) 75.0%
Show Answer

Correct: (D)

Theoretical yield = (5.00/138) × 180 = 6.52 g. Percentage yield = (4.89/6.52) × 100 = 75.0%.

Q9Calcium oxide is made by heating calcium carbonate: CaCO3 → CaO + CO2. What is the atom economy for making CaO? (Mr: CaCO3 = 100, CaO = 56)

(A) 56%
(B) 44%
(C) 100%
(D) 64%
Show Answer

Correct: (A)

Atom economy = (56/100) × 100 = 56%.

Q10A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?

(A) CHO
(B) C2H4O2
(C) CH2O
(D) CH4O
Show Answer

Correct: (C)

Moles in 100 g: C = 40.0/12 = 3.33; H = 6.7/1 = 6.7; O = 53.3/16 = 3.33. Ratio 1 : 2 : 1, so the empirical formula is CH2O.

Q11A 20 cm^3 sample of a gaseous hydrocarbon is burned in excess oxygen. The products are 60 cm^3 of carbon dioxide and 80 cm^3 of water vapour, all volumes measured at the same temperature and pressure. What is the formula of the hydrocarbon?

(A) C3H6
(B) C3H8
(C) C2H6
(D) C4H10
Show Answer

Correct: (B)

Each molecule gives 60/20 = 3 molecules of CO2, so x = 3. For water, 80/20 = 4 = y/2, so y = 8. The formula is C3H8.

🔗 Bonding & Structure

Q12Which species has a tetrahedral shape?

(A) XeF4
(B) BF3
(C) SF4
(D) NH4^+
Show Answer

Correct: (D)

NH4^+ has four bonding pairs and no lone pairs around nitrogen, so it is tetrahedral. XeF4 is square planar, BF3 is trigonal planar and SF4 is see-saw shaped.

Q13Which substance has hydrogen bonding between its molecules?

(A) HF
(B) HCl
(C) H2S
(D) CH4
Show Answer

Correct: (A)

Fluorine is small and highly electronegative, so HF molecules are held together by hydrogen bonds, giving HF an unusually high boiling point.

Q14Use the data to calculate the lattice enthalpy of formation of sodium chloride. Enthalpy of formation of NaCl = −411 kJ mol^-1; enthalpy of atomisation of Na = +107; first ionisation energy of Na = +496; enthalpy of atomisation of Cl = +121; electron affinity of Cl = −349 kJ mol^-1.

(A) −642 kJ mol^-1
(B) −858 kJ mol^-1
(C) −786 kJ mol^-1
(D) −494 kJ mol^-1
Show Answer

Correct: (C)

Born-Haber cycle: lattice enthalpy = −411 − 107 − 496 − 121 − (−349) = −786 kJ mol^-1.

Q15Which compound has the most ionic character?

(A) AlCl3
(B) CsF
(C) NaCl
(D) MgO
Show Answer

Correct: (B)

Caesium and fluorine have the largest difference in electronegativity of the pairs listed, so CsF is the most ionic.

Q16Which species contains a dative (coordinate) covalent bond?

(A) H3O^+
(B) O2
(C) CH4
(D) CO2
Show Answer

Correct: (A)

In H3O^+, the oxygen atom donates a lone pair to an H^+ ion, forming a dative covalent bond.

Q17Which statement about graphite is correct?

(A) Each carbon atom is bonded to four other carbon atoms
(B) Graphite does not conduct electricity
(C) The layers are held together by strong covalent bonds
(D) Each carbon atom forms three covalent bonds, and delocalised electrons allow graphite to conduct electricity along the layers
Show Answer

Correct: (D)

Each carbon forms three covalent bonds within a layer; the fourth electron is delocalised, so graphite conducts along the layers. The layers are held together by weak van der Waals forces.

🔥 Energetics & Thermodynamics

Q18When 0.0100 mol of ethanol is burned completely, it raises the temperature of 100 g of water by 27.3 °C. What is the enthalpy of combustion of ethanol? (Specific heat capacity of water = 4.18 J g^-1 K^-1)

(A) −572 kJ mol^-1
(B) −1141 kJ mol^-1
(C) −114 kJ mol^-1
(D) −2282 kJ mol^-1
Show Answer

Correct: (B)

q = mcΔT = 100 × 4.18 × 27.3 = 11411 J = 11.41 kJ. ΔH = −11.41 / 0.0100 = −1141 kJ mol^-1.

Q19Use the enthalpies of combustion to find the enthalpy of formation of methane. ΔHc / kJ mol^-1: C = −393; H2 = −286; CH4 = −890.

(A) −965 kJ mol^-1
(B) +75 kJ mol^-1
(C) −185 kJ mol^-1
(D) −75 kJ mol^-1
Show Answer

Correct: (D)

ΔHf = −393 + 2(−286) − (−890) = −393 − 572 + 890 = −75 kJ mol^-1.

Q20Use the bond enthalpies to calculate the enthalpy change for H2 + Cl2 → 2HCl. Bond enthalpies / kJ mol^-1: H−H = 436; Cl−Cl = 242; H−Cl = 431.

(A) −184 kJ mol^-1
(B) +184 kJ mol^-1
(C) −92 kJ mol^-1
(D) −862 kJ mol^-1
Show Answer

Correct: (A)

Bonds broken = 436 + 242 = 678 kJ mol^-1. Bonds formed = 2 × 431 = 862 kJ mol^-1. ΔH = 678 − 862 = −184 kJ mol^-1.

Q21In which reaction is the entropy change of the system positive?

(A) H2O(g) → H2O(l)
(B) N2(g) + 3H2(g) → 2NH3(g)
(C) CaCO3(s) → CaO(s) + CO2(g)
(D) 2H2(g) + O2(g) → 2H2O(l)
Show Answer

Correct: (C)

A gas is produced from a solid, so the disorder increases and ΔS is positive. In each of the other reactions, gas is converted into fewer moles of gas or into a liquid.

Q22A reaction has ΔH = −100 kJ mol^-1 and ΔS = −200 J K^-1 mol^-1. At what temperature is ΔG = 0 for this reaction?

(A) 200 K
(B) 500 K
(C) 400 K
(D) 1000 K
Show Answer

Correct: (B)

ΔG = ΔH − TΔS = 0 when T = ΔH/ΔS = 100000/200 = 500 K. The reaction is feasible below 500 K.

Q23A reaction has ΔH = +50 kJ mol^-1 and ΔS = +100 J K^-1 mol^-1. What is ΔG at 298 K, and is the reaction feasible?

(A) ΔG = −20.2 kJ mol^-1; feasible
(B) ΔG = 0; at equilibrium
(C) ΔG = −79.8 kJ mol^-1; feasible
(D) ΔG = +20.2 kJ mol^-1; not feasible
Show Answer

Correct: (D)

ΔG = 50000 − (298 × 100) = 50000 − 29800 = +20200 J mol^-1 = +20.2 kJ mol^-1. A positive ΔG means the reaction is not feasible.

⏱ Kinetics & Rate Equations

Q24In a series of experiments, doubling [A] doubles the rate, and doubling [B] quadruples the rate. What is the rate equation?

(A) rate = k[A][B]^2
(B) rate = k[A]^2[B]
(C) rate = k[A][B]
(D) rate = k[A]^2[B]^2
Show Answer

Correct: (A)

The reaction is first order with respect to A and second order with respect to B.

Q25For a reaction that is second order overall, what are the units of the rate constant k?

(A) s^-1
(B) mol dm^-3 s^-1
(C) mol^-1 dm^3 s^-1
(D) mol^-2 dm^6 s^-1
Show Answer

Correct: (C)

k = rate / [concentration]^2, so the units are (mol dm^-3 s^-1) / (mol dm^-3)^2 = mol^-1 dm^3 s^-1.

Q26A plot of ln k against 1/T for a reaction is a straight line with gradient −7500 K. What is the activation energy? (R = 8.31 J K^-1 mol^-1)

(A) 6.23 kJ mol^-1
(B) 623 kJ mol^-1
(C) 0.902 kJ mol^-1
(D) 62.3 kJ mol^-1
Show Answer

Correct: (D)

The gradient equals −Ea/R, so Ea = 7500 × 8.31 = 62325 J mol^-1 = 62.3 kJ mol^-1.

Q27A first-order reaction has a half-life of 20 minutes. What fraction of the reactant remains after 60 minutes?

(A) 1/4
(B) 1/8
(C) 1/6
(D) 1/3
Show Answer

Correct: (B)

60 minutes is three half-lives, so the fraction remaining is (1/2)^3 = 1/8.

Q28A temperature increase changes the Maxwell-Boltzmann distribution of molecular energies. Which statement is correct?

(A) The total area under the curve increases
(B) The peak of the curve moves to a lower energy
(C) The total area stays the same, and a larger proportion of molecules has energy at or above the activation energy
(D) The activation energy decreases
Show Answer

Correct: (C)

The number of molecules is unchanged so the area is constant; the peak shifts to the right and flattens, so more molecules have energy at least equal to Ea.

⚖ Chemical Equilibria (Kc & Kp)

Q29The equilibrium N2O4(g) ⇌ 2NO2(g) is set up with 1.00 mol of N2O4 in a 1.00 dm^3 flask. At equilibrium the mixture contains 0.20 mol of NO2. What is the value of Kc?

(A) 0.022 mol dm^-3
(B) 0.044 mol dm^-3
(C) 0.40 mol dm^-3
(D) 22.5 mol dm^-3
Show Answer

Correct: (B)

Formation of 0.20 mol of NO2 means 0.10 mol of N2O4 has dissociated, leaving 0.90 mol. Kc = [NO2]^2 / [N2O4] = (0.20)^2 / 0.90 = 0.044 mol dm^-3.

Q30What is the expression for Kp for the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g)?

(A) Kp = p(SO3) / (p(SO2) × p(O2))
(B) Kp = p(SO2)^2 × p(O2) / p(SO3)^2
(C) Kp = p(SO3)^2 × p(O2) / p(SO2)^2
(D) Kp = p(SO3)^2 / (p(SO2)^2 × p(O2))
Show Answer

Correct: (D)

Kp is the partial pressures of the products, each raised to its stoichiometric coefficient, divided by the same for the reactants.

Q31For the exothermic equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g), what happens to the yield of ammonia and to Kc when the temperature is increased?

(A) Yield decreases and Kc decreases
(B) Yield increases and Kc increases
(C) Yield decreases and Kc is unchanged
(D) Yield increases and Kc decreases
Show Answer

Correct: (A)

Raising the temperature shifts the equilibrium in the endothermic (reverse) direction, lowering the yield; Kc falls because the reactant concentrations rise relative to the product concentration.

Q32For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium partial pressures are: PCl5 = 60 kPa, PCl3 = 40 kPa and Cl2 = 40 kPa. What is the value of Kp?

(A) 0.0375 kPa^-1
(B) 0.667 kPa
(C) 26.7 kPa
(D) 1600 kPa
Show Answer

Correct: (C)

Kp = (p(PCl3) × p(Cl2)) / p(PCl5) = (40 × 40) / 60 = 26.7 kPa.

Q33What is the effect of adding a catalyst to a reversible reaction that is at equilibrium?

(A) Kc increases
(B) Kc is unchanged and equilibrium is reached faster
(C) The equilibrium shifts towards the products
(D) Kc decreases
Show Answer

Correct: (B)

A catalyst speeds up the forward and reverse reactions equally, so the position of equilibrium and the value of Kc do not change.




🔋 Redox & Electrode Potentials

Q34What is the oxidation state of chromium in K2Cr2O7?

(A) +2
(B) +3
(C) +7
(D) +6
Show Answer

Correct: (D)

2(+1) + 2(Cr) + 7(−2) = 0, so 2Cr = +12 and Cr = +6.

Q35Calculate the EMF of a Zn|Zn^2+(aq)||Cu^2+(aq)|Cu cell. E°(Zn^2+/Zn) = −0.76 V; E°(Cu^2+/Cu) = +0.34 V.

(A) +1.10 V
(B) +0.42 V
(C) −1.10 V
(D) −0.42 V
Show Answer

Correct: (A)

EMF = E°(reduction electrode) − E°(oxidation electrode) = 0.34 − (−0.76) = +1.10 V.

Q36Using E°(Cl2/Cl^-) = +1.36 V and E°(Br2/Br^-) = +1.07 V, which reaction is feasible?

(A) Br2(aq) + 2Cl^-(aq) → Cl2(g) + 2Br^-(aq)
(B) I2(aq) + 2Br^-(aq) → Br2(aq) + 2I^-(aq)
(C) Cl2(g) + 2Br^-(aq) → Br2(aq) + 2Cl^-(aq)
(D) Br2(aq) + 2F^-(aq) → F2(g) + 2Br^-(aq)
Show Answer

Correct: (C)

Chlorine is the stronger oxidising agent (more positive E°), so it oxidises bromide ions; E°cell = 1.36 − 1.07 = +0.29 V, which is positive and therefore feasible.

Q37A current of 0.50 A is passed through copper(II) sulfate solution for 1930 s. What mass of copper is deposited at the cathode? (F = 96500 C mol^-1; Ar of Cu = 63.5)

(A) 0.159 g
(B) 0.318 g
(C) 0.635 g
(D) 3.18 g
Show Answer

Correct: (B)

Q = It = 0.50 × 1930 = 965 C, which is 0.0100 mol of electrons. Cu^2+ + 2e^- → Cu, so moles of Cu = 0.00500 and mass = 0.00500 × 63.5 = 0.318 g.

Q38Which statement about standard electrode potentials is correct?

(A) A positive E° always means that the metal is a strong reducing agent
(B) E° values are measured with gases at 1 kPa pressure
(C) The standard hydrogen electrode has E° = +1.00 V
(D) The more negative the E°, the stronger the reducing agent on the right-hand side of the half-equation
Show Answer

Correct: (D)

Species in systems with very negative E° values lose electrons readily, so they are powerful reducing agents. The standard hydrogen electrode is assigned E° = 0 V, and standard pressure is 100 kPa.

🧫 Acids, Bases & Buffers

Q39What is the pH of 0.0100 mol dm^-3 hydrochloric acid?

(A) 2.00
(B) 1.00
(C) 10.00
(D) 12.00
Show Answer

Correct: (A)

Hydrochloric acid is a strong acid and is fully ionised, so [H^+] = 0.0100 mol dm^-3 and pH = −log(0.0100) = 2.00.

Q40What is the pH of a 0.100 mol dm^-3 solution of a weak acid with Ka = 1.8 × 10^-5 mol dm^-3?

(A) 1.00
(B) 2.87
(C) 2.37
(D) 4.74
Show Answer

Correct: (B)

[H^+] = sqrt(Ka × c) = sqrt(1.8 × 10^-5 × 0.100) = 1.34 × 10^-3 mol dm^-3. pH = −log(1.34 × 10^-3) = 2.87.

Q41What is the pH of 0.100 mol dm^-3 sodium hydroxide solution? (Kw = 1.00 × 10^-14 mol^2 dm^-6 at 298 K)

(A) 1.00
(B) 7.00
(C) 12.00
(D) 13.00
Show Answer

Correct: (D)

[H^+] = Kw / [OH^-] = (1.00 × 10^-14) / 0.100 = 1.00 × 10^-13 mol dm^-3, so pH = 13.00.

Q42A buffer solution contains 0.100 mol dm^-3 ethanoic acid (Ka = 1.8 × 10^-5 mol dm^-3) and 0.0500 mol dm^-3 sodium ethanoate. What is the pH of the buffer?

(A) 4.74
(B) 5.04
(C) 4.44
(D) 3.74
Show Answer

Correct: (C)

[H^+] = Ka × [acid] / [salt] = 1.8 × 10^-5 × (0.100 / 0.0500) = 3.6 × 10^-5 mol dm^-3, so pH = 4.44. (Equivalently, pH = pKa + log([salt]/[acid]) = 4.74 − 0.30.)

Q4325.0 cm^3 of 0.100 mol dm^-3 hydrochloric acid is mixed with 50.0 cm^3 of 0.100 mol dm^-3 sodium hydroxide solution. What is the pH of the resulting solution?

(A) 12.52
(B) 7.00
(C) 11.52
(D) 13.00
Show Answer

Correct: (A)

Moles of H^+ = 0.00250; moles of OH^- = 0.00500; excess OH^- = 0.00250 mol in 75.0 cm^3, so [OH^-] = 0.0333 mol dm^-3. [H^+] = (1.00 × 10^-14) / 0.0333 = 3.0 × 10^-13, giving pH = 12.52.

⚗ Inorganic Chemistry

Q44Which statement about the Group 2 elements and their compounds is correct?

(A) The solubility of the Group 2 sulfates increases down the group
(B) The solubility of the Group 2 hydroxides increases down the group
(C) The Group 2 carbonates become less thermally stable down the group
(D) The first ionisation energy of the Group 2 elements increases down the group
Show Answer

Correct: (B)

Hydroxide solubility increases down Group 2 (Mg(OH)2 is sparingly soluble while Ba(OH)2 is soluble). In contrast, sulfate solubility decreases, carbonate thermal stability increases, and first ionisation energy decreases down the group.

Q45Chlorine reacts with cold, dilute sodium hydroxide solution, forming sodium chlorate(I), NaClO. What is the oxidation state of chlorine in NaClO?

(A) −1
(B) 0
(C) +1
(D) +5
Show Answer

Correct: (C)

In NaClO: +1 (Na) + Cl + (−2) (O) = 0, so Cl = +1. Chlorine disproportionates, being oxidised from 0 to +1 and reduced from 0 to −1 in NaCl.

Q46Excess concentrated hydrochloric acid is added to a blue solution containing [Cu(H2O)6]^2+ ions, and [CuCl4]^2- ions are formed. What is the colour of the [CuCl4]^2- solution?

(A) Yellow
(B) Colourless
(C) Deep blue
(D) Pink
Show Answer

Correct: (A)

Ligand substitution by Cl^- ions gives the tetrahedral [CuCl4]^2- complex, which is yellow (a mixture of blue and yellow complexes can appear green).

Q47Which ion forms a brown precipitate when aqueous sodium hydroxide is added?

(A) Cu^2+
(B) Zn^2+
(C) Fe^2+
(D) Fe^3+
Show Answer

Correct: (D)

Fe^3+ ions form a brown precipitate of Fe(OH)3. Cu^2+ gives a blue precipitate, Fe^2+ a green precipitate, and Zn^2+ a white precipitate that redissolves in excess NaOH.

Q48In the Contact process, sulfur dioxide is converted into sulfur trioxide. Which catalyst is used, and what type of catalysis is involved?

(A) Iron; homogeneous
(B) Platinum; heterogeneous
(C) Vanadium(V) oxide; heterogeneous
(D) Vanadium(V) oxide; homogeneous
Show Answer

Correct: (C)

Solid vanadium(V) oxide catalyses the gas-phase reaction 2SO2 + O2 ⇌ 2SO3. Because the catalyst is in a different phase from the reactants, it is heterogeneous catalysis.

🛢 Organic Chemistry I

Q49Which of the following is a propagation step in the free-radical substitution of methane by chlorine?

(A) Cl2 → 2Cl•
(B) CH3• + Cl• → CH3Cl
(C) CH3• + CH3• → C2H6
(D) Cl• + CH4 → CH3• + HCl
Show Answer

Correct: (D)

In a propagation step, a radical reacts with a molecule to form a product and a new radical, keeping the chain going. (A) is initiation, and (B) and (C) are termination steps.

Q50Which halogenoalkane is hydrolysed fastest by aqueous sodium hydroxide?

(A) CH3I
(B) CH3F
(C) CH3Cl
(D) CH3Br
Show Answer

Correct: (A)

The C−I bond is the weakest carbon-halogen bond, so iodomethane reacts fastest. Bond strength increases in the order C−I < C−Br < C−Cl < C−F.

Q51Propene reacts with hydrogen bromide. What is the major product?

(A) 1-bromopropane
(B) 2-bromopropane
(C) 1,2-dibromopropane
(D) 2,2-dibromopropane
Show Answer

Correct: (B)

Electrophilic addition follows Markovnikov's rule: the hydrogen adds to the carbon that already has more hydrogens, via the more stable secondary carbocation, giving 2-bromopropane.

Q52Propan-1-ol is heated under reflux with excess acidified potassium dichromate(VI). What is the organic product?

(A) Propanal
(B) Propanone
(C) Propanoic acid
(D) Propene
Show Answer

Correct: (C)

Under reflux with excess oxidising agent, a primary alcohol is oxidised fully to the carboxylic acid, propanoic acid. Distilling the product off as it forms would give propanal instead.

Q532-bromopropane is heated with ethanolic potassium hydroxide. What is the major organic product?

(A) Propan-2-ol
(B) Propane
(C) 1-bromopropane
(D) Propene
Show Answer

Correct: (D)

Hot ethanolic KOH causes elimination of HBr from the halogenoalkane, forming the alkene propene.

Q54What volume of oxygen is needed for the complete combustion of 10 cm^3 of propane gas, all volumes measured at the same temperature and pressure? C3H8 + 5O2 → 3CO2 + 4H2O

(A) 50 cm^3
(B) 30 cm^3
(C) 10 cm^3
(D) 5 cm^3
Show Answer

Correct: (A)

From the equation, 1 volume of propane reacts with 5 volumes of oxygen, so 10 cm^3 of propane needs 50 cm^3 of oxygen.

🔬 Organic Chemistry II & Analysis

Q55Which compound gives a silver mirror when warmed with Tollens' reagent?

(A) Propanone
(B) Ethanal
(C) Ethanoic acid
(D) Ethanol
Show Answer

Correct: (B)

Tollens' reagent oxidises aldehydes, depositing a silver mirror, but it does not oxidise ketones.

Q56Ethanoic acid reacts with propan-1-ol in the presence of an acid catalyst. What is the name of the ester formed?

(A) Ethyl propanoate
(B) Methyl propanoate
(C) Propyl methanoate
(D) Propyl ethanoate
Show Answer

Correct: (D)

An ester is named as an alkyl alkanoate: the alcohol provides the "propyl" part and the acid provides the "ethanoate" part.

Q57In the nitration of benzene, which species acts as the electrophile?

(A) NO2^-
(B) HNO3
(C) NO2^+
(D) SO3
Show Answer

Correct: (C)

Concentrated sulfuric acid protonates the nitric acid, generating the nitronium ion, NO2^+, which attacks the benzene ring.

Q58Which of the following is the strongest base?

(A) Ethylamine
(B) Ammonia
(C) Phenylamine
(D) Ethanamide
Show Answer

Correct: (A)

The ethyl group pushes electron density onto the nitrogen atom (positive inductive effect), making its lone pair more available to accept a proton. In phenylamine, the lone pair is delocalised into the benzene ring, making it a much weaker base.

Q59A compound shows a broad absorption at 2500–3300 cm^-1 and a sharp absorption at about 1700 cm^-1 in its infrared spectrum. What is the compound most likely to be?

(A) An ester
(B) An alcohol
(C) A carboxylic acid
(D) An amine
Show Answer

Correct: (C)

The very broad O−H absorption of a carboxylic acid (2500–3300 cm^-1), together with the C=O stretch near 1700 cm^-1, identifies a carboxylic acid.

Q60How many peaks are there in the ^13C NMR spectrum of butan-2-one, CH3COCH2CH3?

(A) 2
(B) 4
(C) 3
(D) 5
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Correct: (B)

All four carbon atoms are in different environments, so the spectrum has four peaks.

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