A-Level Chemistry Practice Test 1

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 neon contains the isotopes ^20Ne, ^21Ne and ^22Ne with percentage abundances of 90.5%, 0.3% and 9.2% respectively. What is the relative atomic mass of this sample of neon?

(A) 20.00
(B) 20.19
(C) 20.90
(D) 21.02
Show Answer

Correct: (B)

Ar = (20 × 90.5 + 21 × 0.3 + 22 × 9.2) / 100 = (1810 + 6.3 + 202.4) / 100 = 20.19.

Q2The first four ionisation energies of an element X are 578, 1817, 2745 and 11 578 kJ mol^-1. How many electrons does an atom of X have in its outer shell?

(A) 2
(B) 4
(C) 3
(D) 5
Show Answer

Correct: (C)

There is a large jump between the 3rd and 4th ionisation energies, so the 4th electron is removed from an inner shell; X has 3 outer-shell electrons.

Q3Which ion has the electronic configuration 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6?

(A) Fe^2+
(B) Fe^3+
(C) Co^2+
(D) Cr^3+
Show Answer

Correct: (A)

An iron atom is [Ar] 3d^6 4s^2; forming Fe^2+ removes the two 4s electrons, leaving 3d^6.

Q4Which trend is correct across Period 3 from sodium to argon?

(A) The atomic radius increases
(B) The elements become stronger reducing agents
(C) The first ionisation energy decreases steadily
(D) The bonding in the oxides changes from ionic to covalent
Show Answer

Correct: (D)

Sodium oxide is ionic, whereas the oxides of the later, more electronegative elements (such as SO2 and the oxides of chlorine) are covalent.

Q5In a time-of-flight mass spectrometer, a singly charged ion travels along a flight tube of length 2.00 m in a time of 1.20 × 10^-5 s. What is the speed of the ion?

(A) 6.00 × 10^4 m s^-1
(B) 1.67 × 10^5 m s^-1
(C) 2.40 × 10^5 m s^-1
(D) 1.20 × 10^6 m s^-1
Show Answer

Correct: (B)

Speed = distance / time = 2.00 / (1.20 × 10^-5) = 1.67 × 10^5 m s^-1.

⚖ Amount of Substance

Q6What volume, in dm^3, is occupied by 0.250 mol of an ideal gas at 298 K and a pressure of 100 kPa? (R = 8.31 J K^-1 mol^-1)

(A) 0.619
(B) 5.92
(C) 6.19
(D) 61.9
Show Answer

Correct: (C)

V = nRT / p = (0.250 × 8.31 × 298) / 100 000 = 6.19 × 10^-3 m^3 = 6.19 dm^3.

Q7A 25.0 cm^3 sample of sodium hydroxide solution is exactly neutralised by 23.40 cm^3 of 0.100 mol dm^-3 hydrochloric acid. What is the concentration of the sodium hydroxide solution?

(A) 0.0936 mol dm^-3
(B) 0.0855 mol dm^-3
(C) 0.107 mol dm^-3
(D) 0.936 mol dm^-3
Show Answer

Correct: (A)

Moles of HCl = 0.100 × 23.40 / 1000 = 2.34 × 10^-3 mol = moles of NaOH; concentration = 2.34 × 10^-3 / 0.0250 = 0.0936 mol dm^-3.

Q8A compound contains 40.0% carbon, 6.67% hydrogen and 53.3% oxygen by mass. What is its empirical formula?

(A) CH4O
(B) CH2O2
(C) C2H4O
(D) CH2O
Show Answer

Correct: (D)

Moles: C = 40.0/12 = 3.33; H = 6.67/1 = 6.67; O = 53.3/16 = 3.33. Ratio 1 : 2 : 1, so the empirical formula is CH2O.

Q9In a synthesis, the theoretical yield of a product is 40.0 g but only 30.0 g of pure product is obtained. What is the percentage yield?

(A) 25.0%
(B) 75.0%
(C) 80.0%
(D) 133%
Show Answer

Correct: (B)

Percentage yield = (30.0 / 40.0) × 100 = 75.0%.

Q10What is the atom economy for the production of ethanol by the reaction C2H4 + H2O → C2H5OH?

(A) 46%
(B) 78%
(C) 100%
(D) 61%
Show Answer

Correct: (C)

This is an addition reaction with a single product; total Mr of reactants = 28 + 18 = 46 = Mr of ethanol, so atom economy = (46/46) × 100 = 100%.

Q11What mass of sodium carbonate, Na2CO3 (Mr = 106), is needed to make 250 cm^3 of a 0.200 mol dm^-3 solution?

(A) 5.30 g
(B) 2.65 g
(C) 10.6 g
(D) 21.2 g
Show Answer

Correct: (A)

Moles = 0.200 × 250/1000 = 0.0500 mol; mass = 0.0500 × 106 = 5.30 g.

🔗 Bonding & Structure

Q12Which molecule has a bond angle of approximately 107°?

(A) CH4
(B) BF3
(C) BeCl2
(D) NH3
Show Answer

Correct: (D)

Ammonia is trigonal pyramidal: the lone pair repels the three bonding pairs more strongly, reducing the angle from 109.5° to about 107°.

Q13Hydrogen fluoride has a much higher boiling point than hydrogen chloride. This is because

(A) the H-F bond is weaker than the H-Cl bond
(B) HF molecules form hydrogen bonds but HCl molecules do not
(C) fluorine atoms are larger than chlorine atoms
(D) HF is an acid but HCl is not
Show Answer

Correct: (B)

Fluorine is small and highly electronegative, so HF forms strong hydrogen bonds between molecules; HCl has only permanent dipole-dipole forces.

Q14Use the data: ΔHf(NaCl) = −411 kJ mol^-1; enthalpy of atomisation of sodium = +108 kJ mol^-1; first ionisation energy of sodium = +496 kJ mol^-1; enthalpy of atomisation of chlorine (per mole of Cl atoms) = +121 kJ mol^-1; electron affinity of chlorine = −349 kJ mol^-1. What is the lattice enthalpy of sodium chloride?

(A) −376 kJ mol^-1
(B) −411 kJ mol^-1
(C) −787 kJ mol^-1
(D) −1574 kJ mol^-1
Show Answer

Correct: (C)

ΔHf = 108 + 496 + 121 − 349 + ΔHlatt; −411 = +376 + ΔHlatt, so ΔHlatt = −411 − 376 = −787 kJ mol^-1.

Q15Which substance consists of a giant covalent structure?

(A) Silicon(IV) oxide
(B) Ice
(C) Iodine
(D) Sodium chloride
Show Answer

Correct: (A)

In SiO2 every atom is held in a three-dimensional network by strong covalent bonds; iodine and ice are molecular, and NaCl is ionic.

Q16Which bond is the most polar?

(A) C-H
(B) O-H
(C) C-Cl
(D) H-F
Show Answer

Correct: (D)

Polarity increases with the difference in electronegativity; F (4.0) and H (2.1) give the largest difference, 1.9.

Q17Ionic compounds conduct electricity when molten but not when solid because

(A) the ions are smaller in the molten state
(B) in the liquid the ions are free to move, but in the solid they are held in fixed positions in the lattice
(C) electrons become delocalised when the compound melts
(D) the ionic bonds are broken completely on melting
Show Answer

Correct: (B)

Charge is carried by the movement of ions, which is only possible when the lattice breaks down on melting (or dissolving).

🔥 Energetics & Thermodynamics

Q18In a calorimetry experiment, burning 0.200 mol of a liquid fuel raised the temperature of 500 g of water by 15.0 °C. What is the enthalpy of combustion of the fuel? (Specific heat capacity of water = 4.18 J g^-1 K^-1)

(A) −31.4 kJ mol^-1
(B) −78.4 kJ mol^-1
(C) −157 kJ mol^-1
(D) −313 kJ mol^-1
Show Answer

Correct: (C)

q = mcΔT = 500 × 4.18 × 15.0 = 31 350 J = 31.35 kJ; ΔH = −31.35 / 0.200 = −157 kJ mol^-1.

Q19Use the enthalpies of combustion: C = −394 kJ mol^-1; H2 = −286 kJ mol^-1; CH4 = −890 kJ mol^-1. What is the enthalpy of formation of methane?

(A) −76 kJ mol^-1
(B) +76 kJ mol^-1
(C) −1856 kJ mol^-1
(D) −17 kJ mol^-1
Show Answer

Correct: (A)

ΔHf(CH4) = ΔHc(C) + 2ΔHc(H2) − ΔHc(CH4) = −394 + 2(−286) − (−890) = −76 kJ mol^-1.

Q20Which process is accompanied by an increase in entropy?

(A) H2O(l) → H2O(s)
(B) H2O(g) → H2O(l)
(C) 2SO2(g) + O2(g) → 2SO3(g)
(D) CaCO3(s) → CaO(s) + CO2(g)
Show Answer

Correct: (D)

A gas is produced from a solid, so the number of particles and the disorder of the system increase; the other processes all reduce disorder or reduce the number of gas molecules.

Q21A 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) 0.50 K
(D) 1000 K
Show Answer

Correct: (B)

At ΔG = 0, T = ΔH / ΔS = 100 000 / 200 = 500 K.

Q22Use the mean bond enthalpies: C-H = 412; O=O = 496; C=O = 743; O-H = 463 kJ mol^-1. What is the enthalpy change for CH4 + 2O2 → CO2 + 2H2O?

(A) +698 kJ mol^-1
(B) −2640 kJ mol^-1
(C) −698 kJ mol^-1
(D) −349 kJ mol^-1
Show Answer

Correct: (C)

Bonds broken = 4(412) + 2(496) = 2640 kJ; bonds formed = 2(743) + 4(463) = 3338 kJ; ΔH = 2640 − 3338 = −698 kJ mol^-1.

Q23The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions is the

(A) lattice enthalpy of formation
(B) enthalpy of atomisation
(C) enthalpy of hydration
(D) enthalpy of solution
Show Answer

Correct: (A)

This is the definition of the (exothermic) lattice enthalpy of formation.

⏱ Kinetics & Rate Equations

Q24In some experiments on the reaction P + Q → products, doubling [P] (with [Q] constant) doubled the initial rate, and doubling [Q] (with [P] constant) quadrupled the initial rate. What is the rate equation?

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

Correct: (D)

The rate is first order with respect to P (doubling [P] doubles the rate) and second order with respect to Q (doubling [Q] gives 2^2 = 4 times the rate).

Q25The units of the rate constant for an overall second-order reaction are

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

Correct: (B)

k = rate / concentration^2, giving (mol dm^-3 s^-1) / (mol dm^-3)^2 = mol^-1 dm^3 s^-1.

Q26A first-order reaction has a frequency factor A = 1.0 × 10^11 s^-1 and an activation energy of 80 kJ mol^-1. What is the value of the rate constant at 300 K? (R = 8.31 J K^-1 mol^-1)

(A) 1.16 s^-1
(B) 8.62 × 10^-11 s^-1
(C) 1.16 × 10^-3 s^-1
(D) 3.11 × 10^-2 s^-1
Show Answer

Correct: (C)

k = A e^(-Ea/RT); Ea/RT = 80 000 / (8.31 × 300) = 32.1; e^-32.1 = 1.16 × 10^-14; k = 1.0 × 10^11 × 1.16 × 10^-14 = 1.16 × 10^-3 s^-1.

Q27A first-order reaction has a half-life of 20 s. Starting from a given concentration, how long does it take for the concentration to fall to one eighth of its initial value?

(A) 60 s
(B) 40 s
(C) 160 s
(D) 6.7 s
Show Answer

Correct: (A)

Falling to 1/8 means three half-lives (1 → 1/2 → 1/4 → 1/8), so time = 3 × 20 = 60 s.

Q28The mechanism for the reaction 2NO2 + CO → 2NO + CO2 is proposed as: Step 1 (slow): NO2 + NO2 → NO3 + NO; Step 2 (fast): NO3 + CO → NO2 + CO2. What is the rate equation for this reaction?

(A) rate = k[NO2][CO]
(B) rate = k[NO3][CO]
(C) rate = k[NO2]
(D) rate = k[NO2]^2
Show Answer

Correct: (D)

The rate is determined by the slow step, which involves two molecules of NO2 and does not involve CO, so rate = k[NO2]^2.

⚖ Chemical Equilibria (Kc & Kp)

Q29For the equilibrium N2O4(g) ⇌ 2NO2(g), 2.00 mol of N2O4 was placed in a 2.00 dm^3 vessel. At equilibrium the mixture contained 0.800 mol of NO2. What is the value of Kc?

(A) 0.100 mol dm^-3
(B) 0.200 mol dm^-3
(C) 0.400 mol dm^-3
(D) 5.00 mol dm^-3
Show Answer

Correct: (B)

NO2 formed = 0.800 mol, so N2O4 used = 0.400 mol and N2O4 remaining = 1.60 mol. [N2O4] = 1.60/2.00 = 0.800; [NO2] = 0.800/2.00 = 0.400 mol dm^-3. Kc = (0.400)^2 / 0.800 = 0.200 mol dm^-3.

Q30For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g), an equilibrium mixture contains 0.20 mol PCl5, 0.30 mol PCl3 and 0.30 mol Cl2 at a total pressure of 100 kPa. What is the value of Kp?

(A) 0.563 kPa
(B) 5.63 kPa
(C) 56.3 kPa
(D) 563 kPa
Show Answer

Correct: (C)

Total moles = 0.80. Partial pressures: PCl5 = (0.20/0.80) × 100 = 25 kPa; PCl3 = Cl2 = (0.30/0.80) × 100 = 37.5 kPa. Kp = (37.5 × 37.5) / 25 = 56.3 kPa.

Q31In the Haber process, N2 + 3H2 ⇌ 2NH3 (ΔH = −92 kJ mol^-1), which change increases both the rate of reaction and the equilibrium yield of ammonia?

(A) Increasing the pressure
(B) Increasing the temperature
(C) Adding an iron catalyst
(D) Decreasing the pressure
Show Answer

Correct: (A)

Higher pressure shifts the equilibrium towards the side with fewer gas molecules (more NH3) and also increases the frequency of collisions, so the rate increases. Higher temperature increases rate but lowers yield; a catalyst affects rate only.

Q32For the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), the units of Kc are

(A) mol dm^-3
(B) no units
(C) mol^2 dm^-6
(D) mol^-1 dm^3
Show Answer

Correct: (D)

Kc = [SO3]^2 / ([SO2]^2 [O2]); units = (mol dm^-3)^2 / (mol dm^-3)^3 = mol^-1 dm^3.

Q33Which statement about a system in dynamic equilibrium is correct?

(A) The concentrations of the reactants and products are equal
(B) The rate of the forward reaction is equal to the rate of the reverse reaction
(C) Both reactions have stopped
(D) The concentration of products is always greater than that of reactants
Show Answer

Correct: (B)

In dynamic equilibrium the forward and reverse reactions continue at equal rates, so concentrations remain constant (but are not necessarily equal).




🔋 Redox & Electrode Potentials

Q34What is the oxidation state of sulfur in Na2S2O3?

(A) −2
(B) +4
(C) +2
(D) +6
Show Answer

Correct: (C)

2(+1) + 2(S) + 3(−2) = 0, so 2(S) = +4 and the (average) oxidation state of sulfur is +2.

Q35Use the standard electrode potentials: Fe^2+ + 2e^- → Fe, E = −0.44 V; Cu^2+ + 2e^- → Cu, E = +0.34 V. What is the EMF of the cell Fe | Fe^2+ || Cu^2+ | Cu under standard conditions?

(A) +0.78 V
(B) −0.78 V
(C) +0.10 V
(D) −0.10 V
Show Answer

Correct: (A)

EMF = E(more positive) − E(more negative) = +0.34 − (−0.44) = +0.78 V.

Q36Use the standard electrode potentials: Fe^3+ + e^- → Fe^2+, E = +0.77 V; Cl2 + 2e^- → 2Cl^-, E = +1.36 V; Br2 + 2e^- → 2Br^-, E = +1.07 V; I2 + 2e^- → 2I^-, E = +0.54 V. Which species is the strongest reducing agent?

(A) Cl^-
(B) Fe^3+
(C) Br^-
(D) I^-
Show Answer

Correct: (D)

A reducing agent is most easily oxidised when its electrode potential is the least positive; I^- is oxidised most readily.

Q37A current of 2.00 A is passed through copper(II) sulfate solution for 16.1 minutes. What mass of copper is deposited at the cathode? (Faraday constant = 96 500 C mol^-1; Ar of Cu = 63.5)

(A) 0.318 g
(B) 0.635 g
(C) 1.27 g
(D) 6.35 g
Show Answer

Correct: (B)

Q = It = 2.00 × (16.1 × 60) = 1932 C; moles of electrons = 1932 / 96 500 = 0.0200 mol; Cu^2+ + 2e^- → Cu, so moles of Cu = 0.0100; mass = 0.0100 × 63.5 = 0.635 g.

Q38Use the standard electrode potentials: Zn^2+ + 2e^- → Zn, E = −0.76 V; Cu^2+ + 2e^- → Cu, E = +0.34 V; 2H^+ + 2e^- → H2, E = 0.00 V; Fe^3+ + e^- → Fe^2+, E = +0.77 V; I2 + 2e^- → 2I^-, E = +0.54 V. Which reaction is NOT feasible under standard conditions?

(A) Zn + Cu^2+ → Zn^2+ + Cu
(B) 2Fe^3+ + 2I^- → 2Fe^2+ + I2
(C) Cu + 2H^+ → Cu^2+ + H2
(D) Zn + 2H^+ → Zn^2+ + H2
Show Answer

Correct: (C)

For this reaction EMF = 0.00 − (+0.34) = −0.34 V, which is negative, so the reaction is not feasible; the other three all have positive EMF values.

🧫 Acids, Bases & Buffers

Q39What is the pH of 0.0500 mol dm^-3 hydrochloric acid?

(A) 1.30
(B) 1.00
(C) 2.30
(D) 0.70
Show Answer

Correct: (A)

HCl is fully ionised, so [H^+] = 0.0500 mol dm^-3; pH = −log(0.0500) = 1.30.

Q40A weak acid HA has Ka = 1.8 × 10^-5 mol dm^-3. What is the pH of a 0.100 mol dm^-3 solution of HA?

(A) 1.00
(B) 3.74
(C) 4.74
(D) 2.87
Show Answer

Correct: (D)

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

Q41A buffer solution contains 0.100 mol dm^-3 ethanoic acid and 0.200 mol dm^-3 sodium ethanoate. Ka for ethanoic acid = 1.74 × 10^-5 mol dm^-3. What is the pH of this buffer?

(A) 4.46
(B) 5.06
(C) 4.76
(D) 5.36
Show Answer

Correct: (B)

pH = −log Ka + log([salt]/[acid]) = 4.76 + log(0.200/0.100) = 4.76 + 0.30 = 5.06.

Q42At 50 °C, Kw = 5.5 × 10^-14 mol^2 dm^-6. What is the pH of pure water at this temperature?

(A) 7.00
(B) 7.26
(C) 6.63
(D) 13.26
Show Answer

Correct: (C)

In pure water [H^+] = square root of Kw = square root of (5.5 × 10^-14) = 2.35 × 10^-7 mol dm^-3; pH = 6.63. The water is still neutral because [H^+] = [OH^-].

Q43Which mixture will act as a buffer solution?

(A) Ethanoic acid and sodium ethanoate
(B) Hydrochloric acid and sodium chloride
(C) Sodium hydroxide and sodium chloride
(D) Nitric acid and potassium nitrate
Show Answer

Correct: (A)

A buffer needs a weak acid and its conjugate base (the salt) together; the other mixtures contain a strong acid or a strong base with no weak conjugate partner.

⚗ Inorganic Chemistry

Q44Which statement about Group 2 chemistry is correct?

(A) Magnesium hydroxide is more soluble in water than barium hydroxide
(B) Barium sulfate is soluble in water
(C) The reactivity of the Group 2 elements decreases down the group
(D) The solubility of the Group 2 hydroxides increases down the group
Show Answer

Correct: (D)

Hydroxide solubility increases down Group 2 (Mg(OH)2 is sparingly soluble, Ba(OH)2 is much more soluble), whereas sulfate solubility decreases.

Q45When chlorine reacts with cold, dilute sodium hydroxide solution, the products include sodium chloride and sodium chlorate(I). In this reaction, chlorine is

(A) only oxidised
(B) both oxidised and reduced
(C) only reduced
(D) neither oxidised nor reduced
Show Answer

Correct: (B)

Chlorine starts at oxidation state 0; some atoms are reduced to −1 (in NaCl) and others oxidised to +1 (in NaClO). This is disproportionation.

Q46When excess concentrated ammonia solution is added to aqueous copper(II) sulfate, a deep blue solution forms. The deep blue colour is due to the complex ion

(A) [Cu(H2O)6]^2+
(B) [CuCl4]^2-
(C) [Cu(NH3)4(H2O)2]^2+
(D) [Cu(NH3)6]^2+
Show Answer

Correct: (C)

Ammonia replaces four of the water ligands: [Cu(H2O)6]^2+ + 4NH3 → [Cu(NH3)4(H2O)2]^2+ + 4H2O, giving an intense deep blue colour.

Q47A solution of a metal ion gives an orange-brown precipitate when sodium hydroxide solution is added, and the precipitate does not dissolve in excess sodium hydroxide. The metal ion is

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

Correct: (A)

Fe^3+(aq) + 3OH^-(aq) → Fe(OH)3(s), an orange-brown precipitate. Fe^2+ gives a green precipitate, Cu^2+ a blue one, and Al^3+ a white one that dissolves in excess NaOH.

Q48Which property is shown by transition metals but not by the Group 1 metals?

(A) They are shiny when cut
(B) They conduct electricity
(C) They are soft and easily cut
(D) They form compounds with variable oxidation states
Show Answer

Correct: (D)

Variable oxidation states (and coloured compounds and catalytic activity) arise because transition metals can use their partially filled d sub-shell; Group 1 metals form only +1 colourless compounds.

🛢 Organic Chemistry I

Q49In the chlorination of methane in ultraviolet light, the initiation step involves

(A) heterolytic fission of the Cl-Cl bond to form ions
(B) homolytic fission of the Cl-Cl bond to form free radicals
(C) electrophilic attack by Cl^+ on methane
(D) nucleophilic attack by Cl^- on methane
Show Answer

Correct: (B)

UV light provides the energy for homolytic fission: Cl2 → 2Cl•, producing the chlorine free radicals that start the chain reaction.

Q50Which halogenoalkane is hydrolysed most rapidly by aqueous sodium hydroxide?

(A) Fluoromethane
(B) Chloromethane
(C) Iodomethane
(D) Bromomethane
Show Answer

Correct: (C)

The rate of hydrolysis follows the strength of the carbon-halogen bond: the C-I bond is the weakest, so iodomethane reacts fastest (I > Br > Cl > F).

Q51Propene reacts with hydrogen bromide to give a mixture of products. The major product is

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

Correct: (A)

Electrophilic addition follows Markovnikov's rule: the H attaches to the carbon that already has more hydrogens, giving the more stable secondary carbocation and hence 2-bromopropane as the major product.

Q52Propan-1-ol is heated under reflux with excess acidified potassium dichromate(VI). The main organic product is

(A) propanal
(B) propanone
(C) propane
(D) propanoic acid
Show Answer

Correct: (D)

Under reflux with excess oxidising agent, a primary alcohol is oxidised fully to the carboxylic acid. (Propanal is obtained only if it is distilled off as it forms.)

Q53Decane, C10H22, is cracked to give ethene and one other product: C10H22 → C2H4 + X. What is the formula of X?

(A) C8H16
(B) C8H18
(C) C6H14
(D) C8H20
Show Answer

Correct: (B)

Balance the atoms: carbon 10 − 2 = 8; hydrogen 22 − 4 = 18, so X = C8H18 (octane).

Q54When butan-2-ol is dehydrated by heating with concentrated phosphoric acid, the major alkene product is

(A) but-1-ene
(B) 2-methylpropene
(C) but-2-ene
(D) cyclobutane
Show Answer

Correct: (C)

Elimination can remove H from either adjacent carbon, but the more substituted (and more stable) alkene, but-2-ene, is the major product.

🔬 Organic Chemistry II & Analysis

Q55Which reagent would distinguish between propanal and propanone?

(A) Tollens' reagent
(B) Sodium hydrogencarbonate solution
(C) Sodium metal
(D) Calcium carbonate
Show Answer

Correct: (A)

Propanal (an aldehyde) reduces Tollens' reagent to give a silver mirror; propanone (a ketone) does not react. Neither compound reacts with the other reagents.

Q56Ethyl ethanoate is hydrolysed by refluxing with dilute sulfuric acid. The organic products are

(A) ethanoic acid and methanol
(B) ethanol and methanoic acid
(C) ethene and water
(D) ethanol and ethanoic acid
Show Answer

Correct: (D)

Acid hydrolysis of an ester is the reverse of esterification: CH3COOCH2CH3 + H2O ⇌ CH3COOH + CH3CH2OH.

Q57In the nitration of benzene using concentrated nitric acid and concentrated sulfuric acid, the electrophile is

(A) NO2
(B) NO2^+
(C) HNO3
(D) NO^+
Show Answer

Correct: (B)

The sulfuric acid protonates the nitric acid, which loses water to form the nitronium ion, NO2^+, the electrophile that attacks the benzene ring.

Q58Which compound is the strongest base?

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

Correct: (C)

The ethyl group pushes electron density onto the nitrogen (+I effect), making the lone pair more available to accept a proton; in phenylamine the lone pair is delocalised into the benzene ring, and in ethanamide it is delocalised onto oxygen.

Q59A compound with molecular formula C3H6O shows a strong infrared absorption at about 1720 cm^-1 but no broad absorption between 3200 and 3600 cm^-1. The compound is most likely

(A) propanone
(B) propan-1-ol
(C) propene
(D) chloropropane
Show Answer

Correct: (A)

The peak at 1720 cm^-1 is the C=O stretch of a carbonyl group, and the absence of a broad O-H peak rules out an alcohol; propanone fits both observations.

Q60In the mass spectrum of a compound with Mr = 58, there is a peak at m/z = 43. This peak corresponds to the loss, from the molecular ion, of

(A) an OH group (mass 17)
(B) a C2H5 group (mass 29)
(C) an H2O molecule (mass 18)
(D) a CH3 group (mass 15)
Show Answer

Correct: (D)

58 − 43 = 15, which is the mass of a CH3 group; the fragment ion remaining is at m/z 43.

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