A-Level Chemistry Practice Test 12

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 the isotopes Mg-24 (79.0%), Mg-25 (10.0%) and Mg-26 (11.0%). The relative atomic mass of magnesium in the sample is

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

Correct: (A)

Ar = (24 × 0.790) + (25 × 0.100) + (26 × 0.110) = 18.96 + 2.50 + 2.86 = 24.32 ≈ 24.3.

Q2The successive ionisation energies of an element are 738, 1451, 7733, 10 543 and 13 637 kJ mol^-1. This element is

(A) sodium
(B) aluminium
(C) magnesium
(D) silicon
Show Answer

Correct: (C)

The very large increase between the second and third ionisation energies shows two outer-shell electrons, so the element is in Group 2 (magnesium).

Q3The electronic configuration of a chromium atom is unusual. It is

(A) [Ar] 4s^2 3d^4
(B) [Ar] 4s^0 3d^6
(C) [Ar] 4s^2 3d^6
(D) [Ar] 4s^1 3d^5
Show Answer

Correct: (D)

Chromium is [Ar] 4s^1 3d^5: a half-filled 3d subshell with one 4s electron gives extra stability.

Q4A sample of bromine contains Br-79 (50.7%) and Br-81 (49.3%). The relative atomic mass of bromine is

(A) 81.0
(B) 79.5
(C) 80.5
(D) 79.9
Show Answer

Correct: (D)

Ar = (79 × 0.507) + (81 × 0.493) = 40.05 + 39.93 = 79.98 ≈ 79.9.

Q5Which oxide of a Period 3 element is amphoteric?

(A) Al2O3
(B) MgO
(C) Na2O
(D) SO2
Show Answer

Correct: (A)

Aluminium oxide reacts with both acids and alkalis, so it is amphoteric; sodium and magnesium oxides are basic and sulfur dioxide is acidic.

⚖ Amount of Substance

Q6A 0.500 g sample of an unknown gas occupies 240 cm^3 at 100 kPa and 298 K. (R = 8.31 J K^-1 mol^-1.) The molar mass of the gas is approximately

(A) 26 g mol^-1
(B) 13 g mol^-1
(C) 104 g mol^-1
(D) 52 g mol^-1
Show Answer

Correct: (D)

n = pV/RT = (100 000 × 2.40 × 10^-4) / (8.31 × 298) = 24.0 / 2476 = 9.69 × 10^-3 mol. Mr = 0.500 / 9.69 × 10^-3 = 51.6 ≈ 52 g mol^-1.

Q7A 25.0 cm^3 portion of 0.150 mol dm^-3 hydrochloric acid is neutralised by 22.5 cm^3 of calcium hydroxide solution. The concentration of the calcium hydroxide solution is

(A) 0.167 mol dm^-3
(B) 0.0833 mol dm^-3
(C) 0.333 mol dm^-3
(D) 0.0417 mol dm^-3
Show Answer

Correct: (B)

Moles HCl = 0.150 × 0.0250 = 3.75 × 10^-3. Ca(OH)2 reacts with 2 HCl, so moles Ca(OH)2 = 1.875 × 10^-3. Concentration = 1.875 × 10^-3 / 0.0225 = 0.0833 mol dm^-3.

Q8Magnesium carbonate decomposes: MgCO3 → MgO + CO2. (Mr: MgCO3 = 84, MgO = 40.) The atom economy for producing MgO is

(A) 52.4%
(B) 100%
(C) 47.6%
(D) 84.0%
Show Answer

Correct: (C)

Atom economy = (40 / 84) × 100 = 47.6%.

Q9Methane burns in oxygen: CH4 + 2O2 → CO2 + 2H2O. If 0.50 mol of CH4 is mixed with 0.80 mol of O2, the maximum amount of CO2 that can form is

(A) 0.40 mol
(B) 0.80 mol
(C) 0.50 mol
(D) 0.25 mol
Show Answer

Correct: (A)

0.50 mol CH4 would need 1.00 mol O2, so O2 is the limiting reagent. 0.80 mol O2 reacts with 0.40 mol CH4 to give 0.40 mol CO2.

Q10The concentration of a 0.250 mol dm^-3 solution of sulfuric acid, expressed in g dm^-3, is (Mr H2SO4 = 98)

(A) 245 g dm^-3
(B) 24.5 g dm^-3
(C) 2.45 g dm^-3
(D) 49.0 g dm^-3
Show Answer

Correct: (B)

Mass per dm^3 = 0.250 × 98 = 24.5 g dm^-3.

Q11How many molecules are there in 0.88 g of carbon dioxide? (Mr CO2 = 44; Avogadro constant = 6.02 × 10^23 mol^-1)

(A) 1.20 × 10^22
(B) 6.02 × 10^21
(C) 1.20 × 10^24
(D) 3.01 × 10^22
Show Answer

Correct: (A)

Moles = 0.88 / 44 = 0.0200 mol. Molecules = 0.0200 × 6.02 × 10^23 = 1.20 × 10^22.

🔗 Bonding & Structure

Q12What is the shape of an SF6 molecule and its bond angle?

(A) Octahedral, 90°
(B) Tetrahedral, 109.5°
(C) Trigonal planar, 120°
(D) Square planar, 90°
Show Answer

Correct: (A)

Sulfur has six bonding pairs arranged octahedrally, with bond angles of 90° (and 180°).

Q13Which of these substances contains hydrogen bonding between its molecules?

(A) CH4
(B) H2S
(C) SiH4
(D) NH3
Show Answer

Correct: (D)

Hydrogen bonding requires H attached directly to a small, highly electronegative atom (N, O or F); only ammonia qualifies here.

Q14For MgCl2: ΔHf = −642 kJ mol^-1; atomisation of Mg = +150; first + second ionisation energies of Mg = +2188; atomisation of Cl per mol of Cl atoms = +121; electron affinity of Cl = −349 kJ mol^-1. The lattice enthalpy of formation of MgCl2 is

(A) −1882 kJ mol^-1
(B) −642 kJ mol^-1
(C) −1240 kJ mol^-1
(D) −2524 kJ mol^-1
Show Answer

Correct: (D)

ΔHlatt = ΔHf − [150 + 2188 + 2(121) + 2(−349)] = −642 − [150 + 2188 + 242 − 698] = −642 − 1882 = −2524 kJ mol^-1.

Q15Which substance has the highest boiling point?

(A) butane
(B) propanal
(C) propan-1-ol
(D) methoxyethane
Show Answer

Correct: (C)

Propan-1-ol has hydrogen bonding between molecules, stronger than the dipole-dipole and van der Waals forces in the others, so it needs the most energy to boil.

Q16Magnesium has a much higher melting point than sodium mainly because

(A) magnesium atoms are smaller than sodium atoms
(B) sodium has a giant covalent structure
(C) magnesium has a full outer shell of electrons
(D) Mg^2+ ions and twice as many delocalised electrons per atom give stronger metallic bonding
Show Answer

Correct: (D)

In magnesium, each atom contributes two delocalised electrons and forms 2+ ions, so the electrostatic attraction is much stronger than in sodium (1+ ions, one electron each).

Q17Which statement comparing diamond and graphite is correct?

(A) Both conduct electricity
(B) Diamond is soft because its layers can slide
(C) In diamond each carbon forms four bonds; in graphite each carbon forms three bonds within a layer
(D) Graphite has a higher density than diamond because its atoms are closer together
Show Answer

Correct: (C)

Diamond is a 3D tetrahedral network with four bonds per atom; graphite has three bonds per atom in flat layers, with the fourth electron delocalised.

🔥 Energetics & Thermodynamics

Q18Burning 0.460 g of ethanol (0.0100 mol) raises the temperature of 100 g of water by 32.6 K. (c = 4.18 J g^-1 K^-1.) The enthalpy change of combustion of ethanol is approximately

(A) −13.6 kJ mol^-1
(B) −136 kJ mol^-1
(C) −2720 kJ mol^-1
(D) −1360 kJ mol^-1
Show Answer

Correct: (D)

q = mcΔT = 100 × 4.18 × 32.6 = 13 627 J for 0.0100 mol. Per mole: 13 627 / 0.0100 = 1.36 × 10^6 J mol^-1, so ΔH ≈ −1360 kJ mol^-1.

Q19Given ΔHf(C2H4) = +52 kJ mol^-1 and ΔHf(C2H6) = −85 kJ mol^-1, the enthalpy change for C2H4 + H2 → C2H6 is

(A) +137 kJ mol^-1
(B) −33 kJ mol^-1
(C) −85 kJ mol^-1
(D) −137 kJ mol^-1
Show Answer

Correct: (D)

ΔH = −85 − (+52) = −137 kJ mol^-1.

Q20Using bond enthalpies (kJ mol^-1): N≡N = 945, H−H = 436, N−H = 391, the enthalpy change for N2 + 3H2 → 2NH3 is

(A) −93 kJ mol^-1
(B) +93 kJ mol^-1
(C) −2346 kJ mol^-1
(D) +2253 kJ mol^-1
Show Answer

Correct: (A)

Bonds broken = 945 + 3(436) = 2253; bonds formed = 6(391) = 2346. ΔH = 2253 − 2346 = −93 kJ mol^-1.

Q21Which change involves a decrease in the entropy of the system?

(A) CaCO3(s) → CaO(s) + CO2(g)
(B) H2O(g) → H2O(l)
(C) NH4NO3(s) dissolving in water
(D) a solid melting
Show Answer

Correct: (B)

Condensation converts a disordered gas into a much more ordered liquid, so entropy decreases.

Q22A reaction has ΔH = −50 kJ mol^-1 and ΔS = −100 J K^-1 mol^-1. The reaction is feasible only at temperatures

(A) above 500 K
(B) below 500 K
(C) above 5000 K
(D) at all temperatures
Show Answer

Correct: (B)

Feasibility requires ΔG = ΔH − TΔS < 0. ΔG = 0 when T = ΔH/ΔS = 50 000 / 100 = 500 K. With both values negative, ΔG is negative only below 500 K.

Q23Using standard entropies (J K^-1 mol^-1): SO2 = 248, O2 = 205, SO3 = 257, the entropy change for 2SO2(g) + O2(g) → 2SO3(g) is

(A) +187 J K^-1 mol^-1
(B) −196 J K^-1 mol^-1
(C) −187 J K^-1 mol^-1
(D) +9 J K^-1 mol^-1
Show Answer

Correct: (C)

ΔS = 2(257) − [2(248) + 205] = 514 − 701 = −187 J K^-1 mol^-1, consistent with a decrease from 3 mol of gas to 2 mol.

⏱ Kinetics & Rate Equations

Q24A reaction has a constant half-life of 40 s at a fixed temperature, whatever the starting concentration. The reaction is

(A) zero order
(B) third order
(C) second order
(D) first order
Show Answer

Correct: (D)

A constant half-life, independent of concentration, is the defining test for a first-order reaction.

Q25In rate experiments, doubling [A] doubled the rate, and doubling [B] had no effect on the rate. The rate equation is

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

Correct: (D)

Rate is proportional to [A] (first order in A) and independent of [B] (zero order in B), so rate = k[A].

Q26For a zero-order reaction, the experimentally measured rate is 4.0 × 10^-5 mol dm^-3 s^-1. The rate constant k is

(A) 4.0 × 10^-5 mol dm^-3 s^-1
(B) 4.0 × 10^-5 s^-1
(C) 4.0 × 10^-5 mol^-1 dm^3 s^-1
(D) impossible to state without the concentration
Show Answer

Correct: (A)

For a zero-order reaction, rate = k, so k equals the rate and has the same units, mol dm^-3 s^-1.

Q27A graph of ln k against 1/T has a gradient of −6000 K. The activation energy of the reaction is (R = 8.31 J K^-1 mol^-1)

(A) 7.22 kJ mol^-1
(B) 49.9 kJ mol^-1
(C) 6000 kJ mol^-1
(D) 0.722 kJ mol^-1
Show Answer

Correct: (B)

From the Arrhenius equation, gradient = −Ea/R, so Ea = 6000 × 8.31 = 49 860 J mol^-1 ≈ 49.9 kJ mol^-1.

Q28A reaction proceeds in two steps. Step 1 (slow): A + B → C. Step 2 (fast): C + B → D. The rate equation for the overall reaction is

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

Correct: (A)

The rate is determined by the slow (rate-determining) step, whose reactants are A and B, so rate = k[A][B].

⚖ Chemical Equilibria (Kc & Kp)

Q29For the esterification CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O, the expression for Kc is

(A) Kc = [ester][H2O] / ([acid][alcohol])
(B) Kc = [acid][alcohol] / ([ester][H2O])
(C) Kc = [ester] / ([acid][alcohol])
(D) Kc = [acid][alcohol] / [ester]
Show Answer

Correct: (A)

Kc is the concentrations of the products (raised to their powers) divided by the concentrations of the reactants.

Q30For 2SO2(g) + O2(g) ⇌ 2SO3(g) at equilibrium, the partial pressures are: SO3 = 40 kPa, SO2 = 20 kPa, O2 = 10 kPa. The value of Kp is

(A) 4.0 kPa^-1
(B) 0.40 kPa^-1
(C) 0.040 kPa^-1
(D) 2.5 kPa
Show Answer

Correct: (B)

Kp = p(SO3)^2 / (p(SO2)^2 × p(O2)) = 40^2 / (20^2 × 10) = 1600 / 4000 = 0.40 kPa^-1.

Q31The forward reaction of an equilibrium is exothermic. As the temperature is increased, the value of Kc will

(A) increase
(B) stay the same
(C) decrease
(D) first increase, then decrease
Show Answer

Correct: (C)

Raising the temperature shifts the equilibrium in the endothermic (reverse) direction, reducing the product concentrations, so Kc decreases.

Q32PCl5(g) ⇌ PCl3(g) + Cl2(g). Initially 2.00 mol of PCl5 is placed in a 1.00 dm^3 vessel; at equilibrium 1.20 mol of PCl5 remains. The value of Kc is

(A) 0.533 mol dm^-3
(B) 1.07 mol dm^-3
(C) 0.267 mol dm^-3
(D) 0.800 mol dm^-3
Show Answer

Correct: (A)

0.80 mol dissociates, so [PCl3] = [Cl2] = 0.80 and [PCl5] = 1.20. Kc = (0.80 × 0.80) / 1.20 = 0.64 / 1.20 = 0.533 mol dm^-3.

Q33Increasing the total pressure on a gaseous equilibrium at constant temperature will

(A) increase the value of Kp
(B) decrease the value of Kp
(C) leave Kp unchanged, though the position of equilibrium may shift
(D) always shift the equilibrium to the right
Show Answer

Correct: (C)

Kp (like Kc) depends only on temperature. Pressure may shift the position of equilibrium, but the constant itself does not change.




🔋 Redox & Electrode Potentials

Q34The oxidation state of chromium in K2Cr2O7 is

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

Correct: (D)

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

Q35When chlorine is dissolved in cold dilute sodium hydroxide, chlorine is both oxidised and reduced. This type of reaction is called

(A) disproportionation
(B) hydrolysis
(C) neutralisation
(D) esterification
Show Answer

Correct: (A)

Disproportionation is the simultaneous oxidation and reduction of the same element (here Cl(0) → Cl(−1) in NaCl and Cl(+1) in NaClO).

Q36Standard electrode potentials: Fe^2+/Fe = −0.44 V and Ag^+/Ag = +0.80 V. The EMF of the cell Fe | Fe^2+ || Ag^+ | Ag is

(A) +0.36 V
(B) +1.24 V
(C) −1.24 V
(D) +1.68 V
Show Answer

Correct: (B)

EMF = E(right) − E(left) = +0.80 − (−0.44) = +1.24 V.

Q37Given half-cell potentials, which is the strongest oxidising agent?

(A) Cl2 (Cl2/Cl^- = +1.36 V)
(B) Fe^3+ (Fe^3+/Fe^2+ = +0.77 V)
(C) Cu^2+ (Cu^2+/Cu = +0.34 V)
(D) H^+ (H^+/H2 = 0.00 V)
Show Answer

Correct: (A)

The species in the half-cell with the most positive electrode potential gains electrons most readily, so Cl2 is the strongest oxidising agent listed.

Q38Concentrated aqueous sodium chloride is electrolysed with inert electrodes. The products are

(A) sodium at the cathode and oxygen at the anode
(B) hydrogen at the cathode and chlorine at the anode
(C) sodium at the cathode and chlorine at the anode
(D) hydrogen at the cathode and oxygen at the anode
Show Answer

Correct: (B)

At the cathode, H^+ is discharged in preference to Na^+ (hydrogen forms); at the anode, the high Cl^- concentration means chloride is discharged as chlorine.

🧫 Acids, Bases & Buffers

Q39The pH of a 0.200 mol dm^-3 solution of nitric acid is

(A) 0.70
(B) 1.70
(C) 0.20
(D) 2.70
Show Answer

Correct: (A)

pH = −log(0.200) = 0.70.

Q4025.0 cm^3 of 0.100 mol dm^-3 HCl is mixed with 25.0 cm^3 of 0.0500 mol dm^-3 NaOH. The pH of the resulting solution is

(A) 1.30
(B) 2.60
(C) 0.60
(D) 1.60
Show Answer

Correct: (D)

Moles H^+ = 2.50 × 10^-3; moles OH^- = 1.25 × 10^-3. Excess H^+ = 1.25 × 10^-3 mol in 50.0 cm^3, so [H^+] = 0.0250 mol dm^-3 and pH = 1.60.

Q41A weak acid HA of concentration 0.100 mol dm^-3 has a pH of 3.00. The value of Ka for the acid is approximately

(A) 1.0 × 10^-6 mol dm^-3
(B) 1.0 × 10^-5 mol dm^-3
(C) 1.0 × 10^-3 mol dm^-3
(D) 1.0 × 10^-4 mol dm^-3
Show Answer

Correct: (B)

[H^+] = 1.0 × 10^-3. Ka = [H^+]^2 / [HA] = (1.0 × 10^-3)^2 / 0.100 = 1.0 × 10^-5 mol dm^-3.

Q42Which indicator is most suitable for the titration of ethanoic acid (a weak acid) with sodium hydroxide solution?

(A) Methyl orange
(B) Phenolphthalein
(C) No indicator works for this titration
(D) Litmus
Show Answer

Correct: (B)

The equivalence point of a weak acid-strong base titration is above pH 7 (about 8-9), so phenolphthalein, which changes colour in that range, is suitable.

Q43A 0.0100 mol dm^-3 solution of ethanoic acid has Ka = 1.74 × 10^-5 mol dm^-3. Its pH is approximately

(A) 2.00
(B) 4.76
(C) 3.38
(D) 2.87
Show Answer

Correct: (C)

[H^+] = √(Ka × c) = √(1.74 × 10^-5 × 0.0100) = √(1.74 × 10^-7) = 4.17 × 10^-4, so pH = 3.38.

⚗ Inorganic Chemistry

Q44Which statement about chlorine and the other halogens is correct?

(A) Chlorine is a weaker oxidising agent than bromine
(B) Fluorine is less reactive than chlorine
(C) Iodine will displace bromine from potassium bromide solution
(D) The oxidising power of the halogens decreases down the group
Show Answer

Correct: (D)

Oxidising power decreases down Group 7 as the atoms get larger and attract an extra electron less strongly.

Q45The solubility of the Group 2 hydroxides in water, going down the group,

(A) decreases steadily
(B) stays the same
(C) increases steadily
(D) increases, then decreases
Show Answer

Correct: (C)

Group 2 hydroxide solubility increases down the group: Mg(OH)2 is almost insoluble, whereas Ba(OH)2 is reasonably soluble. (Sulfate solubility shows the opposite trend.)

Q46When concentrated hydrochloric acid is added to aqueous copper(II) sulfate, the blue solution turns yellow-green because

(A) the copper is oxidised to Cu^3+
(B) water ligands are replaced by chloride ligands, forming [CuCl4]^2-
(C) copper metal is precipitated
(D) the solution becomes more concentrated
Show Answer

Correct: (B)

A ligand substitution occurs: [Cu(H2O)6]^2+ (blue) is converted to the tetrahedral [CuCl4]^2- complex (yellow), and the mixture appears green.

Q47When sodium hydroxide solution is added to a solution containing Fe^2+ ions, the precipitate formed is

(A) orange-brown
(B) white
(C) green
(D) blue
Show Answer

Correct: (C)

Fe^2+ ions give a green precipitate of iron(II) hydroxide with NaOH, which darkens to brown at the surface as air oxidises it to Fe(III).

Q48Which Group 2 carbonate is the most difficult to decompose on heating?

(A) magnesium carbonate
(B) barium carbonate
(C) calcium carbonate
(D) strontium carbonate
Show Answer

Correct: (B)

The thermal stability of the Group 2 carbonates increases down the group, because the larger metal cation polarises the large carbonate ion less, so barium carbonate requires the highest temperature to decompose.

🛢 Organic Chemistry I

Q49In the initiation step of the chlorination of methane, the Cl−Cl bond breaks

(A) heterolytically, forming Cl^+ and Cl^-
(B) homolytically, forming two Cl• radicals under UV light
(C) to form chlorine molecules with a double bond
(D) only when a catalyst is present
Show Answer

Correct: (B)

UV light provides the energy for homolytic fission of Cl2 into two chlorine free radicals, which start the chain reaction.

Q50Which statement about the burning of alkanes is correct?

(A) In a limited supply of oxygen, carbon monoxide and carbon may be formed
(B) Alkanes cannot burn because they are saturated
(C) Complete combustion of an alkane produces only carbon dioxide
(D) Incomplete combustion produces hydrogen
Show Answer

Correct: (A)

Incomplete combustion (limited O2) gives the toxic gas carbon monoxide and/or soot (carbon), as well as water.

Q51When ethene is shaken with bromine, the orange colour disappears because

(A) ethene reduces bromine to hydrogen bromide
(B) a substitution reaction replaces a hydrogen atom
(C) electrophilic addition across the C=C bond forms dibromoethane
(D) the bromine evaporates
Show Answer

Correct: (C)

The π bond in ethene attacks Br2 in an electrophilic addition, forming 1,2-dibromoethane; this is the standard test for a C=C double bond.

Q52Which halogenoalkane undergoes nucleophilic substitution fastest with aqueous alkali?

(A) 1-chlorobutane
(B) a primary bromoalkane
(C) a tertiary iodoalkane
(D) a tertiary chloroalkane
Show Answer

Correct: (C)

Rate increases from primary to tertiary (carbocation stability, SN1) and from chloro to iodo (weaker C−X bond), so a tertiary iodoalkane reacts fastest.

Q532-Methylpropan-2-ol is not oxidised by acidified potassium dichromate(VI) because it is

(A) a primary alcohol
(B) a secondary alcohol
(C) a tertiary alcohol with no hydrogen on the carbon carrying the −OH group
(D) an aldehyde
Show Answer

Correct: (C)

Oxidation requires removal of the hydrogen attached to the carbinol carbon; tertiary alcohols have no such hydrogen, so they resist oxidation.

Q54When 2-bromobutane is refluxed with potassium hydroxide dissolved in ethanol, the major organic product is

(A) butan-2-ol
(B) but-1-ene only
(C) but-2-ene
(D) butane
Show Answer

Correct: (C)

Ethanolic KOH gives elimination (dehydrohalogenation). Removal of H from the more substituted carbon gives but-2-ene as the major product (Saytzeff's rule); aqueous KOH would give substitution instead.

🔬 Organic Chemistry II & Analysis

Q55Brady's reagent (2,4-dinitrophenylhydrazine) gives an orange precipitate with

(A) both aldehydes and ketones
(B) ketones only
(C) aldehydes only
(D) carboxylic acids
Show Answer

Correct: (A)

2,4-DNPH reacts with the carbonyl group in both aldehydes and ketones to form orange crystalline derivatives; it distinguishes carbonyls from other functional groups but not aldehydes from ketones.

Q56Ethanoic acid can be distinguished from ethanol by adding sodium carbonate, because ethanoic acid

(A) turns the mixture blue
(B) gives effervescence of carbon dioxide
(C) forms a silver mirror
(D) has a higher boiling point
Show Answer

Correct: (B)

Carboxylic acids react with carbonates to give CO2, which fizzes and turns limewater milky; ethanol does not react.

Q57When bromine is added to phenylamine dissolved in water, the observation is

(A) no reaction
(B) the bromine is slowly decolourised with no precipitate
(C) an orange solution forms
(D) the bromine is rapidly decolourised and a white precipitate of 2,4,6-tribromophenylamine forms
Show Answer

Correct: (D)

The −NH2 group strongly activates the ring, so phenylamine undergoes electrophilic substitution at the 2, 4 and 6 positions, decolourising bromine and giving a white precipitate.

Q58In a solid amino acid such as glycine (2-aminoethanoic acid), the molecule normally exists as

(A) a neutral molecule, NH2CH2COOH
(B) a zwitterion, ^+H3NCH2COO^-
(C) a radical
(D) a polymer
Show Answer

Correct: (B)

The acidic −COOH group protonates the basic −NH2 group within the same molecule, giving a zwitterion with both a positive and a negative charge.

Q59Terylene (PET) is a polyester formed from a diol and a dicarboxylic acid. This type of polymerisation is called

(A) addition polymerisation
(B) free-radical polymerisation
(C) condensation polymerisation
(D) electrophilic substitution
Show Answer

Correct: (C)

The monomers join with the elimination of a small molecule (water) at each link, which is condensation polymerisation.

Q60A carboxylic acid can be recognised in its infrared spectrum by

(A) a sharp peak at 1720 cm^-1 only
(B) a peak at 3300 cm^-1 identical to that of an alcohol
(C) the absence of any peak above 1600 cm^-1
(D) a very broad absorption in the range 2500-3300 cm^-1 due to the O−H bond, together with a C=O peak near 1700 cm^-1
Show Answer

Correct: (D)

The O−H group of a carboxylic acid gives a characteristically very broad absorption (2500-3300 cm^-1), alongside the C=O stretch at about 1700 cm^-1.

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