Inorganic & Organic: A Level Chemistry questions and answers
10 free A Level Chemistry Inorganic & Organic questions written in the style of Cambridge Chemistry papers, each with its answer and mark scheme. 4 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Transition elements, Arenes & phenol, Amines, amides & amino acids, Polymerisation, Organic synthesis, NMR & chromatography. Work through the questions below, then generate a full Inorganic & Organic paper with a fresh set every time.
This question is about complexes of copper and platinum.
(a)
Define the term transition element.
[1]
(b)
Concentrated HCl is added to aqueous copper(II) sulfate. Describe the colour change and the change in shape of the complex.
[2]
(c)
Explain why [Cu(H₂O)₆]²⁺ is coloured.
[3]
(d)
Cisplatin, Pt(NH₃)₂Cl₂, is square planar. Explain why it shows geometrical isomerism.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
d-block element forming a stable ion with partially filled d subshell
an element that forms one or more stable ions with an incomplete d subshell B1
1
(b)
blue to yellow-green; octahedral to tetrahedral
blue to yellow / green B1 [Cu(H₂O)₆]²⁺ octahedral to [CuCl₄]²⁻ tetrahedral B1
2
(c)
d orbital splitting; absorbs visible light
ligands split the d orbitals into two energy levels M1 an electron is promoted, absorbing light of a particular frequency in the visible range M1 the complementary colour is seen M1
3
(d)
cis and trans arrangements
square planar with two pairs of different ligands M1 the like ligands can be adjacent (cis) or opposite (trans) M1
2
Question 2Structured · 2 marksA* challenge
Phenylamine is a weaker base than ethylamine.
Explain the difference in basicity.
[2]
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Part
Answer and marking points
Marks
lone pair delocalised into ring in phenylamine
lone pair on N in phenylamine is delocalised into the benzene ring B1 less available to accept a proton B1
2
Question 3Structured · 12 marksA* challenge
An iron tablet of mass 1.50 g is dissolved in dilute sulfuric acid and made up to 250 cm3. 25.0 cm3 of this solution requires 24.00 cm3 of 0.0200 mol dm−3 KMnO4 for complete reaction. MnO4− + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O
(a)
Calculate the percentage by mass of iron in the tablet.
[4]
(b)
State why no indicator is needed, and describe the colour change at the end-point.
[2]
(c)
Explain why transition metal complexes are coloured.
[3]
(d)
[Cu(H2O)6]2+ reacts with excess concentrated HCl. Give the formula, shape and colour of the new complex, and the type of reaction.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
89.3%
mol MnO4−C1 × 5 for Fe2+C1 × 10 for 250 cm3C1 89.3 A1
4
(b)
MnO4− is self-indicating; colourless to pale pink (permanent)
self-indicating B1 to (permanent) pale pink B1
2
(c)
ligands split d orbitals into two energy levels; electrons absorb light to move to higher level; complementary colour transmitted
d-orbital splitting by ligands B1 electron promotion absorbing visible light B1 remaining frequencies seen B1
An ester, C₃H₆O₂, has the proton NMR spectrum: δ 8.0 (1H, singlet), δ 4.2 (2H, quartet) and δ 1.3 (3H, triplet).
(a)
Identify the ester, explaining how the splitting patterns support your answer.
[3]
(b)
State why tetramethylsilane is used in NMR.
[1]
(c)
Explain how D₂O could be used to identify an O–H proton in an isomeric acid.
[2]
(d)
In thin-layer chromatography the ester spot moved 4.5 cm and the solvent front moved 7.5 cm. Calculate Rf.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
ethyl methanoate, HCOOCH₂CH₃
ethyl methanoate A1 HCOOCH₂CH₃ A1 CH₂ next to CH₃ gives quartet and triplet (n + 1 rule) M1
3
(b)
reference, δ = 0
reference standard, single peak at δ = 0 B1
1
(c)
O–H peak disappears
D exchanges with the O–H proton M1 so the O–H peak disappears M1
2
(d)
0.60
0.60 B1
1
Question 5Structured · 10 marks
This question is about amides and peptide hydrolysis.
(a)
Write an equation for the formation of N-methylethanamide from ethanoyl chloride and methylamine.
[2]
(b)
Explain why amides are not basic, unlike amines.
[2]
(c)
Describe the hydrolysis of a dipeptide using acid and using alkali, stating the products in each case.
[4]
(d)
Explain how the isoelectric point is used to separate amino acids by electrophoresis.
[2]
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Part
Answer and marking points
Marks
(a)
CH₃COCl + 2CH₃NH₂ → CH₃CONHCH₃ + CH₃NH₃⁺Cl⁻
correct amide B1 second methylamine (or HCl) as by-product B1
2
(b)
N lone pair delocalised into C=O; not available to accept proton
the lone pair on nitrogen is delocalised over the C=O group B1 so it is not available to accept a proton B1
2
(c)
acid: heat with HCl(aq) → amino acids as NH₃⁺ cations; alkali: heat with NaOH → carboxylate salts with NH₂
acid: heat under reflux with dilute HCl B1 products are protonated amino acids (–NH₃⁺, –COOH) B1 alkali: heat with aqueous NaOH B1 products are sodium salts (–NH₂, –COO⁻Na⁺) B1
4
(d)
at buffer pH above pI negative, moves to anode; at pI does not move
at its isoelectric point an amino acid has no net charge and does not move B1 at a pH above/below the pI it is negatively/positively charged and moves to the anode/cathode B1
2
Question 6Structured · 9 marks
A compound Y, C₄H₈O₂, is an ester. Fig. 6.1 shows its proton NMR spectrum; the number above each peak is the relative number of hydrogen atoms.
Fig. 6.1
(a)
Identify the environment responsible for each peak.
[3]
(b)
Deduce the structure of Y and name it.
[2]
(c)
Explain why a CH₂ group bonded to oxygen appears at higher δ than one bonded to C=O.
[2]
(d)
Describe how thin-layer chromatography could show whether a sample of Y contains unreacted propanoic acid.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
δ3.7 OCH₃; δ2.3 CH₂ next to C=O; δ1.1 CH₃ of ethyl
singlet at 3.7: O–CH₃ (no adjacent H) B1 quartet at 2.3: CH₂ next to C=O, adjacent CH₃ B1 triplet at 1.1: CH₃ adjacent to CH₂ B1
3
(b)
CH₃CH₂COOCH₃; methyl propanoate
CH₃CH₂COOCH₃ B1 methyl propanoate B1
2
(c)
O more electronegative directly attached; more deshielded
oxygen is directly bonded and highly electronegative B1 withdrawing electron density so the protons are more deshielded B1
2
(d)
run sample alongside pure acid and ester; extra spot at Rf of acid
spot the sample alongside pure Y and pure propanoic acid B1 a spot at the Rf of the acid shows contamination B1
2
Question 7Structured · 5 marks
Proton NMR spectroscopy is used to identify ethanol, CH₃CH₂OH.
(a)
State the number of peaks and their splitting patterns in the high-resolution spectrum (ignoring O–H coupling).
[3]
(b)
Explain why D₂O is added to identify the O–H peak.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
3 peaks: triplet, quartet, singlet
three peaks B1 CH₃ triplet, CH₂ quartet B1 OH singlet B1
3
(b)
H exchanged for D; peak disappears
the O–H proton exchanges with deuterium B1 the peak disappears B1
2
Question 8Structured · 11 marks
Nylon-6,6 is made from hexanedioic acid, HOOC(CH₂)₄COOH, and 1,6-diaminohexane, H₂N(CH₂)₆NH₂.
(a)
Draw the repeat unit of nylon-6,6.
[2]
(b)
Name the type of polymerisation and the small molecule eliminated.
[2]
(c)
The repeat unit has Mr = 226. A sample of the polymer has a mean Mr of 22600. Calculate the mean number of repeat units in each chain.
[1]
(d)
A laboratory preparation uses hexanedioyl dichloride instead of hexanedioic acid. Suggest why, and name the other product.
[2]
(e)
Nylon is hydrolysed by hot aqueous acid. Give the products of acid hydrolysis.
[2]
(f)
Explain why nylon fibres are strong, in terms of intermolecular forces between chains.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
–NH(CH₂)₆NHCO(CH₂)₄CO–
amide link –CONH– shown B1 complete repeat unit –NH(CH₂)₆NHCO(CH₂)₄CO– with continuation bonds B1
2
(b)
condensation; water
condensation polymerisation B1 water B1
2
(c)
100
22600 ÷ 226 = 100 B1
1
(d)
acyl chloride more reactive, room temperature; HCl
acyl chlorides are much more reactive / react at room temperature B1 hydrogen chloride B1
2
(e)
hexanedioic acid; 1,6-diaminohexane as ammonium salt
hexanedioic acid B1 the diamine as its salt, ⁺H₃N(CH₂)₆NH₃⁺ (allow 1,6-diaminohexane) B1
2
(f)
hydrogen bonds between N–H and C=O of neighbouring chains
hydrogen bonds form between N–H and C=O groups B1 on neighbouring chains, holding chains together B1
2
Question 9Structured · 11 marks
Table 9.1 shows the pKa values of the conjugate acids of three bases.
base
ammonia
ethylamine
phenylamine
pKa of conjugate acid
9.25
10.8
4.6
Table 9.1
(a)
Use Table 9.1 to put the three bases in order of base strength, strongest first.
[1]
(b)
Explain why ethylamine is a stronger base than ammonia.
[2]
(c)
Explain why phenylamine is a much weaker base than ammonia.
[2]
(d)
3.72 g of phenylamine reacts with excess ethanoyl chloride. Write the equation and calculate the mass of N-phenylethanamide formed if the yield is 72%. (Mr: phenylamine 93; N-phenylethanamide 135)
[4]
(e)
Describe how phenylamine is made from nitrobenzene.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
ethylamine > ammonia > phenylamine
ethylamine > ammonia > phenylamine B1
1
(b)
ethyl group electron-donating; N lone pair more available
the ethyl group is electron-donating (positive inductive effect) B1 increasing the electron density on N, so the lone pair accepts a proton more readily B1
2
(c)
lone pair delocalised into ring; less available
the lone pair on N is delocalised into the benzene ring B1 so it is less available to accept a proton B1
tin and concentrated hydrochloric acid, heated under reflux B1 then add NaOH(aq) to release the free amine B1
2
Question 10Structured · 12 marks
This question is about proteins and polyamides.
(a)
Describe the primary, secondary and tertiary structure of a protein.
[4]
(b)
Explain why heating a protein changes its shape but not its primary structure.
[2]
(c)
Compare the structure of a protein with that of nylon-6,6.
[2]
(d)
Explain why poly(lactic acid) is described as biodegradable and why this is an advantage.
[2]
(e)
Draw the repeat unit of the polymer formed from 2-hydroxypropanoic acid, CH₃CH(OH)COOH, and name the type of polymerisation.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
sequence of amino acids; H bonding α-helix / β-sheet; folding by R-group interactions
primary: sequence of amino acids joined by peptide bonds B1 secondary: α-helix or β-pleated sheet held by hydrogen bonds between C=O and N–H B1 tertiary: overall folding held by R-group interactions B1 e.g. disulfide bridges, ionic bonds, hydrogen bonds, van der Waals B1
In the 9701 syllabus, Inorganic & Organic covers Transition elements, Arenes & phenol, Amines, amides & amino acids, Polymerisation, Organic synthesis, NMR & chromatography.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in