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Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
76 original questions across 15 selected topic areas. No account or payment required to practise.
Cambridge International AS & A Level Chemistry (9701), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.
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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Aromatic delocalisation · Amine and amide basicity · Amino-acid electrophoresis · Addition and condensation polymers
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Question 1Explain why benzene generally undergoes electrophilic substitution rather than addition, why chlorobenzene hydrolyses less readily than chloroethane, and why phenol is more acidic than ethanol. Use delocalisation in each explanation; no experimental reagents or procedures are requested.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
Explain why benzene generally undergoes electrophilic substitution rather than addition, why chlorobenzene hydrolyses less readily than chloroethane, and why phenol is more acidic than ethanol. Use delocalisation in each explanation; no experimental reagents or procedures are requested.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked three distinct delocalisation arguments against aromatic structure, aryl-halide reactivity and phenol acidity outcomes; no conflation of bond polarity with conjugate-base stability. This is conceptual placement, not completion of the entire referenced subsection.
Distinguish the delocalised aromatic system from an isolated alkene pi bond. Electrophilic substitution loses a ring hydrogen and can regain aromaticity, rather than keeping an addition product.
For hydrolysis compare the C-Cl bonding environments. For acidity compare the ions after deprotonation: greater stabilisation of phenoxide favours phenol ionisation relative to ethanol.
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Substitution can restore the aromatic delocalised pi system, whereas addition would remove aromatic stabilisation.
Chlorine lone-pair interaction with the ring gives the aryl C-Cl bond partial double-bond character, making its cleavage harder than in chloroethane.
The phenoxide negative charge can be delocalised into the ring, stabilising phenoxide relative to a localised ethoxide ion.
The comparison concerns conjugate-base stability, not the false claim that ethanol cannot lose a proton at all.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2Compare the basicity of aqueous ethylamine, ammonia and phenylamine qualitatively. Explain why an amide is much less basic than an amine. For an amino acid whose isoelectric pH is 6.0, predict its net charge and migration direction in electrophoresis at pH 2.0 and pH 10.0; assume these values lie below and above its relevant ionisation ranges.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
Compare the basicity of aqueous ethylamine, ammonia and phenylamine qualitatively. Explain why an amide is much less basic than an amine. For an amino acid whose isoelectric pH is 6.0, predict its net charge and migration direction in electrophoresis at pH 2.0 and pH 10.0; assume these values lie below and above its relevant ionisation ranges.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Verified aqueous qualitative ranking, amide resonance and electrophoretic sign rather than invented exact amino-acid pKa/charge magnitude. This is conceptual placement, not completion of the entire referenced subsection.
Basicity concerns acceptance of a proton by a lone pair; possessing nitrogen alone is insufficient. The carbonyl in an amide and aromatic ring in phenylamine each reduce lone-pair availability, but they are different chemical contexts.
At pH well below the isoelectric value, protonation dominates and the ion is attracted to the cathode; above it, deprotonation gives attraction to the anode. The given assumption avoids pretending that isoelectric pH alone gives exact charge magnitudes.
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Ethylamine is more basic than ammonia, which is more basic than phenylamine in this comparison.
An alkyl group increases availability of the nitrogen lone pair, whereas phenylamine delocalises that pair into the ring.
An amide delocalises the nitrogen lone pair with the carbonyl, reducing its availability for proton binding.
At pH 2.0 the amino acid is net positive and moves towards the negative electrode; at pH 10.0 it is net negative and moves towards the positive electrode.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3Compare polymers formed from ethene and from HOCH2CH2OH with HOOCCH2CH2COOH. Write a repeat-unit representation for each, identify polymerisation type and linkage, and explain why their hydrolytic degradability differs. No synthesis procedure is requested.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
Compare polymers formed from ethene and from HOCH2CH2OH with HOOCCH2CH2COOH. Write a repeat-unit representation for each, identify polymerisation type and linkage, and explain why their hydrolytic degradability differs. No synthesis procedure is requested.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Counted all six carbons of the polyester repeat and both ester boundaries; distinguished chain repeat notation from finite-chain water accounting and from guaranteed environmental biodegradation. This is conceptual placement, not completion of the entire referenced subsection.
Open each ethene C=C to obtain a saturated carbon chain. For the polyester, retain both two-carbon diol and the entire four-carbon diacid residue in the displayed repeat.
Condensation joins hydroxyl and carboxyl functional groups. Water loss is associated with forming ester links; do not state an exact water count for an unspecified finite-chain length or end-group convention.
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Ethene gives [-CH2-CH2-]n by addition polymerisation with no small-molecule elimination.
The diol and diacid give [-O-CH2-CH2-O-C(=O)-CH2-CH2-C(=O)-]n, a polyester.
Condensation forms ester links with water elimination during chain formation.
Ester links can undergo hydrolysis, whereas the polyethene carbon backbone lacks these hydrolysable links; this does not imply rapid degradation of every polyester in every environment.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
