Edexcel Chemistry (9CH0) Paper 2
Mark Scheme: Advanced Organic &
Physical | Full Answer Key & Examiner
Notes
Question 1 – total 12
(a) Define the term “ester”.
2 marks
● Organic compound formed from (1) a carboxylic acid and an alcohol (1) with loss of water /
by esterification.
(Allow “contains –COO– functional group” for 1 if both acid + alcohol origin not stated.)
(b)(i) Calculate exact mass of C₅H₁₀O₂ to 4 d.p.
1 mark
● 5×12.0000 + 10×1.0078 + 2×15.9949 = 102.0681 (allow 102.0680 – 102.0682).
(ii) Confirm consistency.
1 mark
● Observed 102.0681 matches calculated (within ±0.001), so formula confirmed.
(c) Deduce structure from ¹H NMR.
4 marks (3 for reasoning, 1 for correct structure)
● δ 1.2 t, 3 H → CH₃ adjacent to CH₂
● δ 2.3 q, 2 H → CH₂ adjacent to CH₃ (therefore ethyl group)
● δ 3.7 s, 3 H → OCH₃ (singlet, no adjacent H)
● Only 5 C: must be ethyl ethanoate (ethyl acetate) CH₃COOCH₂CH₃
Structure: CH₃COOCH₂CH₃ (displayed or skeletal) (1)
, (d)(i) Type of reaction with NaOH.
1 mark
● Base-catalysed hydrolysis / saponification (accept “hydrolysis”).
(ii) Two organic products.
2 marks
● CH₃CH₂OH (ethanol) (1)
● CH₃COO⁻Na⁺ (sodium ethanoate) (1)
(Allow ionic structure or formula.)
(e) Advantage of ethyl ethanoate vs CH₂Cl₂ solvent.
1 mark
● Less toxic / non-chlorinated / biodegradable / lower ozone depletion (any one).
Common errors
● (c) Drawing ethyl propanoate (loses 2).
● (d)(ii) Drawing free acid instead of salt (loses 1).
Question 2 – total 14
(a) Explain S 2 classification.
3 marks
● One step / concerted mechanism (1)
● Rate = k[halogenoalkane][nucleophile] (1)
● Involves inversion of configuration at chiral centre (1)
(b) Transition state for S 2.
3 marks
● Shows partial C–Br bond breaking & C–OH bond forming simultaneously (1)
● Correct δ– on Br and δ– on O (or incoming nucleophile) (1)
● Correct 5-centre geometry (trigonal bipyramidal C) (1)
(c)(i) Reagent & conditions.
1 mark
Mark Scheme: Advanced Organic &
Physical | Full Answer Key & Examiner
Notes
Question 1 – total 12
(a) Define the term “ester”.
2 marks
● Organic compound formed from (1) a carboxylic acid and an alcohol (1) with loss of water /
by esterification.
(Allow “contains –COO– functional group” for 1 if both acid + alcohol origin not stated.)
(b)(i) Calculate exact mass of C₅H₁₀O₂ to 4 d.p.
1 mark
● 5×12.0000 + 10×1.0078 + 2×15.9949 = 102.0681 (allow 102.0680 – 102.0682).
(ii) Confirm consistency.
1 mark
● Observed 102.0681 matches calculated (within ±0.001), so formula confirmed.
(c) Deduce structure from ¹H NMR.
4 marks (3 for reasoning, 1 for correct structure)
● δ 1.2 t, 3 H → CH₃ adjacent to CH₂
● δ 2.3 q, 2 H → CH₂ adjacent to CH₃ (therefore ethyl group)
● δ 3.7 s, 3 H → OCH₃ (singlet, no adjacent H)
● Only 5 C: must be ethyl ethanoate (ethyl acetate) CH₃COOCH₂CH₃
Structure: CH₃COOCH₂CH₃ (displayed or skeletal) (1)
, (d)(i) Type of reaction with NaOH.
1 mark
● Base-catalysed hydrolysis / saponification (accept “hydrolysis”).
(ii) Two organic products.
2 marks
● CH₃CH₂OH (ethanol) (1)
● CH₃COO⁻Na⁺ (sodium ethanoate) (1)
(Allow ionic structure or formula.)
(e) Advantage of ethyl ethanoate vs CH₂Cl₂ solvent.
1 mark
● Less toxic / non-chlorinated / biodegradable / lower ozone depletion (any one).
Common errors
● (c) Drawing ethyl propanoate (loses 2).
● (d)(ii) Drawing free acid instead of salt (loses 1).
Question 2 – total 14
(a) Explain S 2 classification.
3 marks
● One step / concerted mechanism (1)
● Rate = k[halogenoalkane][nucleophile] (1)
● Involves inversion of configuration at chiral centre (1)
(b) Transition state for S 2.
3 marks
● Shows partial C–Br bond breaking & C–OH bond forming simultaneously (1)
● Correct δ– on Br and δ– on O (or incoming nucleophile) (1)
● Correct 5-centre geometry (trigonal bipyramidal C) (1)
(c)(i) Reagent & conditions.
1 mark
