2025 OCR AS Level Chemistry A - H032 模擬試題連答案詳解

CaCO\(_3\) + 2HCl \(\rightarrow\) CaCl\(_2\) + H\(_2\)O + CO\(_2\). \(n(\text{CaCO}_3) = 2.50 / 100.1 = 0.0250\) mol. 1:1 ratio so \(n(\text{CO}_2) = 0.0250\) mol. Volume = \(0.0250 \times 24.0 = 0.600\) dm\(^3\) = 600 cm\(^3\).

B (1 mark). Method: n = m/M = 0.0250 mol; 1:1 mole ratio; V = n x 24.0. Common errors: A (halving, wrong ratio), C (doubling using 2HCl ratio), D (dm3/cm3 conversion error).

\(q = mc\Delta T = 50.0 \times 4.18 \times 6.0 = 1254\) J = 1.254 kJ. Per mole: 1.254 / 0.0100 = 125 kJ mol\(^{-1}\). The temperature rises so the process is exothermic and \(\Delta H\) is negative: \(-125\) kJ mol\(^{-1}\).

C (1 mark). Method: q = mc\(\Delta\)T; divide by moles; correct sign for exothermic. Common errors: A (wrong sign), B and D (power of ten errors).

BF\(_3\) is trigonal planar with three identical polar B-F bonds at 120 degrees, so the dipoles cancel. NH\(_3\) (pyramidal) and H\(_2\)O (non-linear) have lone pairs and do not cancel; CHCl\(_3\) is tetrahedral with unlike bonds so does not cancel.

D (1 mark). Key idea: symmetrical shape with identical polar bonds gives dipoles that cancel.

Electrophilic addition of HBr to CH\(_2\)=CHCH\(_2\)CH\(_3\). H adds to the carbon with more hydrogens, giving the more stable secondary carbocation CH\(_3\)CH\(^+\)CH\(_2\)CH\(_3\), then Br\(^-\) attacks to give 2-bromobutane (Markownikoff).

A (1 mark). Secondary carbocation is more stable than primary so is the major route. B is the minor product; C needs Br2; D has a rearranged carbon skeleton.

Propanal has a lower boiling point than propan-1-ol because it has no hydrogen bonding. Distilling it off as it forms removes it from the oxidising mixture, so it is not oxidised further to propanoic acid.

C (1 mark). A, B and D are not the purpose of distillation in this preparation.

A broad absorption at 2500-3300 cm\(^{-1}\) is the O-H of a carboxylic acid and 1710 cm\(^{-1}\) is C=O. Only propanoic acid, C\(_3\)H\(_6\)O\(_2\), has both and fits the formula; propan-1-ol, propanal and propanone have different formulae and lack one of the absorptions.

D (1 mark). Both the broad O-H and the C=O absorptions are needed.

The forward reaction is endothermic, so raising the temperature shifts the position to the right. Increasing pressure (B, D) favours the side with fewer gas moles (N\(_2\)O\(_4\)). A catalyst does not change the position.

A (1 mark). Endothermic direction favoured by higher temperature; fewer moles on the left means pressure increase shifts left; catalyst has no effect on position.

Down Group 2 the atomic radius and shielding increase, outweighing the greater nuclear charge, so the outer electrons are less strongly attracted and ionisation energies decrease. Strontium therefore forms Sr\(^{2+}\) more easily and is more reactive.

B (1 mark). A and C are the reverse of the trend; D is incorrect because ionic radius increases down the group.

H\(_2\)SO\(_4\) + 2NaOH → Na\(_2\)SO\(_4\) + 2H\(_2\)O. n(H\(_2\)SO\(_4\)) = 0.0250 × 0.100 = 2.50 × 10\(^{-3}\) mol. n(NaOH) = 5.00 × 10\(^{-3}\) mol. V = 5.00 × 10\(^{-3}\) / 0.200 = 0.0250 dm\(^3\) = 25.0 cm\(^3\). Answer B.

B (1 mark). Method: use of 1:2 mole ratio and n = cV. Common errors: ignoring the 1:2 ratio gives 12.5 cm\(^3\) (A); using 0.200 as acid concentration gives other distractors. Award the mark for B only.

Bonds broken: 436 + 243 = 679 kJ. Bonds made: 2 × 432 = 864 kJ. \(\Delta H\) = broken − made = 679 − 864 = −185 kJ mol\(^{-1}\). Answer C.

C (1 mark). Method: energy in − energy out. Error: reversing the subtraction gives +185 (A); forgetting the factor of 2 for HCl gives −247 or +247 (B/D).

BF\(_3\) is trigonal planar with three identical polar B–F bonds at 120°, so the dipoles cancel. NH\(_3\) (pyramidal) and H\(_2\)O (non-linear) have lone pairs and are polar; CHCl\(_3\) is tetrahedral but with different atoms attached, so the dipoles do not cancel. Answer C.

C (1 mark). Key idea: symmetrical shape causes bond dipoles to cancel. Distractors are asymmetric molecules with an overall dipole.

(CH\(_3\))\(_2\)C=CHCH\(_3\) is attacked by H\(^+\). Adding H to the CH carbon gives a tertiary carbocation (CH\(_3\))\(_2\)C\(^+\)CH\(_2\)CH\(_3\), more stable than the secondary alternative. Br\(^-\) attacks to give (CH\(_3\))\(_2\)CBrCH\(_2\)CH\(_3\), 2-bromo-2-methylbutane. Answer A.

A (1 mark). Key idea: tertiary carbocation favoured (Markownikoff). B is the minor product via the secondary carbocation; C and D would need a primary carbocation or a rearrangement and are not formed.

The 1715 cm\(^{-1}\) peak indicates C=O. The absence of a broad O–H absorption excludes an alcohol or carboxylic acid. Propanone, CH\(_3\)COCH\(_3\), is C\(_3\)H\(_6\)O with a C=O and no O–H. Propanoic acid is C\(_3\)H\(_6\)O\(_2\) and also has O–H. Answer D.

D (1 mark). Key idea: C=O present, O–H absent. Alcohols A and B would show a broad O–H and no C=O; C has the wrong formula and a broad O–H.

The forward reaction is endothermic, so raising the temperature shifts the position right. There are more moles of gas on the right (2 versus 1), so lowering the pressure favours the forward reaction. Answer B.

B (1 mark). Both effects must be correct: higher temperature (endothermic direction) and lower pressure (side with more gas moles). Catalysts would not alter position.

In Cl\(_2\) chlorine is 0. In NaCl it is −1 (reduced) and in NaClO it is +1 (oxidised). This is disproportionation. Sodium stays +1 throughout. Answer D.

D (1 mark). Key idea: the same element is both oxidised and reduced (disproportionation). Sodium's oxidation number does not change, so C is wrong.

Propanal has no O–H group so lacks hydrogen bonding between its molecules and boils at a lower temperature than propan-1-ol. It distils off as formed, preventing further oxidation to propanoic acid by the oxidising agent. Answer C.

C (1 mark). Key ideas: lower boiling point due to no hydrogen bonding; avoids over-oxidation. A is factually wrong; B and D give irrelevant reasons.

\(\mathrm{CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2}\), so 1 mol CaCO3 gives 1 mol CO2. \(n(\mathrm{CaCO_3}) = 1.50/100.1 = 0.01499\) mol. \(V = 0.01499 \times 24.0 = 0.360\ \mathrm{dm^3} = 360\ \mathrm{cm^3}\). Answer C.

C (1 mark). Method: n = m/M then multiply by 24.0 dm3 mol-1 with 1:1 ratio. Common errors: A uses a 1:2 ratio inverted; B doubles using the 2HCl coefficient; D is a power-of-ten error in unit conversion.

\(q = mc\Delta T = 50.0 \times 4.18 \times 6.5 = 1358.5\) J = 1.36 kJ. Per mole: \(1.3585/0.0100 = 135.85 \approx 136\) kJ per mol. The temperature rises, so the process is exothermic and \(\Delta H = -136\ \mathrm{kJ\ mol^{-1}}\). Answer B.

B (1 mark). Method: q = mcΔT; divide by moles; sign negative as temperature increases. Errors: A omits division by moles; C has the wrong sign; D is a power-of-ten error.

CCl4 is tetrahedral and symmetrical, so the four polar C–Cl bond dipoles cancel and there is no overall dipole. NH3 (pyramidal), H2O (non-linear) and CH2Cl2 (tetrahedral but unsymmetrical) all have overall dipoles. Answer D.

D (1 mark). Reasoning: symmetrical shape leads to cancelling dipoles. A, B and C are all polar overall because their shapes are not symmetrical.

\(\mathrm{(CH_3)_2C{=}CHCH_3}\): H+ adds to the CH carbon, giving the tertiary carbocation \(\mathrm{(CH_3)_2C^+CH_2CH_3}\), which is more stable than the secondary carbocation formed if H+ added to the other carbon. Br− attacks the tertiary carbocation to give 2-bromo-2-methylbutane. Answer A.

A (1 mark). Markownikoff: tertiary carbocation favoured over secondary. B is the minor product via the secondary carbocation; C and D would need primary carbocations or rearrangement and are not formed.

\(q = mc\Delta T = 50.0 \times 4.18 \times 6.50 = 1358.5\) J = 1.3585 kJ. \(\Delta H = -q / n = -1.3585 \div 0.0250 = -54.3\) kJ mol\(^{-1}\). The sign is negative because the temperature rises (exothermic).

FIRST CHECK THE ANSWER ON THE ANSWER LINE. If answer = -54.3 award 2 marks.

Mark 1: q = 50.0 x 4.18 x 6.50 = 1358.5 J (1.36 kJ).
Mark 2: \(\Delta H\) = -1358.5 / 1000 / 0.0250 = -54.3 kJ mol-1 (sign must be negative).

ALLOW -54.34 or -54 for the final mark if the working is correct. If the sign is positive or missing, award 1 mark maximum (ECF). Common error: answer -54300 (J not converted) = 1 mark.

NH\(_3\) has three bonding pairs and one lone pair around nitrogen, so the shape is pyramidal. Lone pair–bonding pair repulsion is greater than bonding pair–bonding pair repulsion, so the bonding pairs are pushed closer together (107\(^\circ\)). CH\(_4\) has four bonding pairs only, giving 109.5\(^\circ\).

Mark 1: pyramidal (ALLOW trigonal pyramidal).
Mark 2: lone pair repels (bonding pairs) more than bonding pairs repel each other / lone pair repulsion is greater (ORA for CH4 having no lone pair).

IGNORE 'lone pair repels more' without reference to bonding pairs only if the comparison is clear. DO NOT ALLOW 'tetrahedral' or 'non-linear' for the shape.

Attack of H\(^+\) on the double bond can give a primary carbocation, CH\(_3\)CH\(_2\)CH\(_2\)\(^+\), or a secondary carbocation, CH\(_3\)CH\(^+\)CH\(_3\). The secondary carbocation is more stable, so it forms in greater amounts. Bromide then attacks it to give 2-bromopropane.

Mark 1: two possible carbocations / intermediates form (primary and secondary) – ALLOW by structure.
Mark 2: the secondary carbocation is more stable (than the primary) so forms more / is the major route, giving 2-bromopropane.

ALLOW Markownikoff's rule stated as an explanation only if linked to carbocation stability. DO NOT ALLOW 'more stable product' without reference to the intermediate.

In reflux, a condenser is fitted vertically above the flask. Volatile liquids (propan-1-ol and the intermediate propanal) vaporise, condense and fall back into the flask. This lets the mixture be heated for a long time without loss, so oxidation continues to the carboxylic acid.

Mark 1: vapours / volatile substances condense (in the condenser) and return to the flask / prevents escape of vapour.
Mark 2: allows prolonged heating so the alcohol/aldehyde is fully oxidised to the carboxylic acid (AW).

IGNORE references to safety only. DO NOT ALLOW 'to keep the temperature constant'.

A very broad absorption at 2500–3300 cm\(^{-1}\) is the O–H stretch of a carboxylic acid, and the strong peak near 1715 cm\(^{-1}\) is the C=O stretch. Both together indicate a carboxylic acid.

Mark 1: carboxylic acid (-COOH).
Mark 2: broad peak is O–H and the 1715 cm-1 peak is C=O (both bonds correctly assigned).

DO NOT ALLOW alcohol or ketone/aldehyde alone for the first mark. ALLOW 'acid' only if the carboxylic nature is clear.

(a) Products over reactants, each concentration raised to the power of its coefficient. (b) There are 2 mol of gas on the right and 1 mol on the left. Increasing pressure favours the side with fewer gas moles, so the position shifts to the left.

(a) 1 mark: Kc = [NO2]^2 / [N2O4]. Square brackets required; DO NOT ALLOW round brackets.
(b) 1 mark: shifts to the left / towards N2O4 because fewer moles (molecules) of gas on the left. Both the direction and reason are needed.

(a) Strontium forms the hydroxide and hydrogen. (b) Down the group, atomic radius and shielding increase, so the attraction between nucleus and outer electrons decreases. The first and second ionisation energies are lower, so Sr loses two electrons more easily than Mg.

(a) 1 mark: Sr + 2H2O \(\rightarrow\) Sr(OH)2 + H2. State symbols are normally ignored; ALLOW Sr(OH)2(s).
(b) 1 mark: Sr has a larger atomic radius / more shielding, so the outer electrons are less attracted and it is easier to lose two electrons / lower ionisation energies (ORA for Mg).

DO NOT ALLOW 'Sr has more shells' without reference to attraction or ease of electron loss.

In the first propagation step a chlorine radical removes a hydrogen atom from methane, forming HCl and a methyl radical. In the second, the methyl radical reacts with Cl\(_2\) to give chloromethane and regenerate a chlorine radical.

Mark 1: CH4 + Cl\(\bullet\) \(\rightarrow\) \(\bullet\)CH3 + HCl.
Mark 2: \(\bullet\)CH3 + Cl2 \(\rightarrow\) CH3Cl + Cl\(\bullet\).

Radical dots must be shown on the radicals. DO NOT ALLOW a dot on the wrong atom (e.g. CH3\(\bullet\) is allowed only if clearly on carbon). Initiation or termination equations earn no marks.

n(Mg) = 2.40 / 24.3 = 0.0988 mol. Mole ratio Mg : H2 = 1 : 1, so n(H2) = 0.0988 mol. Volume = 0.0988 x 24.0 = 2.37 dm^3.

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 2.37 (dm^3) scores 2 marks.
1 mark: n(Mg) = 2.40 / 24.3 = 0.0988 mol (method).
1 mark: V = 0.0988 x 24.0 = 2.37 dm^3 (accuracy, 3 s.f.).
ALLOW ECF from an incorrect n(Mg). DO NOT ALLOW use of a 2 : 1 ratio (loses accuracy mark only).

q = mcΔT = 50.0 x 4.18 x 6.5 = 1358.5 J = 1360 J to 3 s.f.

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 1360 or 1.36 x 10^3 scores 2 marks.
1 mark: correct use of q = mcΔT with 50.0, 4.18 and 6.5 (1358.5 seen).
1 mark: answer correctly rounded to 3 s.f. = 1360 J.
ALLOW 1.36 kJ if units stated. DO NOT ALLOW 1358.5 as the final answer (not 3 s.f.) for the second mark.

NH3 has three bonding pairs and one lone pair around N, giving a pyramidal shape. A lone pair repels more strongly than a bonding pair, pushing the N-H bonds closer together, so the angle is about 107 degrees, smaller than the 109.5 degrees in CH4 where all four pairs are bonding.

1 mark: pyramidal AND 107 degrees (ALLOW 106-107.5).
1 mark: lone pair repels (bonding pairs) more than bonding pairs repel each other / more repulsion from lone pair, so angle smaller than 109.5 degrees in CH4 (ORA).
ALLOW 'trigonal pyramidal'. IGNORE 'lone pair has more electrons'.

The H+ adds to the terminal carbon, giving a secondary carbocation CH3CH(+)CH3, which is more stable than the primary carbocation. Bromide then attacks the secondary carbocation to give 2-bromopropane as the major product.

1 mark: 2-bromopropane.
1 mark: formed via a secondary carbocation / intermediate which is more stable than the primary carbocation (that would give 1-bromopropane) (AW).
DO NOT ALLOW 'carbon radical'. IGNORE reference to Markownikoff's rule alone.

Reflux has a vertical condenser which condenses vapours and returns them to the reaction flask, so volatile ethanol and the intermediate ethanal do not escape. Heating for a long time with excess oxidising agent oxidises ethanol through ethanal to ethanoic acid.

1 mark: vapours condense and return to the flask / prevents loss of volatile reactants or products (while heating).
1 mark: allows complete oxidation of ethanol via ethanal to ethanoic acid (carboxylic acid) / ethanal not distilled off.
IGNORE 'to prevent explosions'. DO NOT ALLOW 'to stop gases escaping' without reference to condensing.

A broad O-H absorption at 2500-3300 cm^-1 (hydrogen-bonded acid O-H) together with C=O near 1715 cm^-1 indicates a carboxylic acid.

1 mark: O–H (broad band) and C=O (at 1715 cm^-1) both identified.
1 mark: carboxylic acid.
ALLOW 'carboxyl'. DO NOT ALLOW alcohol or ketone/aldehyde alone. ECF not available for the second mark if only one bond is correct unless the group named is consistent with it.

(a) Kc is products over reactants, each concentration raised to its stoichiometric power. (b) A catalyst speeds the forward and reverse reactions equally, so the position and Kc are unchanged.

(a) 1 mark: \(K_c = \frac{[SO_3]^2}{[SO_2]^2[O_2]}\). DO NOT ALLOW round brackets or missing squares; ALLOW omission of units.
(b) 1 mark: no effect on position AND no effect on Kc. Both needed.

Down Group 2 atomic radius and shielding increase, so the attraction of the nucleus for the outer 6s electrons in Ba is weaker than for the 3s electrons in Mg. Less energy is needed to remove two electrons, so Ba forms Ba2+ more easily and is more reactive.

1 mark: Ba has larger atomic radius / more shielding so outer electrons are further from nucleus / less attracted (ORA).
1 mark: lower first and second ionisation energies (less energy to remove two electrons) so Ba is oxidised more easily (ORA).
IGNORE 'nuclear charge decreases'. DO NOT ALLOW 'fewer electrons in outer shell'. Comparison needed throughout.

\(n(\mathrm{CaCO_3}) = 1.50 / 100.1 = 0.01499\) mol. The equation shows 1 mol \(\mathrm{CaCO_3}\) gives 1 mol \(\mathrm{CO_2}\), so \(n(\mathrm{CO_2}) = 0.01499\) mol. Volume = \(0.01499 \times 24.0 = 0.360\ \mathrm{dm^3}\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 0.360 (dm^3) scores 2 marks.

Mark 1 (method): amount of CaCO3 = 1.50 / 100.1 = 0.01499 mol (ALLOW 0.0150) [1]
Mark 2 (accuracy): volume = 0.01499 x 24.0 = 0.360 dm^3 [1]

ALLOW 360 cm^3 if units are clearly given. ALLOW ECF from an incorrect amount of CaCO3. DO NOT ALLOW answers not given to 3 significant figures for the second mark (0.36 is acceptable only if 0.360 shown in working). Common error: using 22.4 dm^3 mol-1 gives 0.336 dm^3, 1 mark.

\(q = mc\Delta T = 100.0 \times 4.18 \times 7.2 = 3009.6\) J = 3.0096 kJ. Per mole: \(3.0096 / 0.0200 = 150.5\) kJ mol-1. The temperature rises, so the process is exothermic and \(\Delta H = -150\ \mathrm{kJ\,mol^{-1}}\) (3 s.f.).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: -150 (kJ mol-1) scores 2 marks.

Mark 1 (method): q = 100.0 x 4.18 x 7.2 = 3009.6 J (3.01 kJ) [1]
Mark 2 (accuracy): divide by 0.0200 and give negative sign: -150 kJ mol-1 [1]

ALLOW -151 or -150.5 if working shown. ALLOW ECF for the sign mark only if the magnitude is wrong but the sign is negative. DO NOT ALLOW +150 (1 mark max). Common error: answer of -150000 with units kJ mol-1 (J/kJ confusion) scores 1 mark.

Both molecules have four electron pairs around the central atom, arranged tetrahedrally (\(109.5^\circ\) in \(\mathrm{CH_4}\), with four bonding pairs). In \(\mathrm{NH_3}\) one pair is a lone pair. A lone pair repels more strongly than a bonding pair, so the three N-H bonding pairs are pushed closer together, giving \(107^\circ\).

Mark 1: NH3 has a lone pair (of electrons) on nitrogen / CH4 has no lone pairs (only bonding pairs) [1]
Mark 2: lone pair repels (bonding pairs) more than bonding pair repels bonding pair / lone pair-bonding pair repulsion greater than bonding pair-bonding pair repulsion, so bond angle is smaller [1]

ALLOW AW for 'more strongly'. ALLOW ORA for CH4. IGNORE references to shape names alone. DO NOT ALLOW 'lone pair repels the hydrogen atoms' or 'lone pair repels the bonds' without reference to pairs of electrons (second mark).

The H+ end of HBr adds to the double bond. Adding H to C1 gives a secondary carbocation \(\mathrm{CH_3CH_2CH^+CH_3}\); adding H to C2 gives a primary carbocation. The secondary carbocation is more stable, so it forms in greater amount and Br- attacks it to give 2-bromobutane, \(\mathrm{CH_3CH_2CHBrCH_3}\).

Mark 1: 2-bromobutane [1] (ALLOW correct structural or skeletal formula, CH3CH2CHBrCH3)
Mark 2: formed via the secondary carbocation (intermediate), which is more stable than the primary carbocation [1]

ALLOW ORA for primary carbocation being less stable. ALLOW Markownikoff's rule stated correctly as an explanation. IGNORE 'more stable product' without reference to the carbocation. DO NOT ALLOW 1-bromobutane for the first mark. ECF: if 1-bromobutane is named, no ECF for the second mark.

n(NaOH) = 0.100 × 21.40/1000 = 2.140 × 10\(^{-3}\) mol.
n(H2SO4) = 2.140 × 10\(^{-3}\) / 2 = 1.070 × 10\(^{-3}\) mol.
c(H2SO4) = 1.070 × 10\(^{-3}\) / (25.0/1000) = 0.0428 mol dm\(^{-3}\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 0.0428 gains 3 marks.
1 (method): n(NaOH) = 2.140 × 10^-3 mol.
2 (method): n(H2SO4) = half of n(NaOH) = 1.070 × 10^-3 mol (ECF from mark 1).
3 (accuracy): concentration = 0.0428 mol dm^-3 to 3 s.f. (ECF).
ALLOW 0.04280. Using 1:1 ratio gives 0.0856 and scores 2 marks (marks 1 and 3 by ECF). Unit is not penalised if omitted, but a wrong unit loses the final mark.

n(Mg) = 0.250 / 24.3 = 0.01029 mol, so n(H2) = 0.01029 mol.
V = nRT/p = (0.01029 × 8.314 × 293) / 101000 = 2.48 × 10\(^{-4}\) m\(^3\).
V = 248 cm\(^3\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 248 gains 3 marks.
1 (method): n(H2) = n(Mg) = 0.250/24.3 = 0.01029 mol.
2 (method): V = nRT/p with p in Pa (101000) and T = 293 K, giving 2.48 × 10^-4 m^3.
3 (accuracy): 248 cm^3 (conversion m^3 to cm^3 × 10^6).
ALLOW 247 to 248. 2.48 × 10^-4 m^3 scores 2 marks only if the conversion to cm^3 is not made. Failure to convert kPa to Pa gives an answer 1000 times too large: 2 marks maximum.

N2O4 reacted = 0.240 / 2 = 0.120 mol, so N2O4 at equilibrium = 0.500 − 0.120 = 0.380 mol.
[N2O4] = 0.380/2.00 = 0.190 mol dm\(^{-3}\); [NO2] = 0.240/2.00 = 0.120 mol dm\(^{-3}\).
\(K_c = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = \dfrac{0.120^2}{0.190} = 0.0758\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 0.0758 gains 3 marks.
1 (method): equilibrium amount of N2O4 = 0.380 mol.
2 (method): both equilibrium concentrations calculated (0.190 and 0.120 mol dm^-3) by dividing by 2.00 dm^3.
3 (accuracy): Kc = [NO2]^2/[N2O4] = 0.0758 (ECF).
ALLOW 0.076. Omitting the square gives 0.632 and loses the final mark. Using moles instead of concentrations gives 0.152 and scores 1 mark. ALLOW units mol dm^-3 if given.

Formation: 3C(s) + 4H2(g) + ½O2(g) → C3H7OH(l).
\(\Delta_f H^\ominus\) = Σ\(\Delta_c H\)(reactants) − Σ\(\Delta_c H\)(products)
= [3(−394) + 4(−286)] − (−2021)
= (−1182 − 1144) + 2021 = −305 kJ mol\(^{-1}\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: -305 gains 3 marks.
1 (method): correct multipliers, 3 × (-394) and 4 × (-286).
2 (method): sum of reactants' combustion enthalpies minus that of propan-1-ol = (-2326) - (-2021).
3 (accuracy): -305 kJ mol^-1 including sign.
ALLOW a correct Hess cycle drawn with arrows as method. Answer of +305 (wrong sign) scores 2 marks. Answer of -4347 (wrong subtraction order) scores 1 mark. Answer -2326 + 2021 with wrong multipliers scores at most 1 mark.

Molar mass of \(\mathrm{CaCO_3}\) = 40.1 + 12.0 + 3(16.0) = 100.1 g mol\(^{-1}\).

\(n(\mathrm{CaCO_3}) = 2.50 / 100.1 = 0.02498\) mol. The equation is 1:1, so \(n(\mathrm{CO_2}) = 0.02498\) mol.

Convert the pressure: 101 kPa = 101000 Pa.

\(V = nRT/p = (0.02498 \times 8.314 \times 303) / 101000 = 6.23 \times 10^{-4}\ \mathrm{m^3}\)

\(6.23 \times 10^{-4}\ \mathrm{m^3} \times 10^6 = 623\ \mathrm{cm^3}\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE. If the answer is 623 (cm^3) award 3 marks.

Method and accuracy marks:
- (1) Method: \(n(\mathrm{CO_2}) = n(\mathrm{CaCO_3}) = 2.50/100.1 = 0.0250\) mol (0.02498 to more s.f.). ALLOW 2.50/100 = 0.025.
- (1) Method: correct rearrangement and substitution into \(V = nRT/p\) with pressure in Pa (101000) and T = 303 K. ALLOW ECF from the wrong n.
- (1) Accuracy: 623 cm^3 (volume correctly converted from m^3 by x 10^6) and given to 3 s.f. ALLOW 6.23 x 10^2. ALLOW 624 if n = 0.0250 is used (rounded value).

Common errors: 0.623 or 6.23 x 10^-4 with no conversion to cm^3 or unit stated as cm^3 = 2 marks. Using 101 instead of 101000 (answer 6.23 x 10^5 cm^3) = 2 marks. Using 24.0 dm^3 mol^-1 (answer 599 cm^3) = 1 mark. DO NOT ALLOW answers to fewer than 3 s.f. for the final mark.

The C=C double bond is electron rich and the H-Br bond is polar, H\(^{\delta+}\)-Br\(^{\delta-}\), so HBr acts as an electrophile (electron pair acceptor).

Step 1: a curly arrow from the C=C double bond to the \(\delta+\) hydrogen; a curly arrow from the H-Br bond to Br. The hydrogen bonds to the CH2 carbon, giving the tertiary carbocation \(\mathrm{CH_3CH_2C^+(CH_3)_2}\) and Br\(^-\).

Step 2: a curly arrow from a lone pair on Br\(^-\) to the C\(^+\) forms the C-Br bond, giving \(\mathrm{CH_3CH_2CBr(CH_3)_2}\), 2-bromo-2-methylbutane.

The mechanism is electrophilic addition. Adding H to the CH2 carbon gives a tertiary carbocation. Adding H to the other carbon would give a primary carbocation, which is less stable, so the minor product 1-bromo-2-methylbutane forms in smaller amounts (Markownikoff's rule).

(a) 4 marks:
- Dipole shown on HBr, H\(^{\delta+}\) and Br\(^{\delta-}\), AND curly arrow from the C=C bond to the H\(^{\delta+}\) (1). Arrow must start from the bond, not from an atom.
- Curly arrow from the H-Br bond to Br, AND correct tertiary carbocation intermediate \(\mathrm{CH_3CH_2C^+(CH_3)_2}\) with the positive charge on the correct carbon (1). ALLOW skeletal or displayed formulae.
- Curly arrow from a lone pair (or negative charge) on Br\(^-\) to C\(^+\), with the product 2-bromo-2-methylbutane (1). ECF from a different carbocation drawn earlier.
- Mechanism named as electrophilic addition (1). DO NOT ALLOW nucleophilic addition or electrophilic substitution.

(b) 1 mark:
- 2-bromo-2-methylbutane AND the intermediate is a tertiary carbocation, which is more stable than the primary carbocation that would give the minor product (1). ALLOW AW, ORA. IGNORE explanations of why the tertiary carbocation is more stable (not required).

Total 5 marks.

\( n(\mathrm{Mg}) = 0.250 / 24.3 = 0.01029 \) mol. From the equation, 1 mol Mg gives 1 mol \(\mathrm{H_2}\), so \( n(\mathrm{H_2}) = 0.01029 \) mol. \( V = 0.01029 \times 24.0 = 0.247 \) dm\(^3\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 0.247 (dm^3) scores 1.5 marks.
0.5 mark: n(Mg) = 0.250 / 24.3 = 0.0103 mol (0.01029).
0.5 mark: 1:1 ratio, so n(H2) = n(Mg).
0.5 mark: V = 0.01029 x 24.0 = 0.247 dm^3 (3 s.f.). ALLOW 247 cm^3 with correct unit. ECF from an incorrect n(Mg). DO NOT ALLOW a final answer not to 3 s.f.

\( q = mc\Delta T = 50.0 \times 4.18 \times 15.0 = 3135 \) J = 3.135 kJ. \( n = 1.20/46.0 = 0.02609 \) mol. \( \Delta H = -3.135/0.02609 = -120 \) kJ mol\(^{-1}\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: -120 (kJ mol^-1) scores 1.5 marks.
0.5 mark: q = 50.0 x 4.18 x 15.0 = 3135 J (3.135 kJ).
0.5 mark: n(ethanol) = 1.20 / 46.0 = 0.0261 mol.
0.5 mark: q / n converted to kJ with negative sign = -120. ALLOW -120.2 or -121 (rounding). Answer of +120 loses the final mark. ECF from earlier errors.

Ammonia has three bonding pairs and one lone pair around nitrogen, giving a pyramidal shape with an angle of about 107\(^\circ\). Methane has four bonding pairs only (109.5\(^\circ\)). The lone pair repels more than bonding pairs, squeezing the bond angle.

0.5 mark: pyramidal (ALLOW trigonal pyramidal).
0.5 mark: 107 degrees (ALLOW 106-108).
0.5 mark: lone pair repels bonding pairs more than bonding pairs repel each other (AW, ORA for CH4 having no lone pair). DO NOT ALLOW 'lone pair repels more' without comparison.

HBr adds across the C=C bond by electrophilic addition. The H adds to the terminal carbon, giving the secondary carbocation \(\mathrm{CH_3CH^+CH_3}\), which is more stable than the primary carbocation \(\mathrm{CH_3CH_2CH_2^+}\). Br\(^-\) attacks it to give 2-bromopropane.

0.5 mark: electrophilic addition.
0.5 mark: 2-bromopropane.
0.5 mark: secondary carbocation (intermediate) is more stable than the primary carbocation (AW, ORA). ALLOW Markownikoff's rule stated with reference to carbocation stability. IGNORE 'more stable product' alone.

Cr2O7 2- (orange) is reduced to Cr3+ (green). Propanal, the intermediate, is more volatile than the alcohol and would distil away under distillation. Under reflux the vapours condense and return to the flask, allowing complete oxidation to propanoic acid.

0.5 mark: orange to green.
0.5 mark: vapours/volatile compounds (propanal, propan-1-ol) are condensed and returned to the flask (AW).
0.5 mark: allows the reaction to continue/complete oxidation to the carboxylic acid (AW). DO NOT ALLOW 'to stop gas escaping' without reference to condensing.

A very broad band at 2500-3300 cm\(^{-1}\) is O-H of a carboxylic acid; the strong band near 1715 cm\(^{-1}\) is C=O. Both together indicate a carboxylic acid.

0.5 mark: O-H for the broad absorption.
0.5 mark: C=O for the absorption at 1715 cm^-1.
0.5 mark: carboxylic acid. ALLOW -COOH. DO NOT ALLOW alcohol or aldehyde/ketone alone.

By le Chatelier's principle, the equilibrium shifts in the endothermic direction (forwards) to oppose the temperature rise. More \(\mathrm{NO_2}\) and less \(\mathrm{N_2O_4}\) means \(K_c\) increases.

0.5 mark: shifts to the right / towards products.
0.5 mark: forward reaction is endothermic / absorbs heat, opposing the temperature increase.
0.5 mark: Kc increases. DO NOT ALLOW 'Kc stays the same'.

Down Group 2 atomic radius and shielding increase, so the attraction between the nucleus and outer 6s electrons is weaker than for magnesium's 3s electrons. The first and second ionisation energies are lower, so barium forms Ba\(^{2+}\) more easily.

0.5 mark: barium atom has larger atomic radius / more shells and shielding than magnesium.
0.5 mark: weaker attraction between nucleus and outer electrons (ORA).
0.5 mark: lower first and second ionisation energies, so electrons lost more easily (ORA). Comparisons required.

n(Mg) = 1.20 / 24.3 = 0.04938 mol. The ratio Mg : H2 is 1 : 1, so n(H2) = 0.04938 mol. V = 0.04938 x 24.0 = 1.185 dm^3 = 1.19 dm^3 (3 s.f.).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 1.19 (dm^3) = 1.5 marks.
0.5 mark: n(Mg) = 1.20 / 24.3 = 0.0494 mol (ALLOW 0.04938).
0.5 mark: n(H2) = n(Mg) (1:1 ratio) AW.
0.5 mark: V = n x 24.0 = 1.19 dm^3 to 3 s.f. (ALLOW 1.185 if shown with a correct unrounded value; ECF from the previous points).
DO NOT ALLOW 1.2 (wrong significant figures) for the final mark.

q = mcΔT = 50.0 x 4.18 x 6.8 = 1421 J = 1.421 kJ. n(HCl) = 0.0250 x 1.00 = 0.0250 mol (equal to n(H2O) formed). ΔH = -1.421 / 0.0250 = -56.8 kJ mol^-1 (exothermic because the temperature rose).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: -56.8 (kJ mol^-1) = 1.5 marks.
0.5 mark: q = 50.0 x 4.18 x 6.8 = 1421 J (mass of 50.0 g, not 25.0 g).
0.5 mark: n = 0.0250 mol.
0.5 mark: ΔH = -56.8 kJ mol^-1 with a negative sign (ECF on q and n). ALLOW -56.9 if q rounded to 1.42 kJ.
DO NOT ALLOW a positive value for the final mark. Common error: using 25.0 g for the mass earns a maximum of 1 mark.

BF3 has three bonded pairs and no lone pairs, so the pairs spread as far apart as possible: trigonal planar, 120 degrees. NH3 has three bonded pairs and one lone pair; the lone pair repels the bonded pairs more strongly than bonded pairs repel each other, giving a pyramidal shape with angle about 107 degrees.

0.5 mark: BF3 trigonal planar AND 120 degrees; NH3 pyramidal AND 107 degrees (both shapes and angles needed). ALLOW 106-107 degrees for NH3. IGNORE 'triangular pyramid'.
0.5 mark: NH3 has a lone pair (of electrons) whereas BF3 has none / only bonded pairs.
0.5 mark: lone pair repels more than a bonded pair (ORA), so the bond angle in NH3 is smaller (AW).
DO NOT ALLOW 'lone pair repels bonds more' without the comparison of repulsion strength.

The pi bond prevents rotation about the C=C bond. Each carbon of the double bond carries two different groups (C2: H and CH3; C3: H and CH2CH3). By CIP rules the higher-priority groups are CH3 on C2 and CH2CH3 on C3; in the E isomer they lie on opposite sides of the double bond (a zigzag skeletal chain).

0.5 mark: restricted rotation about the C=C bond (due to the pi bond) AW.
0.5 mark: each carbon of the C=C is attached to two different groups/atoms.
0.5 mark: E isomer shown or described with CH3 and CH2CH3 on opposite sides of the double bond (zigzag skeletal drawing). ALLOW displayed or skeletal formula if unambiguous.
DO NOT ALLOW 'the double bond cannot rotate' alone without a reason for difference of groups for the second mark.

Acidified K2Cr2O7 is the oxidising agent. Heating under reflux keeps the aldehyde intermediate and volatile reactants in the flask until oxidation to the carboxylic acid is complete. Cr2O7^2- (orange) is reduced to Cr^3+ (green). CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O.

0.5 mark: acidified potassium dichromate(VI) (K2Cr2O7 / H2SO4) AW; ALLOW H+/Cr2O7^2-.
0.5 mark: heat under reflux (with excess oxidising agent). DO NOT ALLOW distillation.
0.5 mark: orange to green AND balanced equation CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O. Both parts are needed for this mark; ALLOW structural or displayed formulae. IGNORE state symbols.

A broad absorption at 2500-3300 cm^-1 is the O-H of a carboxylic acid and 1715 cm^-1 is C=O. A carboxylic acid group COOH has mass 45, leaving 29 for the remainder: C2H5. So X is CH3CH2COOH, propanoic acid, Mr = 74.

0.5 mark: carboxylic acid (functional group).
0.5 mark: broad absorption from O-H and sharp absorption from C=O (both bonds needed).
0.5 mark: propanoic acid / CH3CH2COOH with Mr = 74 consistent with the molecular ion. DO NOT ALLOW methyl ethanoate (no broad O-H absorption).

Kc is products over reactants raised to their stoichiometric powers. Raising the temperature favours the endothermic direction to oppose the change (le Chatelier). The forward reaction has ΔH positive, so the equilibrium shifts to the right, giving more NO2.

0.5 mark: Kc = [NO2]^2 / [N2O4] (square brackets needed; DO NOT ALLOW round brackets).
0.5 mark: equilibrium position shifts to the right / towards NO2.
0.5 mark: the forward reaction is endothermic / the system opposes the increase in temperature by absorbing heat AW.

Sr forms Sr(OH)2 and H2. Going down Group 2 the atomic radius and shielding increase, so the attraction between the nucleus and the outer 5s electrons is weaker; less energy is needed to form the 2+ ion (lower first and second ionisation energies), so Sr is more reactive.

0.5 mark: Sr(s) + 2H2O(l) -> Sr(OH)2(aq) + H2(g); ALLOW Sr(OH)2(s). State symbols may be ignored if the species are correct.
0.5 mark: strontium has a larger atomic radius / more shielding than magnesium so the outer electrons are less strongly attracted (ORA).
0.5 mark: lower first and second ionisation energies so electrons are lost more easily (ORA). DO NOT ALLOW 'Sr has fewer electrons'.

\(M_r(CaCO_3) = 40.1 + 12.0 + 3\times16.0 = 100.1\). \(n(CaCO_3) = 2.40/100.1 = 0.0240\) mol (0.02398). 1:1 ratio so \(n(CO_2)=0.02398\) mol. \(V = 0.02398 \times 24.0 = 0.575\ dm^3\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE: 0.575 (dm^3) = 1.5 marks. Allow 575 cm^3 if units given.
0.5 mark: n(CaCO3) = 2.40/100.1 = 0.02398 mol.
0.5 mark: n(CO2) = n(CaCO3) by 1:1 ratio.
0.5 mark: V = n x 24.0 = 0.575 dm^3 (3 s.f.). ECF from an incorrect n. DO NOT ALLOW wrong s.f. for the final 0.5 mark.

\(q = mc\Delta T = 50.0 \times 4.18 \times 6.5 = 1358.5\) J. \(n(H_2O) = 0.0250 \times 1.00 = 0.0250\) mol. \(\Delta H = -1358.5/0.0250 = -54340\) J mol^-1 = -54.3 kJ mol^-1.

FIRST CHECK THE ANSWER: -54.3 = 1.5 marks. Allow -54.4 if q rounded to 1360 J.
0.5 mark: q = 50.0 x 4.18 x 6.5 = 1358.5 J (mass 50.0 g, not 25.0).
0.5 mark: n = 0.0250 mol.
0.5 mark: -54.3 kJ mol^-1; negative sign required (exothermic). Max 1 if sign missing. ECF.

Ammonia has three bonding pairs and one lone pair around N: pyramidal. Methane has four bonding pairs (109.5 degrees). Lone pair-bonding pair repulsion is greater than bonding pair-bonding pair repulsion, squeezing the H-N-H angle.

0.5 mark: pyramidal (allow trigonal pyramidal).
0.5 mark: ammonia has a lone pair (3 bonding pairs + 1 lone pair) whereas methane has only bonding pairs.
0.5 mark: lone pair repels more than bonding pairs (AW), so angle reduced from 109.5 degrees. IGNORE 'lone pair has more electrons'.

Electrophilic addition of HBr: H+ adds to the double bond giving either a primary carbocation (CH3CH2CH2+) or a secondary carbocation (CH3CH+CH3). The secondary is more stable so is formed preferentially; bromide then attacks giving 2-bromopropane as the major product (Markownikoff).

0.5 mark: 2-bromopropane.
0.5 mark: two possible carbocations / intermediates formed (primary and secondary).
0.5 mark: secondary carbocation more stable (than primary), so more of it forms. ALLOW Markownikoff's rule for 0.5 only if stability not mentioned.

Reflux condenses vapours back into the flask so the aldehyde is kept in contact with the oxidant until it becomes the acid. Dichromate(VI) is orange and is reduced to green chromium(III). Alcohol gains one O to make the aldehyde (and water), then another O to make the acid: CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O.

0.5 mark: (heating under) reflux.
0.5 mark: orange to green.
0.5 mark: CH3CH2CH2OH + 2[O] -> CH3CH2COOH + H2O. ALLOW structural or displayed formulae. DO NOT ALLOW a single [O].

Broad 2500-3300 cm^-1 is the O-H of a carboxylic acid; 1715 cm^-1 is C=O. Together these indicate COOH. With C3H6O2 the compound is CH3CH2COOH.

0.5 mark: carboxylic acid.
0.5 mark: O-H for the broad absorption AND C=O for 1715 cm^-1.
0.5 mark: propanoic acid / CH3CH2COOH.

Left side has 1 mol gas, right side 2 mol. Increasing pressure favours the side with fewer gas moles, so the position shifts left (le Chatelier). The enthalpy change is not relevant at constant temperature.

0.5 mark: shifts to the left / towards N2O4.
0.5 mark: fewer (gas) moles / molecules on the left (1 vs 2).
0.5 mark: shift opposes (reduces) the pressure increase. DO NOT ALLOW the answer that Kc changes.

Down Group 2 atomic radius and shielding increase, so the outer 5s/4s/3s electrons are further from the nucleus and held less strongly. First and second ionisation energies decrease, so 2+ ions form more easily and reactivity increases.

0.5 mark: reactivity increases down the group (Mg to Ba).
0.5 mark: first and second ionisation energies decrease (less energy to remove two electrons).
0.5 mark: greater atomic radius / more shielding so weaker attraction between nucleus and outer electrons. ORA accepted.

(b) Mass of water lost = 3.69 - 1.80 = 1.89 g.
M(MgSO4) = 24.3 + 32.1 + 64.0 = 120.4 g/mol, so n(MgSO4) = 1.80 / 120.4 = 0.01495 mol.
M(H2O) = 18.0 g/mol, so n(H2O) = 1.89 / 18.0 = 0.105 mol.
Ratio n(H2O) : n(MgSO4) = 0.105 / 0.01495 = 7.02, so x = 7.
(a) Heat gently then strongly, allow to cool (e.g. in a desiccator), weigh, reheat and reweigh until the mass no longer changes between successive weighings.
(c) If heating was incomplete, some water remained in the residue, so the measured mass loss was too small and x was too low.

FIRST CHECK THE ANSWER: x = 7 gets all 4 marks in (b).
(a) Heat (then cool and) weigh [1]; reheat and reweigh until constant mass / two successive masses the same (AW) [1]. IGNORE 'heat for a long time'.
(b) Mass of water = 1.89 g [1]; n(MgSO4) = 1.80/120.4 = 0.0150 mol [1]; n(H2O) = 1.89/18.0 = 0.105 mol [1]; ratio = 7 / x = 7 [1]. ECF from an incorrect Mr or mass of water. Allow x = 7.0.
(c) Any one error that leaves water in the solid (heating not long enough/not to constant mass) [1] with explanation that mass lost is too small / residue mass too high so x too low [1]. Alternatively: the solid was weighed with the lid off while hot and absorbed water [1] linking to a mass that is too high. DO NOT ALLOW errors that would give a high x (e.g. decomposition of MgSO4).

(b) Formation: 2C(s) + 3H2(g) + 1/2 O2(g) -> C2H5OH(l). Both routes end at 2CO2 + 3H2O.
DH_f = sum DH_c(reactants) - DH_c(product) = 2(-394) + 3(-286) - (-1367) = -788 - 858 + 1367 = -279 kJ/mol.
(c) q = m c DT = 150 x 4.18 x 15.2 = 9530 J = 9.53 kJ.
n(C2H5OH) = 0.92 / 46.0 = 0.0200 mol.
DH_c = -9.53 / 0.0200 = -477 kJ/mol (3 s.f.).
(d) Heat lost to surroundings / beaker; incomplete combustion; non-standard conditions; evaporation of ethanol.
(e) Use a lid / draught shield / copper calorimeter / insulation.

(a) Enthalpy change when one mole of a substance [1] is completely burned in oxygen with all reactants and products in standard states (100 kPa, stated temperature, usually 298 K) [1]. ALLOW 'burnt completely in excess oxygen'.
(b) FIRST CHECK THE ANSWER: -279 gets 3 marks. Correct cycle with elements, ethanol, 2CO2 + 3H2O [1]; use of 2 x (-394) and 3 x (-286) with ethanol -1367 reversed [1]; -279 with sign [1]. Common error: -2155 (forgetting ethanol) gets 1 mark; +279 gets 2 marks.
(c) FIRST CHECK THE ANSWER: -477 (kJ/mol) gets 4 marks. q = 150 x 4.18 x 15.2 = 9530 J [1]; n = 0.0200 mol [1]; 9.53/0.0200 = 476.5 kJ/mol [1]; negative sign and 3 s.f. = -477 [1]. Allow -476 to -477. Using 0.92 g in q or DT = 15.2 C/K both fine; ECF.
(d) Any two: heat loss to surroundings / to beaker / air [1]; incomplete combustion (soot / CO) [1]; evaporation of ethanol [1]; non-standard conditions [1]. IGNORE 'human error'.
(e) Lid on beaker / draught shield / insulation around calorimeter / use of a copper calorimeter closer to flame [1].

(a) N shares both electrons of its lone pair with B; the diagram shows 3 N-H shared pairs, 3 B-F shared pairs and an N-B pair drawn with both electrons from N; nitrogen donates.
(b) BF3: 3 bonded pairs, no lone pairs, trigonal planar, 120 degrees. NH3: 3 bonded pairs + 1 lone pair, pyramidal, about 107 degrees. Lone pairs repel more than bonding pairs, squeezing the bond angle below the tetrahedral 109.5.
(c) F is more electronegative than B so each B-F bond is polar, but the three dipoles are arranged symmetrically (120 degrees) and cancel, giving no overall dipole.
(d) NH3 forms hydrogen bonds between N-H and the lone pair on N of another molecule, because N is highly electronegative. PH3 cannot (P is not sufficiently electronegative), so has only London and weak permanent dipole forces. Hydrogen bonds are stronger and need more energy to break, so the boiling point of NH3 is higher.

(a) Correct diagram: N with 3 bonding pairs to H, a pair to B, B with 3 pairs to F, all with correct electrons (F lone pairs shown) [1]; statement that nitrogen donates both electrons of the pair / lone pair on N [1]. ALLOW arrow N->B.
(b) BF3 trigonal planar 120 [1]; NH3 pyramidal ~107 (allow 106-108) [1]; NH3 has a lone pair (BF3 has none) [1]; lone pair-bond pair repulsion greater than bond pair-bond pair repulsion (AW) [1].
(c) F more electronegative than B so bonds are polar [1]; symmetrical shape so dipoles cancel / no net dipole [1].
(d) NH3 has hydrogen bonding [1]; PH3 has only London (dispersion) forces and/or weaker permanent dipole-dipole forces (no hydrogen bonding) [1]; hydrogen bonds stronger so more energy needed to overcome the intermolecular forces (ORA) [1]. DO NOT ALLOW breaking of covalent bonds.

(a)(i) Restricted rotation about the C=C double bond (pi bond), and each carbon of the C=C has two different groups attached. (ii) Both CH3 and CH2CH3 on the same side of the double bond.
(b)(i) Electrophilic addition: the pi bond attacks the H(delta+) of H-Br; H-Br bond breaks heterolytically with the pair going to Br; a secondary carbocation CH3CH(+)CH3 forms; Br- lone pair attacks C+ to give CH3CHBrCH3.
(ii) The secondary carbocation is more stable than the primary carbocation.
(c)(i) -CH(CH3)-CH2- inside brackets. (ii) Combustion for energy / feedstock recycling / recycling.
(d) Step 1: HBr, room temperature. Step 2: NaOH(aq), heat under reflux.
(e) n(2-bromopropane) = 8.0/123.0 = 0.0650 mol; theoretical mass = 0.0650 x 60.0 = 3.90 g; yield = 2.4/3.90 x 100 = 61.5%.

(a)(i) Restricted rotation about C=C [1]; two different groups/atoms on each carbon of the double bond [1].
(ii) Correct Z isomer with both alkyl groups on same side [1].
(b)(i) Curly arrow from C=C to H of HBr [1]; dipole H(delta+)-Br(delta-) and curly arrow from H-Br bond to Br [1]; correct secondary carbocation intermediate [1]; curly arrow from lone pair of Br- to C+ and product correct; name: electrophilic addition [1]. Arrows must start from the bond or lone pair.
(ii) Secondary carbocation more stable than primary (ORA) [1].
(c)(i) Correct repeat unit with continuation bonds [1]. (ii) Combustion with energy recovery, or feedstock recycling, or recycling (reuse) [1]. IGNORE 'landfill'.
(d) Step 1: HBr [1], room temperature (no catalyst needed) [1]; step 2: NaOH(aq) / aqueous alkali [1], heat (under reflux) [1]. DO NOT ALLOW ethanolic NaOH.
(e) FIRST CHECK THE ANSWER: 61.5% gets 3 marks. n = 0.0650 mol [1]; theoretical mass 3.90 g [1]; 2.4/3.90 x 100 = 61.5% [1]. ECF. Allow 61.5-61.6.

(a) Moles: C 64.9/12.0 = 5.41; H 13.5/1.0 = 13.5; O 21.6/16.0 = 1.35. Divide by 1.35: C 4.0 : H 10 : O 1. Empirical formula C4H10O.
(b) Empirical mass = 48 + 10 + 16 = 74 = M, so molecular formula C4H10O.
(c) Broad O-H absorption at 3300 means an alcohol; no absorption near 1700 means no C=O, so not an aldehyde, ketone or carboxylic acid.
(d) Primary alcohols (butan-1-ol, 2-methylpropan-1-ol): heat with acidified dichromate; distil immediately to get aldehyde (orange to green; IR has C=O at about 1700, no broad O-H), or heat under reflux with excess oxidant to get carboxylic acid (green; IR has C=O and broad O-H). Secondary (butan-2-ol): heating gives a ketone, orange to green; IR shows C=O around 1700, no O-H. Tertiary (2-methylpropan-2-ol): no oxidation, stays orange; IR still has broad O-H and no C=O.
(e) m/z 45 = CH3CH(OH)+ (C2H5O+), from loss of C2H5 (29) from 74.
(f) CH3CH(OH)CH2CH3 + [O] -> CH3COCH2CH3 + H2O; butanone.

(a) FIRST CHECK THE ANSWER: C4H10O gets 3 marks. Division by A_r: 5.41, 13.5, 1.35 [1]; ratio by dividing by smallest 4:10:1 [1]; C4H10O [1].
(b) C4H10O [1].
(c) Alcohol / O-H [1]; no C=O so not aldehyde, ketone or carboxylic acid (any one) [1].
(d) LEVEL OF RESPONSE (6 marks).
Indicative content: A: Primary alcohols oxidise in two stages: distillation gives aldehyde, reflux with excess oxidant gives carboxylic acid; orange to green. B: Secondary alcohols oxidise to ketones, orange to green; tertiary alcohols do not oxidise, stay orange. C: IR: aldehyde/ketone show C=O (about 1630-1820 cm-1) with no broad O-H; carboxylic acid shows C=O plus broad O-H (2500-3300); unoxidised alcohol/tertiary retains broad O-H and no C=O; so products distinguish types.
Level 3 (5-6 marks): all three alcohol types covered with conditions, products, observations and IR; logically structured. Level 2 (3-4 marks): two of A, B, C covered well, or all three with gaps. Level 1 (1-2 marks): some relevant points, e.g. colour change only, with limited links. 0 marks: no relevant content. Communication statement decides the mark within a level.
(e) CH3CH(OH)+ or C2H5O+ [1]; loss of C2H5 / ethyl radical (mass 29) [1]. Charge must be shown on ion.
(f) CH3CH(OH)CH2CH3 + [O] -> CH3COCH2CH3 + H2O [1]; butanone [1]. ALLOW skeletal/displayed formulae.

The forward reaction is exothermic, so a lower temperature moves the equilibrium to the right and increases the yield of methanol; however at low temperature the rate is too slow because fewer molecules have energy of at least the activation energy (Boltzmann distribution). 500 K is a compromise between acceptable yield and rate. There are 3 moles of gas on the left and 1 on the right, so a high pressure moves the position to the right (fewer gas molecules) and increases collision frequency, so increases rate; but very high pressure needs expensive equipment, thick-walled vessels and more energy, so 60 atm is a compromise on cost and safety. The catalyst provides an alternative route with lower activation energy, so a greater proportion of molecules exceed Ea and the rate increases at a lower temperature (saving energy); it does not change the position of equilibrium or Kc.

LEVEL OF RESPONSE (6 marks).
Indicative content: A (temperature): exothermic so lower T favours products (Le Chatelier) but lower rate as fewer molecules exceed Ea (Boltzmann); moderate T is a compromise. B (pressure): 3 mol gas -> 1 mol gas so high p favours products and increases collision frequency/rate; high cost and safety limit the pressure used. C (catalyst): provides alternative route with lower Ea, more molecules exceed Ea, faster rate at lower T / lower energy cost; no effect on position of equilibrium; allows a lower, more economical temperature.
Level 3 (5-6 marks): all of A, B and C covered with both equilibrium and rate linked to energy distribution; clear compromise reasoning. Level 2 (3-4 marks): two areas covered well, or all three with one aspect (rate or equilibrium) missing. Level 1 (1-2 marks): isolated correct statements, e.g. 'catalyst speeds up reaction' without explanation. 0 marks: no relevant content. Communication statement decides the mark within a level.

(a) Mass of water = 2.50 - 1.22 = 1.28 g, so n(H\(_2\)O) = 1.28 / 18.0 = 0.0711 mol. M(MgSO\(_4\)) = 24.3 + 32.1 + 64.0 = 120.4 g mol\(^{-1}\), so n = 1.22 / 120.4 = 0.01013 mol. Ratio = 0.0711 / 0.01013 = 7.02, so x = 7.
(b) Total uncertainty = 0.005 + 0.005 = 0.010 g. Percentage uncertainty = (0.010 / 1.22) x 100 = 0.82%.
(c)(i) MgCO\(_3\) + 2HCl \(\rightarrow\) MgCl\(_2\) + H\(_2\)O + CO\(_2\).
(ii) M(MgCO\(_3\)) = 24.3 + 12.0 + 48.0 = 84.3; n = 1.50 / 84.3 = 0.01779 mol = n(CO\(_2\)). V = nRT/p = (0.01779 x 8.314 x 298) / 101000 = 4.37 x 10\(^{-4}\) m\(^3\) = 437 cm\(^3\).

FIRST CHECK THE ANSWER ON THE ANSWER LINE. Correct final values earn full marks for that part.
(a) [3] M1: mass of water 1.28 g and n(H2O) = 0.0711 (mol) (1). M2: n(MgSO4) = 0.0101 (mol) (1). M3: ratio = 7(.0) so x = 7 (1). ECF from M1 and M2. ALLOW x = 7 only if the ratio is shown; DO NOT ALLOW non-integer x as the final answer.
(b) [2] M1: sum of the two uncertainties = 0.010 g (1). M2: 0.010/1.22 x 100 = 0.82% (1). Common error: 0.41% (one reading only) scores 1 mark. ALLOW 0.8%.
(c)(i) [1] Correct balanced equation; state symbols ignored.
(c)(ii) [2] M1: n(CO2) = 1.50/84.3 = 0.0178 mol (1). M2: V = nRT/p = 4.37 x 10^-4 m3 = 437 cm3 (1). ALLOW 4.37 x 10^-4 m3 only if converted to cm3 on the answer line; common error 0.437 cm3 or 437 m3 (unit conversion) scores 1 mark. ECF from M1.

(a) \(\Delta T\) = 21.5 K; q = 150 x 4.18 x 21.5 = 13480 J = 13.5 kJ. M(C\(_2\)H\(_5\)OH) = 46.0, n = 0.92/46.0 = 0.0200 mol. \(\Delta_c H\) = \(-\)13.48/0.0200 = \(-\)674 kJ mol\(^{-1}\).
(b) Heat lost to the surroundings / not all heat transferred to the water; incomplete combustion; heat absorbed by the copper can; evaporation of ethanol from the burner (any two).
(c) Cycle: elements (2C + 3H\(_2\) + 3\(\tfrac12\)O\(_2\)) \(\rightarrow\) ethanol + 3O\(_2\) via formation; and elements \(\rightarrow\) 2CO\(_2\) + 3H\(_2\)O via combustion; ethanol + 3O\(_2\) \(\rightarrow\) 2CO\(_2\) + 3H\(_2\)O by combustion. \(\Delta_f H\) = 2(\(-\)394) + 3(\(-\)286) \(-\) (\(-\)1367) = \(-\)788 \(-\) 858 + 1367 = \(-\)279 kJ mol\(^{-1}\).
(d) Bonds broken = 614 + 436 = 1050; bonds formed = 347 + 2(413) = 1173; \(\Delta H\) = 1050 \(-\) 1173 = \(-\)123 kJ mol\(^{-1}\).
(e) Reactants higher than products (exothermic); uncatalysed hump with Ea labelled from reactants to top; lower catalysed hump; \(\Delta H\) labelled from reactants to products.

FIRST CHECK THE ANSWER ON THE ANSWER LINE. -674 (kJ mol-1) scores 3 marks.
(a) [3] M1: q = 150 x 4.18 x 21.5 = 13480 J (13.5 kJ) (1). M2: n = 0.92/46.0 = 0.0200 mol (1). M3: -674 with negative sign and kJ conversion (1). ALLOW -673 to -675. Common errors: +674 scores 2; -0.674 (J not converted) scores 2; using 0.92 g as mass of water scores 0.
(b) [2] Any two from: heat loss to surroundings (1); incomplete combustion (1); heat absorbed by can / thermometer (1); evaporation of ethanol (1); non-standard conditions (1). IGNORE human error; IGNORE 'impurities' unless specified.
(c) [3] M1: correct cycle with arrows in correct directions (1). M2: 2(-394) + 3(-286) = -1646 (1). M3: -279 (1). ECF for M3. Common error: -2013 (wrong sign for ethanol combustion) scores 2 marks.
(d) [2] M1: 1050 and 1173 (1). M2: -123 (1). ALLOW 1 for +123.
(e) [2] M1: exothermic profile with reactants above products, delta H labelled (1). M2: two humps with the catalysed lower and Ea labelled for uncatalysed route (1).

(a) N in the centre with four shared pairs, one pair made of two electrons from N (the dative pair) and three pairs each with one N and one H electron; four H around; 8 electrons around N; charge shown.
(b)(i) NH\(_3\): pyramidal, 107\(^\circ\). NH\(_4^+\): tetrahedral, 109.5\(^\circ\).
(ii) NH\(_3\) has three bonded pairs and one lone pair; a lone pair repels more strongly than a bonded pair, so the bond angle is reduced. NH\(_4^+\) has four bonded pairs which repel equally.
(c) NH\(_3\) forms hydrogen bonds between molecules (N is electronegative, with a lone pair, and N\(-\)H is polar). PH\(_3\) has only weaker permanent dipole-dipole and London forces. More energy is needed to overcome hydrogen bonding, so NH\(_3\) has the higher boiling point.
(d) Octahedral; 90\(^\circ\).

(a) [2] M1: four bonded pairs around N shown with 8 electrons, N and H electrons distinguished (1). M2: dative pair shown as two N electrons shared with H+ (1) and the + charge on the brackets. ALLOW any combination of dots and crosses that shows a dative bond.
(b)(i) [2] NH3 pyramidal AND 107 (allow 106-107) (1); NH4+ tetrahedral AND 109.5 (1).
(b)(ii) [2] M1: NH3 has a lone pair and NH4+ has none / NH3 has 3 bonded pairs and 1 lone pair (1). M2: lone pair repels more than bonded pair (so angle smaller) ORA (1). DO NOT ALLOW 'lone pair repels more' without a comparison.
(c) [3] M1: hydrogen bonds between NH3 molecules (1). M2: PH3 has only weaker (permanent dipole-dipole / London) forces (1). M3: more energy needed to break hydrogen bonds in NH3 ORA (1). DO NOT ALLOW breaking of N-H bonds / covalent bonds. ALLOW 'P is less electronegative so cannot form hydrogen bonds'.
(d) [2] octahedral (1); 90 degrees (1).

(a)(i) Electrophilic addition. Curly arrow from the C=C bond to the H of H\(^{\delta+}\)\(-\)Br\(^{\delta-}\); curly arrow from H\(-\)Br bond to Br; secondary carbocation CH\(_3\)CH\(^+\)CH\(_2\)CH\(_3\); curly arrow from a lone pair on Br\(^-\) to C\(^+\) giving CH\(_3\)CHBrCH\(_2\)CH\(_3\).
(ii) The secondary carbocation is more stable than the primary one, so it forms preferentially.
(b) CH\(_3\)CH=CHCH\(_2\)CH\(_3\). On C2 the higher priority group is CH\(_3\) (over H); on C3 it is C\(_2\)H\(_5\) (over H). E: higher priority groups on opposite sides; Z: same side. Priority is by higher atomic number of the atom attached to the C=C carbon (then the next atoms along the chain).
(c) Orange/brown to colourless; 1,2-dibromohexane.
(d) \(-\)CH\(_2\)\(-\)CH(C\(_2\)H\(_5\))\(-\) with continuation bonds.
(e) Method: add the alcohol to acidified dichromate in a flask (or add dropwise); heat gently and distil the propanal out as it forms, because propanal has a lower boiling point than propan-1-ol (no hydrogen bonding) and would otherwise be further oxidised to propanoic acid; use a condenser with water flowing in at the bottom and a receiver, possibly in ice, to condense the volatile propanal. Calculation: n(propan-1-ol) = 6.00/60.0 = 0.100 mol; theoretical mass of propanal = 0.100 x 58.0 = 5.80 g; % yield = 3.19/5.80 x 100 = 55.0%.

(a)(i) [4] M1: curly arrow from C=C to H of HBr (1). M2: dipole H(delta+)-Br(delta-) shown and curly arrow from H-Br bond to Br (1). M3: correct secondary carbocation intermediate (1). M4: curly arrow from lone pair on Br- to C+ and correct product (1). Name: electrophilic addition must be given for M2 or M4 (if absent, max 3). DO NOT ALLOW arrow starting from the H atom.
(a)(ii) [1] Secondary carbocation more stable than primary (Markownikoff) ORA.
(b) [3] E and Z drawn correctly (1 each, labels must match drawings) (1)+(1). CIP: higher atomic number / higher priority groups on the same side = Z, opposite sides = E (1). ECF on labels.
(c)(i) [1] orange / brown (bromine water) to colourless. DO NOT ALLOW 'clear'; DO NOT ALLOW decolourised alone from 'red'.
(c)(ii) [1] 1,2-dibromohexane.
(d) [1] Repeat unit with ethyl side group and two open bonds; ALLOW displayed or skeletal. DO NOT ALLOW a C=C remaining.
(e)* [6] Level of response.
Level 3 (5-6 marks): full method with reasons (A, B) AND a correct calculation (C). Communication is clear and logical.
Level 2 (3-4 marks): two of A, B and C addressed; may have an error in the calculation or incomplete explanation.
Level 1 (1-2 marks): some relevant points from A, B or C with limited reasoning.
0 marks: no relevant response.
Indicative content: A: heat gently / distil off propanal as it forms because it has a lower boiling point than the alcohol (no hydrogen bonding), preventing oxidation to propanoic acid (rather than reflux). B: use of condenser with water entering at the bottom, collecting in a cooled receiver to reduce loss of volatile aldehyde; excess alcohol / controlled oxidant amount; thermometer at the still head to check the boiling range. C: n = 0.100 mol; theoretical mass 5.80 g; yield 55.0% (ALLOW 55%).

(a) C: 54.5/12.0 = 4.54; H: 9.1/1.0 = 9.1; O: 36.4/16.0 = 2.28. Dividing by 2.28: C 1.99, H 3.99, O 1. Empirical formula = C\(_2\)H\(_4\)O.
(b) Empirical formula mass = 44; M\(_r\) from the molecular ion = 88, so the molecular formula is 88/44 = 2 times the empirical formula: C\(_4\)H\(_8\)O\(_2\).
(c) Carboxylic acid; broad absorption is O\(-\)H (in COOH), 1710 cm\(^{-1}\) is C=O.
(d) CH\(_3\)CH\(_2\)CH\(_2\)COOH (butanoic acid) and (CH\(_3\))\(_2\)CHCOOH (2-methylpropanoic acid), drawn displayed.
(e)(i) Butan-1-ol, primary.
(ii) Acidified potassium dichromate(VI) (K\(_2\)Cr\(_2\)O\(_7\)/H\(_2\)SO\(_4\)); orange to green.
(iii) CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)OH + 2[O] \(\rightarrow\) CH\(_3\)CH\(_2\)CH\(_2\)COOH + H\(_2\)O.
(iv) Reflux prevents the loss of volatile alcohol / aldehyde intermediate so that complete oxidation to the carboxylic acid occurs.

FIRST CHECK THE ANSWER ON THE ANSWER LINE where there is one.
(a) [3] M1: ratios 4.54 : 9.1 : 2.28 (1). M2: divide by smallest value, 2 : 4 : 1 (1). M3: C2H4O (1). ECF from M1.
(b) [2] M1: C4H8O2 (1). M2: M+ peak (88) gives Mr, which is twice the empirical formula mass of 44 (1).
(c) [2] carboxylic acid (1); O-H (broad) and C=O (1). Both bonds needed for the second mark.
(d) [2] one mark each: butanoic acid and 2-methylpropanoic acid, displayed with all bonds shown. DO NOT ALLOW skeletal/structural formulae for the full mark but ALLOW 1 mark for correct structural formulae. DO NOT ALLOW an ester or hydroxy-aldehyde (IR does not fit).
(e)(i) [2] butan-1-ol (1); primary (1).
(e)(ii) [2] acidified potassium dichromate(VI) / Cr2O7 2- with H+ (1); orange to green (1).
(e)(iii) [2] reactants and products correct (1); balanced with 2[O] and H2O (1). ALLOW 1 for 1 mark if the aldehyde was written as the product.
(e)(iv) [1] prevents escape/loss of volatile aldehyde or alcohol (AW), allowing complete oxidation to the carboxylic acid.

Temperature: the forward reaction is exothermic, so a lower temperature moves the equilibrium to the right and gives a higher yield of SO\(_3\). However, a low temperature gives a slow rate because fewer molecules have energy above the activation energy and collisions are less frequent, so about 700 K is a compromise.
Pressure: there are 3 moles of gas on the left and 2 on the right, so higher pressure moves the position to the right, giving a higher yield; higher pressure also increases the rate by increasing collision frequency. However, high pressure needs expensive equipment and the yield is already high, so a pressure slightly above atmospheric is used.
Catalyst: V\(_2\)O\(_5\) provides a route with lower activation energy, increasing the rate, but does not change the equilibrium position or yield. It allows a lower temperature and so saves energy, giving an acceptable rate at 700 K.

Level of response (6 marks).
Level 3 (5-6 marks): all three variables addressed (temperature, pressure, catalyst) with effects on both yield and rate, and the compromise explained. Clear, logical communication.
Level 2 (3-4 marks): two variables addressed well, or all three with partial explanation of yield or rate; compromise may be missing or weak.
Level 1 (1-2 marks): some correct statements about one or two factors with little explanation.
0 marks: no relevant response.
Indicative content: A (temperature): exothermic, so lower T favours forward reaction/higher yield; lower T gives slower rate (fewer molecules above Ea, fewer collisions); compromise of 700 K for acceptable yield and rate. B (pressure): 3 mol to 2 mol gas so higher p favours SO3; higher p increases collision frequency/rate; cost and safety limit the pressure used. C (catalyst): lower Ea route, faster rate, no change in position of equilibrium/yield; lowers energy costs. ALLOW ORA throughout. DO NOT ALLOW the catalyst increasing yield; DO NOT ALLOW 'equilibrium is reached faster' as the reason for a higher yield. The communication statement decides the mark within a level.

Butane: non-polar molecules (C-C and C-H bonds have very small electronegativity differences). Only induced dipole-dipole (London) forces, arising from temporary fluctuations in electron density that induce dipoles in neighbouring molecules. These are weak, so little energy is needed to separate the molecules: boiling point -1 °C.

Propanone: the C=O bond is polar because oxygen is much more electronegative than carbon, so the molecule has a permanent dipole. Molecules attract through permanent dipole-dipole interactions in addition to London forces (similar in size since Mr is the same). These are stronger overall than London forces alone, so more energy is needed: 56 °C. There is no H atom bonded to N, O or F, so no hydrogen bonding between propanone molecules.

Propan-1-ol: contains an O-H bond, highly polar. A lone pair on the O of one molecule attracts the delta-plus H of another, forming hydrogen bonds. Hydrogen bonding is the strongest of the intermolecular forces here, so the most energy is needed to overcome them: 97 °C.

Covalent bonds within molecules are not broken on boiling; only intermolecular forces are overcome.

Level of response (6 marks). Indicative content:
A Butane: non-polar / no permanent dipole; only induced dipole-dipole (London, van der Waals) forces from temporary fluctuations in electron density; weakest, least energy needed, lowest boiling point.
B Propanone: polar C=O (O more electronegative) giving a permanent dipole; permanent dipole-dipole interactions in addition to London forces; stronger than London forces alone so higher boiling point than butane; no hydrogen bonding as no O-H/N-H/F-H.
C Propan-1-ol: O-H bond polar; hydrogen bonding between delta+ H and lone pair on O of a neighbouring molecule; strongest force, most energy needed, highest boiling point.

Level 3 (5-6 marks): all three compounds explained with correct force type, its origin and a correct comparison of strengths/energy needed. Well-developed, logically structured, no significant errors or omissions.
Level 2 (3-4 marks): correct force identified for at least two compounds with some explanation of origin or comparison of strength; some gaps or minor errors.
Level 1 (1-2 marks): some correct statements about intermolecular forces with limited explanation or comparison.
Level 0 (0 marks): no response worthy of credit.
The communication/structure of the answer decides the mark within a level.

ALLOW van der Waals' forces, dispersion forces, London forces for induced dipole-dipole. ALLOW ORA comparisons. DO NOT ALLOW answers that state covalent bonds break on boiling (cap at Level 2). IGNORE comments on Mr being the same other than as context.

Oxidation: warm the acidified dichromate(VI) and add the propan-1-ol dropwise from a dropping funnel, with distillation apparatus set up so propanal is distilled off as it forms. Propanal has a lower boiling point than propan-1-ol (no hydrogen bonding), so it distils out immediately and cannot be oxidised further to propanoic acid. Use of reflux would give the acid.

Purification: the distillate (propanal plus water and some acid/alcohol) is shaken in a separating funnel and the aqueous layer is run off and discarded. The organic layer is dried with anhydrous magnesium sulfate (or calcium chloride) to remove traces of water, then filtered/decanted and redistilled, collecting the fraction at the boiling point of propanal (about 49 °C), which separates it from impurities with different boiling points.

Calculation: n(propan-1-ol) = 5.00 / 60.0 = 0.0833 mol. 1:1 ratio so theoretical n(propanal) = 0.0833 mol. Theoretical mass = 0.0833 x 58.0 = 4.83 g. Percentage yield = 2.10 / 4.83 x 100 = 43.4%.

Level of response (6 marks). Indicative content:
A Oxidation: heat acidified dichromate(VI) and add alcohol slowly/dropwise; distil off the propanal as it forms because it has a lower boiling point than the alcohol (no hydrogen bonding), so it is not further oxidised to propanoic acid (not reflux).
B Purification: separating funnel, discard aqueous layer; dry organic layer with anhydrous MgSO4/CaCl2/Na2SO4 to remove water; redistil and collect fraction at the boiling point of propanal.
C Calculation: n(propan-1-ol) = 5.00/60.0 = 0.0833 mol (1 mark); theoretical mass of propanal = 0.0833 x 58.0 = 4.83 g (1 mark); percentage yield = 2.10/4.83 x 100 = 43.4% (1 mark).

FIRST CHECK THE ANSWER: 43.4% (ALLOW 43.5% from rounding; ALLOW ECF for one slip) scores the calculation marks.

Level 3 (5-6 marks): both the oxidation and purification described with reasons and the percentage yield correct or with one error only. Logically structured.
Level 2 (3-4 marks): oxidation and/or purification described with some reasons, and/or a partly correct calculation; some gaps.
Level 1 (1-2 marks): some correct points on procedure or calculation with little explanation.
Level 0 (0 marks): no response worthy of credit.
The communication/structure of the answer decides the mark within a level.

ALLOW anhydrous calcium chloride, sodium sulfate. DO NOT ALLOW reflux for the oxidation step. IGNORE state symbols.

On each carbon of the double bond the higher CIP priority group is CH3 on C2 (over H) and CH2CH3 on C3 (over H). If these two groups are on opposite sides the isomer is E; on the same side it is Z. All C-H bonds and the C-C bonds are drawn in the displayed formulae.

1 mark: correct displayed formula of E isomer (CH3 and C2H5 on opposite sides of C=C), all bonds shown, labelled E.
1 mark: correct displayed formula of Z isomer (CH3 and C2H5 on same side of C=C), all bonds shown, labelled Z.
The label on each structure must match its drawing to score.
DO NOT ALLOW skeletal or structural formulae for the structures (displayed required) - max 1 mark if both are not fully displayed but are correctly arranged.
DO NOT ALLOW a missing H on a carbon.

Four electron pairs repel to a tetrahedral arrangement; one is a lone pair, so the molecular shape is pyramidal. The lone pair repels the bonded pairs more than bonded pairs repel each other, so the bond angle is smaller than 109.5 degrees (about 107 degrees in the specification model).

1 mark: pyramidal 3-D diagram with P bonded to three H, using at least one wedge and one dashed bond, lone pair shown on P (dots or pair of electrons).
1 mark: bond angle labelled between 90 and 109 degrees (ALLOW 107 degrees; ALLOW 93 degrees).
DO NOT ALLOW 109.5 degrees or 120 degrees.
DO NOT ALLOW a flat (planar) diagram for the first mark.

The pi electrons in C=C attack the delta+ hydrogen of H-Br. The H-Br bond breaks heterolytically, both electrons going to Br to form Br-. H adds to the terminal CH2, leaving a positive charge on the central carbon. This gives the more stable secondary carbocation CH3CH(+)CH3 (Markownikoff).

1 mark: curly arrow from the C=C double bond to the H atom of H-Br AND curly arrow from the H-Br bond to Br; dipole delta+ on H and delta- on Br shown. Arrows must start at the bond and finish at an atom.
1 mark: correct secondary carbocation CH3CH(+)CH3 with the + on the central carbon (Br- may be shown).
DO NOT ALLOW the primary carbocation CH3CH2CH2(+).
IGNORE the name of the mechanism.

A primary haloalkane has the carbon bonded to Br attached to only one other carbon. With four carbons: unbranched BrCH2CH2CH2CH3 (1-bromobutane) and branched BrCH2CH(CH3)CH3 (1-bromo-2-methylpropane). The other two isomers, 2-bromobutane (secondary) and 2-bromo-2-methylpropane (tertiary), are not primary.

1 mark: skeletal formula of 1-bromobutane (four-carbon zigzag chain with Br on a terminal carbon).
1 mark: skeletal formula of 1-bromo-2-methylpropane (three-carbon chain with a methyl on the middle carbon and Br on a terminal carbon).
ALLOW displayed or structural formulae if unambiguous and correct (skeletal asked, so allow only if clear).
DO NOT ALLOW 2-bromobutane or 2-bromo-2-methylpropane.
DO NOT ALLOW the same isomer drawn twice.

(i) The C=C opens and each carbon bonds to the next monomer, leaving a two-carbon backbone with an ethyl side group on one carbon. (ii) For a repeat unit with a two-carbon backbone and a methyl side group, the double bond is restored: propene, CH2=CHCH3, with all bonds and hydrogens drawn.

1 mark: repeat unit with two backbone carbons, ethyl group on one carbon, with continuation bonds extending through both brackets/ends.
1 mark: displayed formula of propene with C=C and all C-H bonds shown.
DO NOT ALLOW a repeat unit with a C=C retained.
DO NOT ALLOW missing continuation bonds for the first mark.
DO NOT ALLOW structural or skeletal formula for (ii).

Butan-1-ol is a primary alcohol; with distillation the aldehyde is removed as formed so the product is butanal, CH3CH2CH2CHO. Butan-2-ol is a secondary alcohol and is oxidised to the ketone butanone, CH3COCH2CH3; it cannot be oxidised further.

1 mark: butanal drawn as a four-carbon skeletal chain with C=O at the end carbon (aldehyde).
1 mark: butanone drawn as a four-carbon skeletal chain with C=O on carbon 2.
DO NOT ALLOW butanoic acid for (i) (requires reflux).
ALLOW displayed or structural formula if unambiguous.
DO NOT ALLOW a ketone drawn for (i) or an aldehyde for (ii).