Unit 4: Organic 2 & Physical 2 Exam Practice

(a)

• The power to which the concentration of a reactant is raised in the rate equation. [1] (Accept: "The power of concentration term in rate equation").

(b)

• Order wrt S₂O₈²⁻ = 1 [1]
• Comparing Exp 1 and Exp 2: [S₂O₈²⁻] doubles while [I⁻] remains constant; initial rate doubles (1.20 × 10⁻⁴ to 2.40 × 10⁻⁴), hence first order. [1]
• Order wrt I⁻ = 1 [1]
• Comparing Exp 1 and Exp 3: [I⁻] triples while [S₂O₈²⁻] remains constant; initial rate triples (1.20 × 10⁻⁴ to 3.60 × 10⁻⁴), hence first order. [1]

(c)

• Rate = k [S₂O₈²⁻][I⁻] [1] (Allow error carried forward from part b).

(d)

• k = Rate / ([S₂O₈²⁻][I⁻]) = (1.20 × 10⁻⁴) / (0.010 × 0.015) [1]
• k = 0.80 (or 0.8) [1]
• Units = dm³ mol⁻¹ s⁻¹ [1] (Allow ECF for units matching candidate rate equation).

(e)

• ln k = (-E_a/R)(1/T) + ln A ⇒ gradient = -E_a / R [1]
• -6250 = -E_a / 8.31 ⇒ E_a = 6250 × 8.31 = 51,937.5 J mol⁻¹ [1]
• E_a = +51.9 kJ mol⁻¹ (allow +52 or range 51.9 to 52.0) [1] (Must divide by 1000 for kJ mol⁻¹; + sign preferred but do not penalise if omitted).

(f)

• Many more molecules possess kinetic energy greater than or equal to the activation energy (E ≥ E_a). [1]
• A much greater proportion / fraction of collisions are successful per second (collision frequency increases). [1]

(a)

• Moles of CO = 1.50 - 0.930 = 0.570 mol [1]
• Moles of H₂ = 3.00 - 2(0.930) = 3.00 - 1.860 = 1.14 mol [1] (Must account for 1:2 stoichiometry).

(b)

• Total moles = 0.570 + 1.14 + 0.930 = 2.64 mol. Mole fraction x(H₂) = 1.14 / 2.64 = 0.432 (allow 0.43 or 0.4318) [1]
• Partial pressure p(H₂) = 0.4318 × 190 = 82.0 kPa (allow 82.0 to 82.1 kPa) [1] (Allow ECF from part a).

(c)

• K_p = p(CH₃OH) / [p(CO) × p(H₂)²] [1] (Do NOT accept square brackets; must use partial pressure p(X) notation).

(d)

• p(CO) = (0.570 / 2.64) × 190 = 41.0 kPa; p(CH₃OH) = (0.930 / 2.64) × 190 = 66.9 to 67.0 kPa [1]
• Substitution into expression: K_p = 67.0 / [41.0 × (82.0)²] [1]
• Denominator = 41.0 × 6724 = 275,684 [1]
• Numerical value: K_p = 2.43 × 10⁻⁴ (allow 2.4 × 10⁻⁴) [1]
• Units: kPa⁻² [1]

(e)

• As temperature increases, the numerical value of K_p decreases. [1]
• By Le Chatelier's principle, an increase in temperature favours the endothermic direction, shifting equilibrium to the left (reactants). [1]
• Therefore, the forward reaction must be exothermic (ΔH is negative). [1]

(a)

• Type of reaction: Nucleophilic addition (also accept reduction). [1]
• Nucleophile: Hydride ion / :H⁻ (reject "H" or "hydrogen"). [1]

(b)

• M1: Curly arrow from the lone pair on :H⁻ to the carbonyl carbon (>C(δ+)). [1]
• M2: Curly arrow from the C=O double bond to the oxygen atom (O(δ−)). [1]
• M3: Correct structure of the tetrahedral intermediate alkoxide ion CH₃CH₂C(H)(O⁻)CH₃ showing a lone pair and negative charge on oxygen. [1]
• M4: Curly arrow from a lone pair on the intermediate oxygen to H⁺ (or H on H₂O) to form butan-2-ol. [1]

(c) (i)

• Optical isomerism: Molecules that exist as non-superimposable mirror images of each other. [1]
• Chiral carbon: A carbon atom bonded to four different atoms or groups of atoms. [1] (Reject "four different bonds").

(c) (ii)

• The carbonyl group (C=O carbon) is planar. [1]
• The hydride nucleophile (:H⁻) has an equal probability of attacking the planar carbonyl from above or below the plane. [1]
• An equimolar (50:50) mixture of both enantiomers is produced; their equal and opposite rotations of plane-polarised light cancel out completely. [1]

(d)

• The carbonyl (C=O) peak at 1680-1750 cm⁻¹ will completely disappear. [1]
• A broad alcohol (O-H) peak will appear at 3230-3550 cm⁻¹. [1]
• Comparison of the fingerprint region (below 1500 cm⁻¹) against a reference spectrum of pure butan-2-ol confirms purity. [1]

(a)

• HNO₃ + H₂SO₄ ⇌ H₂NO₃⁺ + HSO₄⁻ [1]
• H₂NO₃⁺ ⇌ NO₂⁺ + H₂O [1]
• (Or accept overall single equation: HNO₃ + 2H₂SO₄ ⇌ NO₂⁺ + H₃O⁺ + 2HSO₄⁻ for 2 marks).

(b)

• M1: Curly arrow from the delocalised pi ring of methylbenzene to the nitrogen atom of NO₂⁺. [1]
• M2: Correct structure of the arenium intermediate carbocation showing a horseshoe open at carbon 4 with the positive charge located inside the cavity. [1]
• M3: Curly arrow from the C-H bond at carbon 4 back into the ring to restore the aromatic pi sextet. [1]
• M4: Correct structure of 4-nitromethylbenzene (1-methyl-4-nitrobenzene) and regenerated H⁺ (or HSO₄⁻ + H⁺ → H₂SO₄). [1]

(c)

• Electrophilic substitution. [1] (Reject "nucleophilic" or "addition").

(d)

• The methyl group is an electron-donating group. [1]
• It releases electron density into the benzene pi electron cloud through a positive inductive effect (+I). [1]
• The higher electron density makes the ring more attractive / more susceptible to attack by the electrophile (NO₂⁺). [1]

(e)

• Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl), followed by NaOH. [1] (Reject LiAlH₄).
• Equation: C₇H₇NO₂ + 6[H] → C₇H₇NH₂ + 2H₂O (or CH₃C₆H₄NO₂ + 6[H] → CH₃C₆H₄NH₂ + 2H₂O) [2] (1 mark for correct organic formulae, 1 mark for balanced 6[H] and 2H₂O).

(a) (i)

• Repeating unit: [-CO-CH₂-CH₂-CO-NH-(CH₂)₆-NH-] [2]
• 1 mark for correct carbon backbone from butanedioic acid and hexane-1,6-diamine; 1 mark for correct amide linkage (-CO-NH-) with open bonds at both ends.

(a) (ii)

• Water / H₂O. [1]

(b)

• Polyamides contain hydrogen bonds between adjacent polymer chains. [1]
• Hydrogen bonds form between the partially positive H on N-H and the lone pair on the carbonyl oxygen (C=O) of adjacent chains. [1]
• Polyalkenes only have much weaker Van der Waals / temporary dipole-dipole forces between chains, which require less energy to overcome. [1]

(c)

• Polyamides contain polar carbon-nitrogen (C-N) amide bonds. [1]
• These polar bonds are susceptible to nucleophilic attack and can undergo hydrolysis by water, acid, or alkalis. [1] (Polyalkenes have non-polar, inert C-C backbones).

(d) (i)

• Zwitterion: CH₃CH(NH₃⁺)COO⁻ [2] (Must show positive charge on N and negative charge on O).

(d) (ii)

• In acid: CH₃CH(NH₃⁺)COOH [2] (The carboxylate group accepts a proton; the amine remains protonated).

(d) (iii)

• In base: CH₃CH(NH₂)COO⁻ [2] (The ammonium group loses a proton to form -NH₂; the carboxylate remains -COO⁻).

(a)

• Step 1: Oxidation
  Reagent: Potassium dichromate(VI) (K₂Cr₂O₇) and dilute sulfuric acid (H₂SO₄). [1]
  Conditions: Heat under reflux (to ensure complete oxidation to 2-methylpropanoic acid). [1]
• Step 2: Esterification
  Reagents: Methanol (CH₃OH) and concentrated sulfuric acid catalyst (conc. H₂SO₄). [1]
  Conditions: Heat / warm under reflux. [1]

(b) (i)

• Sodium hydrogencarbonate: NaHCO₃ neutralises and removes acidic impurities (unreacted 2-methylpropanoic acid and the sulfuric acid catalyst). [1]
• Anhydrous MgSO₄: a drying agent that removes water. [1]

(b) (ii)

• Record the temperature at which the liquid distils / measure its boiling point. [1]
• A pure liquid distils at a constant temperature equal to the data value, while an impure one distils over a range. [1]

(c)

• Peak 1 (-CH(CH₃)₂): Integration = 6 [1]
• Peak 2 (-CH-): Integration = 1 [1]
• Peak 3 (-OCH₃): Integration = 3 [1]
• Correct ratio 6 : 1 : 3 correctly linked to the three proton environments. [1]