Zadanie 3c

(i) Step 2: curly arrow from the C=C double bond to the hydrogen of H₃O⁺ AND a second arrow from the O–H bond to oxygen so that H₂O leaves, giving the secondary carbocation CH₃CH⁺CH₂CH₂CH₃ ✔ ✔ ✔

Step 3: curly arrow from a lone pair on the oxygen of H₂O to the positively charged carbon, giving the oxonium ion CH₃CH(OH₂⁺)CH₂CH₂CH₃ ✔ ✔

Step 4: curly arrow from the O–H bond of the oxonium ion to oxygen AND a lone pair on water (or on the acid anion) taking the proton, giving pentan-2-ol and regenerating H₃O⁺ ✔ ✔

any four of the seven marking points: arrow from C=C to H of H₃O⁺; arrow showing water leaving; structure of the secondary carbocation; arrow from the lone pair on water to the carbocation; structure of the oxonium ion; arrow showing the O–H electrons moving onto oxygen; lone pair from water/acid anion attacking the hydrogen to reform the catalyst

(ii) the frequency of collisions increases «the particles move faster, so more collisions per unit time» ✔

a greater proportion/fraction of the particles has energy ≥ E_a, so a greater proportion of collisions is successful ✔

(iii) pentan-2-ol is more soluble in water than pent-1-ene AND it has an –OH «hydroxyl» group which forms hydrogen bonds with water ✔

pent-1-ene is less soluble/not soluble AND is non-polar/has only London (dispersion) forces ✔

(iv) three repeating units of −CH₂−CH(C₃H₇)⁻ joined by single C–C bonds in one chain, e.g. …CH₂−CH(C₃H₇)−CH₂−CH(C₃H₇)−CH₂−CH(C₃H₇)⋅⋅⋅, with propyl side chains «not methyl», no C=C bonds left and continuation bonds shown at both ends ✔

(v) benzene-1,4-dicarboxylic acid «HOOC−C₆H₄−COOH, the two acid groups at positions 1 and 4» ✔

butane-1,4-diol «HOCH₂CH₂CH₂CH₂OH» ✔

(i) [4] any four of the seven marking points listed in the key; the carbocation must be the secondary one (the product is pentan-2-ol); an arrow must start from a bond or a lone pair and end at an atom; curly arrows may be shown in the structures or described unambiguously. (ii) [2] M1: increased frequency of collisions; a reference to frequency or to time is required, so do not accept "chance" or "probability" without mention of time; M2: increased collision energy, so a greater proportion of collisions are successful, OR a greater fraction of molecules with E ≥ Ea. Do not accept answers about the equilibrium shifting. (iii) [2] M1: pentan-2-ol more soluble AND the –OH (hydroxyl) group forms hydrogen bonds with water; M2: pent-1-ene less/not soluble AND non-polar/only London forces. Do not accept hydroxide for hydroxyl. Award [1] for "pentan-2-ol is polar AND pent-1-ene is non-polar" alone; award [1] for "pentan-2-ol is soluble/more soluble AND pent-1-ene is not/less soluble" alone. (iv) [1] correct repeating unit with the propyl side chain, three units joined, no double bonds, continuation bonds shown; side chains may be drawn on opposite sides or head to tail; penalize missing H atoms once in the paper; ignore square brackets or n; accept a skeletal formula. (v) [2] M1: benzene-1,4-dicarboxylic acid; M2: butane-1,4-diol. Accept skeletal formulas. Accept benzene-1,4-dicarbonyl dichloride (two Cl in place of two OH) for M1, but not for both monomers.