Zadanie 3c, Paper 2, listopad 2025, TZ1, Chemia IB HL
Wzorowane na: Chemistry HL Paper 2, November 2025, TZ1, pytanie 3(c). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
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Zadanie
Wspólne informacje do wiązki zadań
Many esters are still known by trivial names. Propyl ethanoate, CH3COOCH2CH2CH3, which has the odour of pears, is manufactured from propan-1-ol and ethanoic acid in the presence of an acid catalyst.
CH3CH2CH2OH(l) + CH3COOH(l) ⇌ CH3COOCH2CH2CH3(l) + H2O(l)
Both reactants can be synthesized from alkanes: propan-1-ol from propane via 1-bromopropane, and ethanoic acid from ethane via bromoethane and ethanol. The overall equation for the synthesis of the ester from the two alkanes is
Polecenie
(i) Concentrated sulfuric acid, H2SO4, is used as the catalyst in the manufacture of propyl ethanoate. Explain how the catalyst affects the reaction.
(ii) The conversion of propane into 1-bromopropane,
C3H8(g) + Br2(g) → CH3CH2CH2Br(g) + HBr(g)
begins with an initiation step:
Br2 → 2 Br ∙
State the condition needed for this initiation step to take place.
(iii) Warming 1-bromopropane with aqueous sodium hydroxide, NaOH(aq), converts it into propan-1-ol. Explain the mechanism of this reaction, using curly arrows to show the movement of electron pairs.
(iv) Deduce, for the high-resolution NMR spectrum of propan-1-ol, the number of signals expected, the chemical shift range and integration of each, and the splitting pattern shown by each one. Use section 21 of the data booklet.
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Pokaż rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) increases the rate of reaction «and is not used up» ✔
provides an alternative pathway with a lower activation energy / a greater proportion of collisions have energy greater than the activation energy ✔
(ii) ultraviolet light / UV «radiation» ✔
(iii) the lone pair (negative charge) on the oxygen of HO− attacks the carbon bonded to Br, shown by a curly arrow running from that lone pair to the carbon ✔
a second curly arrow runs from the C–Br bond to the Br atom as it leaves ✔
the transition state «in square brackets» carries a negative charge and shows the partial bonds HO···C···Br ✔
CH3CH2CH2OH forms as the organic product, together with Br− ✔
(iv) Number of signals: 4 ✔
Chemical shift ranges: 0.9–1.0 «ppm, CH3» AND 1.3–1.4 «ppm, middle CH2» AND 3.3–3.7 «ppm, CH2–O» AND 1.0–6.0 «ppm, OH» ✔
Integration traces: 3 AND 2 AND 2 AND 1 ✔
Splitting patterns: triplet «CH3» AND multiplet / sextet «middle CH2» AND triplet «CH2–O» AND «broad» singlet «OH» ✔
Schemat punktowania
(i) M1: increases the rate; M2: lower activation energy through an alternative pathway, or more particles with energy at least equal to the activation energy. (ii) [1] for UV light; accept sunlight or heat; do not accept light alone. (iii) M1: arrow from the nucleophile's lone pair to C; M2: arrow from the C–Br bond to Br (accept it drawn in the transition state); M3: transition state with partial bonds and the negative charge; M4: both products. Award [3 max] for an SN1 mechanism; do not award M3 if a full HO–C bond is drawn in the transition state. (iv) [1] for each line: number of signals, chemical shift ranges, integration, splitting; accept the signals in any order; accept any three correct details for a single signal for [1] each; accept 'multiplet' for the middle CH2.
Komentarz. 1-Bromopropane is a primary halogenoalkane, so the concerted SN2 pathway with a single transition state is expected; drawing a carbocation intermediate (SN1) loses one mark.