Zadanie 1a
Wzorowane na: Chemistry HL Paper 2, May 2023, TZ2, pytanie 1(a). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 1a-1b.
Analytical and spectroscopic techniques are used to determine the formulas and structures of organic compounds.
Polecenie
An unknown organic compound, X, contains only carbon, hydrogen and oxygen. Elemental analysis shows that it contains 54.5 % carbon and 36.3 % oxygen by mass.
(i) Determine the empirical formula of X.
The mass spectrum of X is shown.
(ii) Identify the species responsible for the peaks at m/z 88 and m/z 73. Use section 22 of the data booklet.
The infrared spectrum of X is shown.
(iii) Identify the bond mainly responsible for each of the absorptions A and B. Use section 20 of the data booklet.
(iv) X has an unbranched carbon chain. Deduce the structural formula of X.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «n(C) =» = 4.54 «mol» AND «n(O) =» = 2.27 «mol» ✔
«% H = 100 − 54.5 − 36.3 =» 9.2 % so «n(H) =» = 9.1 «mol» ✔
«ratio C : H : O = 2 : 4 : 1, so empirical formula =» C2H4O ✔
(ii) m/z 88: molecular ion / M+ / C4H8O2+ ✔
m/z 73: loss of 15 «from the molecular ion» / «M − CH3»+ / C3H5O2+ ✔
(iii) A: O-H «in a carboxylic acid» ✔
B: C=O ✔
(iv) CH3CH2CH2COOH «butanoic acid, with all bonds shown» ✔
Schemat punktowania
(i) M1: amounts of C and O; M2: percentage and amount of H by difference; M3: simplest ratio and formula. Award [2] for the correct ratio 2 : 4 : 1 without the formula. (ii) [1] each; accept loss of CH3 for m/z 73 and molecular ion for m/z 88. (iii) [1] each; A must be O-H (not N-H and not C-H), B must be C=O. (iv) [1] for a structural formula; do not accept the molecular formula C4H8O2.
Komentarz
The empirical formula C2H4O has a formula mass of 44, so the molecular ion at m/z 88 shows that the molecular formula is C4H8O2. The broad band together with C=O points to a carboxylic acid.