Zadanie 5d
Wzorowane na: Chemistry HL Paper 2, May 2022, TZ1, pytanie 5(d). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Polecenie
Bromoethane can be converted into ethanol:
CH3CH2Br + HO− → CH3CH2OH + Br−
(i) Identify the type of reaction.
(ii) Outline what must be true of a collision between reacting particles for a reaction to occur.
(iii) Explain the mechanism of this reaction, drawing curly arrows to show how electron pairs move.
(iv) Carbon-halogen bonds are polar, which facilitates attack by HO−. State, giving a reason, whether the C−Cl bond or the C−I bond is more polar. Use section 9 of the data booklet.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) «nucleophilic» substitution ✔
OR
SN2 ✔
(ii) energy of the collision ≥ activation energy / Ea ✔
correct orientation «of the reacting particles» ✔
(iii) curly arrow from the lone pair / negative charge on O of HO− to the carbon atom ✔
curly arrow from the C−Br bond to Br «Br leaving as Br−» ✔
transition state shown with HO and Br partially bonded to carbon, a negative charge and square brackets «HO and Br on opposite sides of the carbon» ✔
(iv) C−Cl AND chlorine is more electronegative «3.2» than iodine «2.7» / the electronegativity difference with carbon «2.6» is larger for Cl «0.6» than for I «0.1» ✔
Schemat punktowania
(i) [1]; accept hydrolysis, SN2 or bimolecular nucleophilic substitution. (ii) [1] each. (iii) M1: arrow from the lone pair or charge on O to C; M2: arrow showing Br leaving; M3: transition state with partial bonds, charge and brackets. An arrow that starts on the H of HO- is not accepted. HO and Br need not be drawn exactly 180° apart. M3 is lost if the O-C bond is drawn as a full line. A two-step SN1 mechanism earns at most [2]. (iv) [1] for the bond with a reason based on electronegativity values.