Zadanie 7b
Wzorowane na: Chemistry HL Paper 2, May 2023, TZ2, pytanie 7(b). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 7a-7f.
Pent-1-ene, CH2=CHCH2CH2CH3, reacts with hydrogen chloride, HCl.
Polecenie
Two products are possible in this reaction.
(i) Explain the mechanism for the formation of the major product, using curly arrows to show the movement of electron pairs.
(ii) Explain why the mechanism results in one product being formed in greater quantity than the other.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz i punktacja
Odpowiedź z klucza
(i) curly arrow from the C=C bond to the H atom of HCl AND curly arrow from the H-Cl bond to Cl ✔
correct carbocation «CH3−CH+−CH2CH2CH3, with the positive charge on carbon 2» ✔
curly arrow from a lone pair on Cl− to the positively charged carbon ✔
correct final product «CH3CHClCH2CH2CH3, 2-chloropentane» ✔
(ii) the secondary carbocation is more stable than the primary carbocation «that leads to the minor product» ✔
two alkyl groups release electron density «positive inductive effect» and stabilize the positive charge more than one alkyl group ✔
Schemat punktowania
(i) Award [3 max] for the correct mechanism of formation of the minor product (primary carbocation, 1-chloropentane). Curly arrows must start at a bond or lone pair and end at an atom or bond. (ii) Do not accept Markovnikov's rule, or the carbon with more hydrogen atoms, without reference to the stability of the carbocation.