Zadanie 4e
Wzorowane na: Chemistry HL Paper 2, May 2026, TZ3, pytanie 4(e). Treść, dane i kontekst są zmienione; sprawdzane umiejętności, format i punktacja jak w oryginale.
Darmowe konto pozwoli wrócić do niego później.
Informacja do zadań 4a-4f.
Organic compounds are made from many different starting materials and by a variety of reactions, and some of them are used as fuels.
Polecenie
(i) Explain why 1,2-dichloroethene, ClCH=CHCl, has two stereoisomers but 1,1-dichloroethene, CH2=CCl2, does not.
(ii) State the hybridization of the carbon atoms in propene, CH3CH=CH2.
(iii) Predict the number of signals and the relative areas under these signals in the 1H NMR spectrum of hex-3-ene, CH3CH2CH=CHCH2CH3.
(iv) Sketch the mechanism for the reaction of hydrogen bromide, HBr, with pent-1-ene, CH2=CHCH2CH2CH3, showing the formation of the major product 2-bromopentane. Use curly arrows to represent the movement of electrons.
(v) Draw the two stereoisomers of 2-bromopentane using wedge-dash representations.
(vi) Outline how to distinguish between the two stereoisomers of 2-bromopentane.
(vii) Cyclohexene decolourizes bromine water quickly, but benzene, C6H6, does not. Discuss why benzene does not react with bromine in a similar manner to cyclohexene.
Sprawdź rozwiązanieUkryj rozwiązanieKlucz CKE i punktacja
Odpowiedź z klucza
(i) 1,2-dichloroethene: each carbon of the C=C bond carries two different groups «H and Cl»
OR
1,1-dichloroethene: one carbon of the C=C bond carries two identical groups «two Cl» ✔
(ii) sp3 AND sp2 ✔
(iii) Number of signals: 3 ✔
Relative areas: 3 : 2 : 1 ✔
(iv) curly arrow from the C=C bond to the H of HBr AND curly arrow from the H–Br bond to Br ✔
secondary carbocation CH3−C+H−CH2CH2CH3 drawn with the charge on carbon-2 ✔
curly arrow from a lone pair / negative charge on Br− to the C+ ✔
(v) two mirror-image structures of CH3CHBrCH2CH2CH3 drawn about carbon-2 with the Br and the H on a wedge and a dash ✔
(vi) use a polarimeter
OR
shine plane-polarized light through each sample ✔
the plane of polarization is rotated in opposite directions by the two isomers ✔
(vii) benzene has a ring of delocalized electrons / an aromatic system ✔
addition would destroy the stabilizing delocalization
OR
breaking the aromatic ring needs more energy than breaking the single π bond of an alkene ✔
Schemat punktowania
(i) [1]: accept a structural drawing as part of the explanation. (ii) [1]: both needed. (iii) [1] for 3 signals; [1] for 3:2:1 in any order (accept 6:4:2). (iv) M1: both arrows; M2: carbocation with the charge on the correct carbon; M3: bromide attack. Do not penalize missing partial charges on H–Br for M1. (v) [1]: wedge-dash required, no 90° drawings. (vi) [1] each. (vii) [1] each; accept "resonance structures" for M1.
Komentarz
Part (iv) needs the carbocation on carbon-2 (secondary, more stable than primary); that choice is why 2-bromopentane is the major product.