2020 DHS-RVHS-TJC H3 Prelim Key

1 © DHS / RVHS / TJC 2020 9813/01 [Turn Over Section A 1 (a)  The electrochemical epoxidation can be ran at ambient conditio ns, compared to the high T and P of the silver–based method.  The current methods may produce CO 2 due to overoxidation, leading to an increased carbon footprint  Water is an abundant / safe / sustainable oxygen source compar ed to flammable mCPBA  Generates useful H 2 (g) side–product  Higher atom efficiency compared to the mCPBA method that produ ces a large amount / stoichiometric amounts of RCOOH sideproduct. ( b ) ( i ) (ii) No. of mol of cyclooctene oxide = ହ଴ ଵଶ଺݈݋݉ No. of mol of electrons req = ൈ 2 ൈ ଵ଴଴ ଷ଴ ݈݋݉ Total charge = ܥ c ) ( i ) The Mr of the product epoxide can be determined using mass spectrometry. The molecular ion peak is expected to increase from 126 to 128 if the oxygen source was H218O. (ii)  Use renewable electricity/ development of renewable energy sou rces/ low cost technology that generate electricity e.g. solar cells  Use renewable chemical feedstock e.g. CO 2 / create renewable ethene instead of fossil–fuel derived ethene  Optimising conditions to improve yield of epoxide over ketone  Understanding mechanism for ketone formation  Increase the yield of epoxide by increasing the selectivity to epoxide with which it’s created.  Changing a catalyst to improve selectivity for epoxide over ke tone  Modify the electrolyte to improve the interaction between ethe ne and the anode. ( d ) Since water is in large excess, the reaction is pseudo–zero ord er with respect to [H2O]. Let n be the order of reaction with respect to [cyclooctene]. rate = k’[cyclooctene] n Taking logarithms on both sides of the equation, ሿ

2 © DHS / RVHS / TJC 2020 9813/01 ൌ െ2.56 െ ሺെ3.52ሻ െ0.8 െ ሺെ2.16ሻ ൌ 0.7059 Hence the order of reaction with respect to [cyclooctene] is 0.7. ( e ) ( i ) Ring strain in a 3–membered ring, where the bond angle is constrained to 60° instead of 109.5° f o r t h e s p3–hybridised atoms / additional repulsion between bond pairs destabilises the ring (ii) CH2=CH2 (g) + H2O (l) → H2 (g) + (g) H = ∑BEreactants – ∑BEproducts = BE(C=C) + 2(O–H) + Hvap – [ BE(H–H) + BE (C–C) + 2BE(C–O) – 105] = 610 + 2(460) + 40.8 – [ 436 + 350 + 2(360) – 105 ] = +169.8 kJ mol‒1 (iii) S > 0 as the number of gaseous particles increases from 1 to 2 m ol as the reaction proceeds. Since G = H – TS and |TΔS| increases as temperature increases, G becomes less positive /more negative, reaction becomes more spontaneous. [Total: 16] -3.6 -3.4 -3.2 -3 -2.8 -2.6 -2.4 -2.2 -2 -1.8 -1.6 -1.4 -1.2 -1 -0.8 -0.6 lg (rate) lg [cyclooctene]

3 © DHS / RVHS / TJC 2020 9813/01 [Turn Over 2 (a) (i) Compare the angles of rotation of plane–polarised light by the enantiomerically pure reactant and the stereochemical products using a polarimeter. (ii) SN1 / Unimolecular nucleophilic substitution Compound X forms a stable carbocation that is resonance stabilised as the positive charge can delocalise into the benzene ring. For S N2, the bulky phenyl ring provides steric hindrance to the approach of the nucleophile from the side opposite the leaving group. Moreover for S N1, ratio of % retention to % inversion is about 1:1 / there is only a slight excess of inversion. There is equal probability o f the nucleophile approach either face of the trigonal planar carbocation intermediate. (iii) This minimises the formation of side–products as a result of the attack by the polar solvent molecules which could potentially act as nucleophiles. ( i v ) The negatively charged N 3 i s a stronger/better nucleophile than the uncharged CH3OH molecule so there is a greater extent of SN2. OR The linear N3 anion can approach the side opposite the leaving group with less steric hindrance than the CH3OH molecule so there is a greater extent of SN2. OR The negatively charged N3 is repelled by the negatively charged leaving group so there is lower probability of attack from the same side whe re the leaving group leaves. (v) Formation of ion–dipole interactions between water molecules and the OPNB anion/ leaving group could facilitate the capture of water molecules/ correct orientation of water molecules for attack at the same side of the carbocation, which results in retention of configuration.

4 © DHS / RVHS / TJC 2020 9813/01 ( b ) ( i ) progress of reaction Energy/ kJ mol1 H < 0 Ea1 Ea2 OPNB + CH3CO2  + OCOCH 3 + OPNB + CH3CO2  + OPNB Label  A x e s  E a1, Ea2 and Ea1 > Ea2  H < 0  Reactants, products and intermediate (ii) Yes. Since the first step has a higher activation energy than the second step / the first step is slow while the second step is fast, the carbocation is consumed in the second step as quickly as it is generated. Thus, the concentration of the carbocation intermediate is low and constant for most of the reaction, and the rate of change of concentration of the carbocation intermediate can be approximated to be zero. (iii) The carbocation formed from compound Y is more stable than that formed from compound X due to the presence of the electron–donating methyl group on the phenyl ring which helps to disperse the positive charge. Remark: While both carbocation intermediates are resonance stabilised, the carbocation intermediate formed from Y has the positively charg ed carbon in one of its resonance structures adjacent to the electron–donati ng methyl group. + + + + Using Hammond’s Postulate, since the first step is endothermic, the transition state is product-like. The late transition state for the first step will be of lower energy since it should be similar in energy / geometry to the carbocation intermediate. Hence the rate of reaction is faster with a lower activation energy.

5 © DHS / RVHS / TJC 2020 9813/01 [Turn Over 3 (a) kH/kD > 1 means that the rate of the elimination of HBr from 1-bromo-2- phenylethane is faster than elimination of DBr from 2-bromo-1,1-dideuterio-1- phenylethane, due to less energy needed to break the weaker C–H bond. Since breaking of C–H bond affects the overall rate, this process is involved in the rate determining step. Hence 1–bromo–2–phenylethane is likely to und ergo an E2 mechanism. ( b ) In the gaseous phase, solvent effects can be ignored, so the reaction mechanism is only affected by structure of alkyl chloride and the nucleophile. ( c ) ( i ) S N2 E 2 Note: stereochemistry needed (ii) SN2: the transition state does not involve breaking the C-H or C-D bond, so the strength of bond has minimal effect on the overall rate of reaction, so kH ൎ kD  deuterium KIE ൎ 1. E2: rate determining step involves breaking the C-D bond which is stronger than C-H bond, so kH > kD  deuterium KIE > 1. (iii) As the alkyl chlorides become more substituted from primary to secondary to tertiary, the deuterium KIE increases, implies that E2 reaction becomes the dominant mechanism over SN2. This is because as the alkyl chlorides becomes more sterically hindered around the electrophilic C, it becomes increasingly difficult for the nucleophile to approach and attack the electron deficient carbo n and subsequently easier for the base to abstract a H from the adjac ent C. Hence the probability of the alkyl chloride undergoing S N2 decreases and E2 increases. ( d ) With CH3COO–, deuterium KIE will be lower than 1.71. Since the acid strength of ethanoic acid is higher than HClO, CH3COO– is weaker conjugate base than ClO –. CH3COO– will have a lower tendency to abstract H than ClO –, so S N2 reaction would be favoured over E2 OR the extent of E2 reaction would be l