2020 ASRJC-NYJC H3 Prelim Key

A COOLABORATION BETWEEN ANDERSON SERANGOON JUNIOR COLLEGE & NANYANG JUNIOR COLLEGE H3 Chemistry Preliminary Examination 2020 Suggested Solutions 1 (a) The active site bind specifically to one enantiomer better than the other enantiomer. The enantiomers of a drug can have different effect on the body. (b) The Lewis acid catalyst interact with the reactant and its ster ically bulky groups effectively shields on one of the stereotopic faces of the reactant, leavin g the other face available for reaction to occur. (c) (i) H 2 gas is evolved at the Pt electrode (cathode). 2H+ + 2e‒ → H2 (ii) N N Ph O Ph + D + 2H+ + 2e Ph O N N Ph O Ph O Ph (iii) N N Ph O Ph + D + H2Ph O N N Ph O Ph O Ph ( i v ) Hr = 2×BE(C‒H) ‒ BE(C‒C) ‒ BE(H‒H) = 2(410) ‒ 350 ‒ 436 = +34 kJ mol ‒1  ( v ) The increase in the total number of gaseous molecules in the sy stem leads to and increase in the entropy of the system. S is positive. Since H is positive and −T S is negative, G is negative only at high temperature. Hence, the reaction is spontaneous at high temperature. ( v i ) Amount of D (77 % yield) = 1.0 406.0 = 2.463 × 10‒3 mol Theoretical Amount of D = 2.463 × 10‒3 × 100 77 = 3.199 × 10‒3 mol Amount of limiting reactant = 3.199 × 10‒3 mol Amount of catalyst = 3.199 × 10‒3 × 5 % = 1.599 × 10‒4 mol Mass of catalyst = 1.599 × 10‒4 × 857.3 = 0.137 g (vii) Theoretical amount of electrons transferred = 2 × 3.199 × 10‒3 = 6.398 × 10‒3 mol I × 11 × 60 × 60 = 6.398 × 10‒3 × 96500 I = 0.0156 A

2 (d) From the Data Booklet, Cu2+ + 2e‒ ⇌ Cu (+0.34 V) Zn2+ + 2e‒ ⇌ Zn (‒0.76 V) Zn → Zn2+ + 2e‒ Since the Eʅ value of Zn is the most negative, Zn will be preferentially oxidized. (e) From Abstract 1, -Rh1 produces R configuration product while -Rh2 produces S configuration product. Hence, -Rh2 is used in reaction 1 as D has a S configuration. N N Ph O O E: must show S configuration (f) 1st order reaction. Rate of formation of bubbles is proportional to the concentration of A. The concentration of B and (CH3)3SiCl are in large excess relative to A, and -Rh1/-Rh2 is a catalyst. Hence, concentration of B, (CH 3)3SiCl and -Rh1/-Rh2 remain constant throughout the reaction. It is a pseudo first order reaction. (g) O Ph [Rh] N O Ph [Rh] N Ph OSi(CH3)3 OSi(CH3)3 Ph 2 (a) Ph CH2Cl HHHH Ph CH2Cl H H H H Ph CH2Cl H HHH I: eclipsed (syn) (most unstable) II: gauche III: eclipsed Ph CH2Cl H H H H Ph CH2Cl H HHH Ph CH2Cl H H H H IV: staggered (anti) (most stable) V: eclipsed VI: gauche

3 (b) (i) In an external applied magnetic field, all protons have their m agnetic moments either parallel with the field or antiparallel to it. As a result, the degenerate spin states split into two states of unequal energy. The energy difference between the two levels is small and falls within the radio frequency region. The nuclear magnetic resonance phenomenon occ urs when nuclei parallel with an applied magnetic field are induced to absorb e nergy and change their spin orientation such that it is antiparallel to the applied magnetic field. (ii) OH (iii) Cl F O O OH G O - Na + O OH O Cl O step 1 step 2 step 3 step 4step 5 step 6 * *** H step 1: ethanolic NaOH, heat under reflux * step 2: H2O(g), H3PO4 catalyst, 300oC, 60 atm or conc. H 2SO4, room temp followed by H2O(l), heat * step 3: I2(aq), NaOH(aq), warm * step 4: HCl(aq) or H2SO4(aq), room temp * step 5: PCl5 or SOCl2, room temp * step 6: phenol in NaOH, room temp * 0o 6 0 o 120 o 180 o 240 o 360 o I II III IV V VI Energy / kJ mol‒1 angle of rotation

4 3 (a) (i) (ii) Bond order of F2 = 1 2 (10  8) = 1 (iii)  molecular orbital: * molecular orbital: (b) (i) Ratio of relative rates of production of J and L for bromine is 1 : 4.03. (accept 0.248) Ratio of relative rates of production of J and L for chlorine is 1 : 3.42. (accept 0.293) Bromine has a larger atomic size than chlorine. Hence, bromine would favour the substitution at the 4-position as there will be less steric hindrance as compared to being substituted at the 2-posi tion where it will be in close proximity with the methyl group. LUMO HOMO

5 (ii) Using the equation Ea = A + BH, For the production of J for bromine: +84.3 = A + B(14.7) -------(1) For the production of K for bromine: +95.9 = A + B(+1.75) -------(2) Taking (1)  (2): (+84.3  (+95.9)) = B(14.7  (+1.75)) 11.6 = 16.45B B = 0.7051 Hence, +84.3 = A + 0.7051(14.7) A = 94.66 For the production of L for bromine: +80.8 = 94.66 + 0.7051H H = 19.7 kJ mol1 (iii) 4 (a) (i) Organic molecules that contain atoms with lone pairs of electro ns,  bonds or conjugated  bond systems can absorb energy in the UV region of the spectrum. Their electrons can be excited from a bonding molecul ar orbital to an anti-bonding molecular orbital (or HOMO to LUMO). For UV absorption, only the following transitions:   *, n  * and n  * are allowed. Molecules with more conjugated  bond systems will give rise to a smaller energy gap between the HOMO and LUMO. Hence, electrons will abs orb UV at longer wavelengths and they may appear coloured. (ii) p-courmaric acid has higher degree of conjugation with additiona l double bond as compared with gallic acid. This decreases the gap between HOMO and LUMO, thus shifting its absorptions to a longer wavelength. (iii) Using Absorbance (A)= cl Given that absorbance 265 nm = 1.50 c = = 1.74 x 10‒4 mol dm‒3 (b) (i) M and N are diastereomers. 1.5 8640 1 A l   CH3 Br + CH3 Cl + 19.7 –20.8 Energy Reaction Coordinate +72.8 +80.8

6 (ii) Cl C(CH3)3 H CH3 (iii) C(CH3)3 CH3 ( i v ) M (more stable conformation) CH3 (CH3)3C H Cl H (less stable conformation) anti-periplanar Cl C(CH3)3 H CH3 CH3O- CH3 C(CH3)3 -HCl (v) M only has only one hydrogen that is in anti-periplanar arrangem ent to the chlorine. Hence, M only forms one E2 product. M can only undergo E2 elimination when it ring flip to the less stable chair conformation. This causes the bulky –C(CH 3)3 group to be in the axial position which results in 1,3-diaxial steric interaction. ( v i ) CH3 C(CH3)3 CH3 C(CH3)3 Q R Alkene Q is a more highly substituted alkene, and hence it is more stab le (Zaistev product) as compared to alkene R.

7 a b c d e f Hg Hg Hg 5 (a) n = Hence, there are 11 carbons in Compound S. From the NMR spectrum, there are 17 H in Compound S. 195 – (12 x 11) = 63 mass units for H+N+O 63 –17= 46 mass units for N and O. Hence, there is only 1 N and 2 O to give mass units of 46. Formula of compound S is therefore C 11H17NO2. (b)  / ppm splitting number of protons deductions 1.2 Triplet 3 Three protons (Ha) belong to -CH3 adjacent to a -CH2. 2.0 singlet 1 One proton (Hb) belong to the labile proton in NH group, which undergo a proton exchange with the solvent (spin decoupling), hence it is a singlet. 2.6 Singlet 3 Three protons (Hc) belong to -CH3 with no adjacent H and next to an electronegative atom N. The peaks shifted more downfield due to the inductive effect of the adjacent electronegative nitrogen, causing the protons to be more deshiel