2023 RI H3 Chemistry Prelim Key Answer

© Raffles Institution 2023 9813/01/S/23 2023 Y6 H3 Chemistry Preliminary Exams – Suggested Solutions 1(a) It refers to the balance between high energy content with low sensitivities/good thermal stability (safety)/OWTTE. 1(b) The integration approach relies on the known properties of different types of structures/moiety, thus combining the advantages of each moiety in the design of the compound (OWTTE). Or The integration approach may accelerate the development of advanced EMs instead of developin g a molecule from scratch. 1(c)(i) AEMs need to have high energy content so that large amounts of energy can be released when detonated. 1(c)(ii) 2C(s) + N2(g) + H2(g) + ½O2(g) (g) 2 Hatomo of C BE(C–C) + 2 BE(C–H) + BE(N≡N) + BE(H–H) + 2 BE(N–O) + 2 BE(C=N) + BE(O=O) 2C(g) + 2N(g) + 2H(g) + 2O(g) Hf = [2 Hatomo of C + BE(N ≡N) + BE(H –H) + ½BE(O=O)] – [BE(C –C) + 2 BE(C –H) + 2 BE(N–O) + 2 BE(C=N)] = [(2x715) + 944 + 436 + ½(496)] – [350 + (2x410) + (2x201) + (2x610)] = +266 kJ mol–1 1(c)(iii) Standard enthalpy change of vaporisation of (liquid) 1,2,5-oxadiazole. 1(d) Tricyclic structure in compound A provides larger π-conjugated systems which increases the stability of the molecule. Hfo

© Raffles Institution 2023 9813/01/S/23 1(e) 1(f)(i) Condensation / addition-elimination 1(f)(ii) 1(f)(iii) (CH3CO)2O acts as a Lewis acid as it accepts an electron pair from HNO 3 leading to the formation of acetyl nitrate. H 2SO4 acts as a Bronsted-Lowry acid as it protonates HNO 3 leading to the formation of the NO2+ electrophile. 2(a)(i)  Amide J is less basic than amine H due to the delocalisation of the lone pair of electrons on the N atom into the C=O. H ence, the lone pair is not available to accept a proton.  Amidine K is more basic than amine H as the delocalisation of the nitrogen lone pair in the NH 2 group increases the electron density on the NH group. Accept “positive charge on conjugate acid is delocalised over both N atoms, stabilising the conjugate acid”.

© Raffles Institution 2023 9813/01/S/23 2(a)(ii) L is more basic than K as the additional NH can donate electron density onto the underlined NH, increasing electron density on the underlined N atoms. 2(a)(iii) The reaction of K is slower. The amide carbon in J is more electron deficient than the amidine carbon in K as oxygen is more electronegative than nitrogen, and is hence more susceptible towards nucleophilic attack. 2(a)(iv) The C=O bond in J absorbs at higher frequency than C=N bond in K as the C=O bond is stronger (740 kJ mol−1) than the C=N bond (610 kJ mol−1). Can accept explanation based on lower atomic mass of N leading to higher frequency of C=N 2(b)(i) Since the electronegative iodine atom is closer to the negativ ely-charged C atom in CH3COCHI− than that in −CH2COCHI and is able to disperse the negative charge to a larger extent, the CH 3COCHI− intermediate is more stable/ lower in energy and formed faster. 2(b)(ii) 2(b)(iii) Using the same concentrations and volumes of iodine, NaOH and the ketones,  Pour iodine and NaOH in a conical flask. Place the flask over a mark.  Add in ketone and start stopwatch immediately.  Stop stopwatch once the yellow ppt obscures the mark completely / time taken for first appearance of yellow ppt. 2(b)(iv) (1) The kinetic product Q is favoured when a bulky base li ke LDA is used. There is less steric hindrance on a primary α-carbon atom compared to a secondary α-carbon atom, hence less E a is required to form the kinetic intermediate, which is formed faster. 2(b)(iv) A shorter reaction time favours the kinetic product which is formed faster and does

© Raffles Institution 2023 9813/01/S/23 (2) not allow time for the thermodynamic product to be formed to an appreciable extent. AND A low temperature favours the kinetic produc t as it does not allow the kinetic product to overcome the Ea barrier to form the thermodynamic product Or Low temperature reduces the proportion of reactant molecules that can overcome the activation energy required to form the thermodynamic product J. 2(c) CHI3 has one molecular ion peak at m/z = 394 corresponding to [CH127I3]+. CHCl3 has four molecular ion peaks: species [CH35Cl3]+ [CH 35Cl237Cl]+ [CH 35Cl37Cl2]+ [CH 37Cl3]+ m/z 118 120 122 124 relative abundance 3 32 7 46 4   27 : 2 312 73 446 4      27 : 2 31 93 44 6 4      9 : 3 11 46 4   1 3(a) 2 electrons in one sp2 orbital, 1 electron in each of remaining orbitals 3(b)(i) The electron density of the pyridine ring is decreased by the electron-withdrawing effect of the electronegative nitrogen atom. 3(b)(ii) p sp2 sp2 sp2 • • • • •

© Raffles Institution 2023 9813/01/S/23 Attack at the C4 position results in the positive charge on the electronegative N atom, giving rise to a very unstable ca nonical form. Hence, substitution at C4 occurs less readily. 3(c)(i) 3(c)(ii) Compound U: 2 doublets and 2 doublet of doublets Compound S: 2 doublets 3(d)(i) The cationic intermediate formed with pyridine re acts faster with alcohol as the partial positive charge on the acyl carbon is intensif ied by the electron-withdrawing positively charged nitrogen atom. 3(d)(ii) Pyridine is protonated in the course of t he reaction and is consumed by the acid-base reaction. 3(d)(iii) 4(a) TMS is commonly used as a standard as  It gives an intense sharp signal even at low concentrations since it has 12 chemically equivalent protons.  It is soluble in most organic solvents.  It is chemically inert and has a low boiling point (26.5 oC) so that it is easily removed from a recoverable sample of a valuable organic compound.  It resonates at a frequency lower than that of any 1H nuclei in common organic molecules and so its peak does not interfere with the other peaks. 4(b) W has 1H NMR peaks present at 6.8 and 7.5 ppm which represent aromatic protons, while V does not. The higher degree of conjugation in W will cause the energy gap of   * to decrease, shifting the UV absorpti on to a longer wavelength. Therefore, W will have a greater λmax than V.

© Raffles Institution 2023 9813/01/S/23 4(c) IR Analysis absorption / cm–1 Functional group present in V 1640 – 1750 absent No C=O stretch, no ketones 3030 C=C stretch, alkene present 3360 (w) N-H stretch, amine present 1H NMR Analysis chemical shift / ppm no. of H multiplicity deductions 4.8 1 singlet NH proton 3.9 2 triplet Two separate –CH groups (Hb) with neighbouring –CH2 group (Ha) Hb shows higher chemical shift than Ha due to closer proximity to the electronegative N atom. 3.2 2 triplet –CH2 group (Ha) with two neighbouring –CH (Hb) groups with chemically equivalent protons. 2.0 6 singlet 2 isolated –CH 3 groups (Hc) 6.8 and 7.5 for W - - Aromatic protons (V could be cyclized from 1,5-diketone and NH3) V: W: 4(d) At 280nm, 0.86 = 12000[V](1.0) + 5600[W](1.0) --- equation (1) At 300nm, 0.41 = 3400[V](1.0) + 14200[W](1.0) --- equation (2) From eqn(1), [V] = (0.86 – 5600[W]) / 12000 substitute into eqn (2) 0.41 = 3400((0.86 – 5600[W]) / 12000) + 14200[W] 0.41 = 0.2833(0.86 – 5600[W]) + 14200[W] 0.41 = 0.2436 – 1586.48[W] + 14200[W] [W] = 1.319  10 –5 mol dm–3 Sub in eqn (1) 0.86 = 12000[V] + 5600(1.319  10–5) 0.86 = 12000[V] + 0.073876 [V] = 6.551  10–5 mol dm–3

© Raffles Institution 2023 9813/01/S/23  [V] = 6.55  10–5 mol dm–3, [W] = 1.3