HCI H3 CHEM P1 KEY

C2 Chemistry H3 Prelim 2011 Answers 9812/01/HCI 1 Answers for 2011 H3 Chemistry Prelims 1 (a) (i)  Disruptions of the synthesis of folic acid  Disruptions of protein synthesis  Plasma membrane disruption  Disruption of nucleic acid transcription Any 2 of the above. (ii) N S R O OH O HN S R O OH O OR RO N SR O OH O O R H H HN SR O OH O O R O H H HN SR O OH O OH O H H + ROH NH S R O OH O O R HN S R O OH O OHRO H (iii) Penicilloic acid does not retain the rigid conformation as that of penicillin and thus will not be able to effectively interact with the D-Ala-D-Ala termini in glycopeptides transpeptidase. Accept if students talk about the fact that the strained -lactam ring is essential in the SAR. (iv) It is an irreversible inhibitor as it covalent ly modifies the enzymes and this inhibition cannot be reversed. (b) O OR OH H2NOCO H3CO CH3 H3C O OR OHH3C CH3 OCH3 OCONH2 The more stable conformation is indicated in the box above. It is more stable as it has minimal 1,3-diaxial interaction (two as compared to four). [Accept if student mention anomeric effect] (c) (i) Pivampicillin is more lipophilic (non-polar) and thus can be better absorbed through the gut wall. (ii) The ester group further away from the penicillin nucleus, is less sterically hindered as compared to the other ester group and thus more susceptible to attack by nucleophiles. Hence it undergoes hydrolysis more easily.

C2 Chemistry H3 Prelim 2011 Answers 9812/01/HCI 2 (iii) esterase O H O (d) (i) Around the molecular ion peak, isoniazid-A and isoniazid-B will give the following pattern respectively: isoniazid-A isoniazid-B MM + 2 M + 4 MM + 2 M + 4 3 : 4 : 1 1 : 2 : 1 (ii) (I) NaCN; nucleophilic substitution (II) H2NNH2 2 (a) (i) Irreversible inhibition of the cyclo-oxygenase (COX) enzyme that it needed to turn arachidonic acid into prostaglandins. As prostaglandins are hormones responsible for the transmission of pain information to the brain, inhibiting the enzyme will reduce the pain felt, hence the analgesic effect. (ii) Since aspirin is less polar than benzoic acid, aspirin has greater affinity for the non-polar stationary phase and it eluted later. Thus peak 3 is due to aspirin. (iii) Peak area of component 1 = 930.02.63.02 1  Peak area of component 2 = 625.05.25.02 1  Peak area of component 3 = 800.06.10.12 1  Ratio of component 1 : 2 : 3 = 0.930 : 0.625 : 0.800

C2 Chemistry H3 Prelim 2011 Answers 9812/01/HCI 3 (b) Number of carbon = (100/1.1)(15.4/100) = 14 Molecular formula = C14H22N2O  Deduction 1.13 (t 6H) 2 CH 3 groups next to CH2 2.29 (q 4H) 2 CH 2 groups next to CH3 2.23 (s 6H) 2 CH 3 groups with no neighbouring proton 3.22 (s 2H) CH2 with no neighbouring proton; deshielded  next to electronegative atom N 7.09 (m 3H) 3 protons of tri-substituted benzene 8.92 (s 1H) Labile proton of NH Wavenumber / cm1 Deduction 3250 NH stretch 1675 C=O stretch 2800 - 3000 sp3 CH stretch 1500 aromatic C=C stretch Fragment ion at m/e 86: [CH2N(CH2CH3)2]+ Structure: N C H O C N CH2CH3 H H CH2CH3 H3C H3C (Can accept the methyl groups to be 2,6 substituted and/or interchange NH with CO in the amide group) (c) (i) Step 1: NaOH (or Na), CH 3Br, heat Step 2: CH3COCH2CH2CO2H (ii) NH CH3O CO2H NH CH3 CH3O CO2H NH CH3 H H+ N H N CH3 CH3O CO2H NH2H D E CH3O NH N CH3 CO2H H C H

C2 Chemistry H3 Prelim 2011 Answers 9812/01/HCI 4 3 (a) (i) Different modes of vibrations in a molecule give rise to different vibrational energy levels, such as stretching and bending. For a vibrational mode to be active in the IR region, there must be a net change in the dipole moment of the molecule when the vibration takes place. (ii) BeCl2 has 2 absorption bands caused by asymmetric stretching and bending. Cl2O has 3 absorption bands caused by symmetr ic stretching, asymmetric stretching and bending. (b) (i) serotonin: 2 characteristic peaks for N-H stretch (~3400 cm−1) [also accept 3 peaks] mCPP: 1 characteristic peak for N-H stretch (~3400 cm−1) serotonin: Presence of peak for O-H stretch (~3300 cm −1) mCPP: Absence of peak for O-H stretch (ii) Benzene ring: dispersion forces N atom beside the benzene ring: hydrogen bonding or ionic interactions The other amine group: hydrogen bonding or ionic interactions (c) (i) Compound G is able to bind to the receptor site via intermolecular forces similar to serotonin. Because it is able to fit better in receptor site, it is able to block the site without changing the shape of the receptor, hence there is no biological effect. (ii) Compounds G and H give one and three reduction products respectively. Product from G: H N N CH3 CH2CH3 CH3 H N CF3 Two of the products from H are: OH CF3 HO N CH3 CH2CH3 CH3 The last product from H can be any of these: OHHO OH H CH3OH (iii) The amide bonds are more resistant to hydrolysis than the ester bonds due to the partial delocalisation of the nitrogen lone pair over the carbonyl group. Hence compound H is more susceptible to hydrolysis. (iv) Solubility: Any suggestion that will form more hydrogen bonding with water (e.g. adding OH and NH2 groups, replacing benzene ring with pyridine). Flexibility: Introduce a ring to the urea functional group to make the structure more rigid (e.g. reacting both NH groups with Cl-CO-CH 2-CO-Cl to form a ring)

C2 Chemistry H3 Prelim 2011 Answers 9812/01/HCI 5 (d) (i) NNN N N N Deduct −0.5 for any mistake. Driving force: Formation of a 6 electrons aromatic system. (ii) Compound J is less conjugated than PTP1B inhibitor, therefore the gap between the energy level is greater. The maximum absorption will occur at a shorter wavelength for Compound J. 4 (a) (i) A stimulant is a drug that acts on the nerve synapses to "wake up" the central nervous system. Accept any 2 of increased heart rate, blood pressure, respiratory rate, blood sugar levels, adrenaline levels etc. (ii) Amphetamine binds to the protein carrier on the surface of the presynaptic nerve, allowing stores of neurotransmitter to leak out of the nerve to the synaptic cleft. It inhibits the re-uptake of noradrenaline and dopamine into the presynaptic nerve. (iii) As a prodrug, Vyvanse is in itself inactive, and time is required to hydrolyse the amide bond to get the active amphetamine, so because there is no euphoric rush, there is much less potential for abuse. (iv) The body builds up a tolerance to the stimulant. Assuming a drug acting as an antagonist at the receptor, depriving the receptor of its natural ligand may induce the cell to synthesise more receptors over time. Hence a higher dose of the antagonist is required to suppress the binding of the natural ligand / for the same biological effect. (to accept explanation based on agonist as well) (v) CH2Ph H CH3 N C H O H2N H CH2CH2CH2CH2NH2 S S (vi) K: O L: NH2 OH NH2 OH or

C2 Chemistry H3 Prelim 2011 Answers 9812/01/HCI 6 (b) N Cl OH N HO + Cl N The intermediate is stable as there are electron-donating ethyl groups attached to the nitrogen / participation of the neighbouring amine group. The OH − nucleophile attacks the less sterically hindered carbon of the intermediate, hence the primary alcohol is formed. (c) (i) Lithium is more electropositive than silicon. Protons of CH3Li more shielded than those in TMS. (ii) CC Ha Ha O H CH3 CC Ha Ha O H CH3 By mesomeric / resonance effect, the oxygen atom increases electron density at the terminal carbon, hence Ha is more shielded and δ is lower. 5 (a) (i) Movement of