HCI Prelim ANS

H3 Chemistry Preliminary Examination 2009 Mark Scheme Setter & Marker: Ms Grace Chua 1 a) Narcotic analgesics work by depressing the CNS , hence affecting the capacity of the brain to appreciate pain [1] Non–narcotic analgesics work on the pain receptors themselves, preventing them from responding normally to pain stimuli. [1] b) Natural products, current drugs etc [2] ci) Mode of action (1) Aspirin Mode of action (2) Ibuprofen Mode of action (3) Indomethacin [1] Mechanism (1) takes place with the formation of a covalent bond with serine. Aspirin can form an ester linkage with the hydroxy residue. [1] Mechanism (2) should take place with substances that mimic arachidonic acid. Ibuprofen has the most similar structure to arachidonic acid. [1] Mechanism (3) should involve electrostatic interactions between the basic group on arginine and hydrogen bonding/ dispersion forces with the tyrosine group for it to be reversible. Indomethacin fits the description. [1] cii) N NS O O H2N CF3 1 m for each reasonable interaction [2] di) Precursor: N HO O CH3 Cl CO2H [1] Reagents and conditions: (1) NaOH(aq) (2) CH3I, heat [1] dii) SN2. [1] If the mechanism were SN1, the carbocation formed is very unstable/ less steric hindrance Hydrogen bonding with tyrosine/ serine Electrostatic forces of interaction with arginine Hydrophobic interactions with t yrosine C2 Chemistry H3 Prelim 2009_answers 9812/01/HCI 1

C2 Chemistry H3 Prelim 2009_answers 9812/01/HCI 2 r primary halide therefore S 2 favoured. [1] , if not no marks awarded. fo N Note: Answer must be accompanied with reason e) OHO OH H N CH3 R S S R S Z [1/2] for each correct assignment. [3] (f) H CH2SHR CH3Ph H H CH2SHR PhH3C H H RHSH2C PhH3C H H RHSH2C CH3Ph H Enantiomers Enantiomers Diastereomers Diastereomers Diastereomers Compound A Diastereomers [3] [Total: 20m] Setter & Marker: Ms Jessie Koh 2 ai) Rate of hydrolysis A<penicillin-G<B [1] A should hydrolyse slower due to the electron-withdrawing C l group, B should hydrolyse faster due to the electron-donating CH3CH2 group. [1] The electron-withdrawing C l group withdraws electrons from the C=O attached to the R group, thus makes the O of this C=O a poorer nucleophile which will have lower tendency to attack the carbonyl carbon of -lactam ring to cause ring-opening. [1]

ii) 1 mark for each equatio n: NHCH3 C Cl H H + NaOH N N H2N NH2 S-Na+ + N N H 2N NH2 SH N N H 2N NH2 S-Na+ N N H 2N NH2 NHCH3 S C H H 2H2O2 N N H 2N NH2 NHCH3 S O O C H H N N H 2N NH2 NHCH3 S C H H [3] bi) Agonists compete for the receptor site, causing the necessary change in shape there. Antagonists block the site without causing the necessary change in shape. [2] bii) Agonist: Compound I, III, IV. Agonist should promote growth hence tumor grows to a larger size. [1] biii) Hydrogen bonding of phenol group [1] Van der Waals forces of attraction of hydrophobic benzene ring or alkyl group [1] (-1 m if student identifies ether) c) IR spectrum for compound III contains broad ─OH stretch at about 3300- 3500 cm -1 that is missing in IR spectrum for compound II. [1] di) TMS is used in 1H NMR spectroscopy because it has 12 hydrogen atoms all of which are in exactly the same environment, i.e. they are equivalent and absorb energy at the same frequency. This produces a single but intense peak. [1] Other accepted answers: It is inert and would not react with the sample. /It is non-toxic, hence, safe to handle. /It is volatile (boiling point = 26.6 0C ) and can be distilled dii) 7.3 7 2.3 1.3 0 ppm 2 Hc 4 Ha 2 Hb 8 Hd TMS HO OH c d a b C2 Chemistry H3 Prelim 2009_answers 9812/01/HCI 3

 Axis and TMS peak [1]  NMR spectrum shows 4 peaks: Ha, Hb, Hc and Hd [1]  NMR spectrum shows correct splitting of the 4 peaks: Hd is multiplet, Hc is singlet, Ha is triplet, Hb is multiplet (quintet) [1]  NMR spectrum shows correct relative area integrations under the peaks: Ha: Hb: Hc: Hd 4 :2: 2 : 8 (or 2:1:1:4) [1] diii) The Hd protons are aromatic protons which are very deshielded in the 7 – 8ppm region, due to anisotropic diamagnetic effect shown by H atoms attached to a benzene ring. [1] Hc protons are also deshielded as it is bond to the electron withdrawing O atom , thus the electron density around the H nucleus is reduced, resulting in Hc protons being deshielded. [1] div) The singlet due to the phenolic –OH will disappear in the presence of deuterated water as it is a labile proton. [Total: 20m] Setter & Marker: Mr Colin Loy 3 ai) As viruses use the host cell's biochemical reacti ons to reproduce themselves, it is difficult to design drugs which target the viruses but do not affect healthy cells as well. [1] aii) Antibody: protein whose synthesis in the bod y in B lymphocytes is triggered by the presence of a specific antigen on the surface of an invading bacterium or virus. [1] Monoclonal: antibody which has been isolated and generated in unlimited quantities from one particular B lymphocyte. [1] aiii) J has an additional NH 2 group which may exist as a charged NH 3 + at physiological pH leading to favourable interactions with water. [1] In catalyzing deamination, the enzyme adenosine deaminase is rather specific. J may now be too different from the natural substrate adenosine/ unable to fit into the active site due to larger size. [1] bi) (1 m for diagram of setup & 1 m for electrophoretogram) [2] electrophoretogram Buffer at pH 5.9 gl ycine cysteine glutamic acid C2 Chemistry H3 Prelim 2009_answers 9812/01/HCI 4

bii) At pH 5.8, glutamic acid and cysteine are positively charged. Serine if present would also be positively charged and there would have been 3 amino acids (3 spots) migrating towards the anode. [1] Glycine is negatively charged and migrated towards the cathode as shown in the expected electrophoretogram. [1] ci) All have cyclic/ring structure with overlapping p orbitals/  electron from carbon atoms or heteroatom. [1] All satisfy Huckel's Rule (4n+2 π electrons) with n=1 for benzene, pyridine and pyrimidine and n=2 for quinoline. [1] cii) Nitrogen is more electronegative than carbon and withdraws  electron density from the ring. [1] Or The nonbonding electrons on the nitrogen are perpendicular to the  system and cannot stabilize the positively charged intermediate effectively. [1] ciii) For 5-bromopyridine: The positive charge in the intermediate does not lie at any of the nitrogen atoms in any of the resonance structures, which would be unfavourable. [1] For 4-bromopyridine: civ) The N in the amide group is least basic (N A) as the lone pair is delocalized over the 3 atoms O-C-N and is hence unavailable. The N in the pyridine is basic (N B) as it is available for coordination to form pyridinium ions. The N in the tertiary amine is most basic (N C) as the availability of its lone pair is enhanced by the electron donating effect of the alkyl groups. cv) [Total: 20m] NH2 N O N C B A C2 Chemistry H3 Prelim 2009_answers 9812/01/HCI 5

Setter & Marker: Ms Liew PC 4 ai) Placebo is a “blank” pill/ injection that contains no active pharmaceutical agent, but should look and taste as similar as possible to the pre