TJC H2 CHEM P1 ans Prelim

2016 Preliminary Examination H2 Chemistry MCQ Solution 1 2 3 4 5 6 7 8 9 10 A B B D A D D C B B 11 12 13 14 15 16 17 18 19 20 D B D C C A C C D A 21 22 23 24 25 26 27 28 29 30 A D A C C B B A A D 31 32 33 34 35 36 37 38 39 40 B C A B D B A B D C

16. A Observation 1: H2SO4 + Cl– HCl + HSO4– HCl bubbled into Br –, no visible reaction is observed and solution remained colourless. Observation 2: Addition of Ag+ into solution containing Cl– and NH3(aq) gives a colourless solution containing [Ag(NH3)2]+ as AgCl is soluble in aq NH3 17. C A: Charge of Group II metal ions is the same but the ionic radius increases down the group. Hence, with decreasing charge density, the magnitude of Hhyd decreases down the group. B: Solubility of Group II sulfate decreases down the group. C: Down the group as the charge density of the metal ion decreases, its polarisation power decreases, thus there is less weakening of the C - O bond resulting in greater stability. D:Tendency to form complexes should be decreasing as the charge density of the metal ion decreases down the group. 18. C Blood red seen is due to the complex formed in the following reaction: [Fe(H2O)6]3+ + SCN– H2O + [Fe(H2O)5(SCN)]2+ Blood red In the presence of an alkali, red- brown precipitate of iron(III) hydroxide is formed as well thus diluting the blood red colour making the letters appear orange brown. [Fe(H 2O)6]3+ + 3OH– 3H2O + [Fe(OH)3(H2O)3](s) red-brown ppt 19. D Energy level diagram indicates in the single reaction step, the reactant(s) undergo bond breaking only ie reactant(s) absorb energy. There is no bond formation taking place. Hence, true for option D only. In options A and C, bond formation only occurs. In option B, there is bond breaking followed by bond formation in the reaction. 20. A Oxo reaction: Similarly for but-2-ene, 21. A 2,4-DNPH and I 2/NaOH(aq) doesn’t react with X and Y. X is an ester, not a carbonyl compound. Y reacts with NaOH(aq), but it doesn’t give observable result. Y reacts with Na to give effervescence of H 2. 22. D CH 3CH2CH=CHCH2CH2OH Cr2O72-/H+ CH 3CH2CH=CHCH2CO2H which goes through hydrogenation to form CH 3CH2CH2CH2CH2CO2H. 23. A Nucleophilic substitution gives CH3CH2OD. B) oxidation to give CH3CO2- C) Elimination to give CH2=CH2 D) Condensation to give CH3CO2CH3

aq NaOH, heat 24. C Z is an aldehyde and contains –OH group. Upon reduction by H2, Z forms a product that contains chiral carbon and no plane of symmetry. A) Does not contain –OH B) Is not an aldehyde C) Contains –OH and aldehyde. Product of reduction reaction is a chiral compound. D) Product from reduction CH 2(OH)CH(CH3)CH2OH does not have chiral carbon. 25. C As shown in the diagram above, there is no sp3 C atom overlapping with sp2 N atom. 26. B Maximum number of stereoisomers = 2n+m = 27+0 There are 7 chiral carbons in compound. There isn’t any C=C that can display cis-trans isomerism. 27. B NC Br CONHCH 3 Br                        28. A The lower the pKb, the stronger the base. Hence, molecules should be arranged from weakest base to the strongest base. is the least basic due to the close proximity of the electron withdrawing Cl atom, which decreases the electron density on the N atom. is the most basic due to the presence of two electron release groups, and the absence of any electron withdrawing groups. 29. A Trypsin treatment: arg arg gly Chemotrypsin treatment: tyr arg tyr arg gly tyr arg tyr arg gly Cyanogen bromide treatment: tyr arg tyr met arg gly tyr arg met tyr arg gly Primary structure: pro–tyr-arg-met-tyr-arg-gly Note: A much faster way would be to analyse the options given, instead of solving in the forward direction. CH3 H * acidification X  *

30. D General rules Taking the neutral form of the dipeptide into account, If pH > pI, –COOH will be deprotonated, resulting in a negatively charged dipeptide If pH < pI, –NH2 will be protonated, resulting in a positively charged dipeptide If pH = pI, the dipeptide is electrically neutral pH 7.4 > pI of carnosine (pI = 6.83)  deprotonation 31. B In the cubic structure, each B atom is bonded covalently to 4 N atoms in a tetrahedral manner, where one of the bonds is a dative bond (where N lone pair of electron is donated to B atom). Like graphite, there are extensive van der Waals forces of attraction between the layers in hexagonal boron nitride. Option 3 is incorrect. The boron-nitrogen bond in hexagonal boron nitride is shorter and stronger than that in cubic boron nitride, due to the pi bond present. Each B atom uses all its three valence electrons to form covalent bonds with three neighbouring nitrogen atoms. Each nitrogen atom still has a lone pair of electrons which it uses to form a dative pi-bond with an adjacent boron atom. 32. C 33. A EƟzn2+/Zn = -0.76 V & EƟCu2+/Cu= +0.34 V For anode, oxidation: select EƟ that is more negative. Hence, Zn is preferentially discharged. 1 is correct. For cathode, reduction: select E Ɵ that is more positive. EƟzn2+/Zn = -0.76 V & EƟH+/H2= +0.00 V Hence, H2 evolved. H+ (aq) + e →½H2(g), effervescence observed. 2 is correct [H +] decreases, pH of solution increases. 3 is correct. 34. B Statement 1 is correct. Increasing the temperature favours the endothermic (forward) reaction, and by Le Chatelier’s principle, the position of equilibrium will shift to the right. Statement 2 is correct. Since Fe 3O4(s) and FeO(s) are both in solid states, they are not part of the KC expression. KC = [CO2(g)] / [CO(g)] Therefore KC has no units. Statement 3 is not correct. Adding a solid reactant or product does not affect the position of equilibrium. 35. D 1. pH of aqueous solution of chlorides decreases from 6.5 (MgCl 2) to 1 (PCl5). 2. maximum oxidation states of the elements in the chlorides increases from MgCl2 to PCl5. 3. All are insulators in the solid state from MgC l2 to PCl5. 36. B Option 1: Due to its greater nuclear charge, Cu has a higher atomic mass and smaller atomic volume. Hence, its density is higher than Ca. Option 2: Metallic bond for Cu is stronger since both 3d and 4s electrons are delocalised compared to the 4s electrons only for Ca. Hence, more

energy is needed during melting accounting for its higher m pt. Option 3: Electrical conductivity for Cu should be higher since it has a greater number of delocalised electrons (both 3d and 4s) to conduct electricity compared to the 4s electrons only for Ca. 37. A Statement 1 is correct. The magnitude of the equilibrium constant is large (>>100), which implies the position of equilibrium lies far to the right. Statement 2 is correct. Since K 2 is larger in magnitude than K1, it implies that en has a higher tendency to replace the water ligands than ammonia. Statement 3 is correct. K stab = __[[Ni(NH3)6]2+]__ [[Ni(H2O)6]2+][NH3]6 K’stab = __[[Ni(en)3]2+]___ [[Ni(H2O)6]2+][en]3 K’’stab = [[Ni(en)3]2+][NH3]6 [[Ni(NH3)6]2+][en]3 = __[[Ni(en)3]2+]__ x [[Ni(H2O)6]2+][NH3]6 [[Ni(H2O)6]2+][en]3 [[Ni(NH3)6]2+] = K’ stab / Kstab 38. B Statements 1 and 2 are correct. Reactant in stage III is saturated, it cannot undergo addition reaction. Stage III is elimination. 39. D Option 1 is correct. Protonation of R-groups of the residues in the tertiary structure disrupts existing ionic interactions and hydrogen bonds between them, which results in denaturation. Option 2 is incorrect. The amide functional group in the peptide linkages are neutral, and will not react with H + Option 3 is incorrect. Primary structure is only affected during hydrolysis. Complete hydrolysis will only occur when ovalbumin is heated for a prolonged period of time under highly acidic