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-t