SRJC_2008 JC2 Prelim H2 Paper1 Solutions

Solution H2 MCQ Prelim 2008 1) Solution: B Amount of H3PO3 used in the experiment = 30 1000 x 0.05 = 1.50 x 10-3 mol Amount of NaOH neutralised = 15 1000 x 0.2 = 3 x 10-3 mol H3PO3 ≡ 2 NaOH 2 mol of NaOH neutralizes 2 mol of H + Ö 1 mol of H + is left (since H3PO3 ≡ 3H+) Ö Salt formed will contain 1 mol of H + 2) Solution: C Large jump in 2nd Ionisation energy, 2nd electron removed is found in inner shell. G has 1 valence electron; G is a Group I element. Thus, the oxide of G has the formula: G 2O 3) Solution: A Using pV = nRT, At constant T, pV = constant, so graph should be a straight line at all density 4) Solution: B O H OC H C H N H C OH O I I I I H H Tetrahedral about C Bent about O Trigonal planar about C Trigonal pyramidal about N 5) Solution: C To show an energy profile diagram with two activation energies and the first activational energy is greater than the 2nd one as the 1st step is the slow step.

6) Solution: B Kc = [] [] [ ]2 2 2 I H HI no units 7) Solution: B Initial pH of 13 indicated that a strong base was used as it dissociates fully to give OH -. The final pH of 2.5 indicates a weak acid buffer region due to presence of CH 3COOH with CH 3COO-Na+. The end point is basic due to hydrolysis of the salt CH3COO-Na+. CH3COO-Na+ (s) Æ CH3COO- (aq) + Na+ (aq) CH 3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq) 8) Solution: B Let [Ba 2+] be w mol dm-3 , [L-] will be 2w mol dm-3 q = (w) (2w) 2 q= 4w3 w = 31 4 q ⎟ ⎠ ⎞⎜ ⎝ ⎛ [L-] = 31 4 q 2 ⎟ ⎠ ⎞⎜ ⎝ ⎛ 9) Solution: D There are 5 C* and number of optical isomers = 25 = 32 10) Solution: C KMnO 4 will cleave the 2 double bonds present as well as oxidised the secondary alcohol to ketone CH 3 CH 2CH=C(CH 3)2 HO Cleave the double bonds 2oROH will be oxidised to ketone

11) Solution: C 2 “OD” will be added across the 2 C=C in Linalo-ol C CH3 H3C OD C H Br C H C H H H C CH 3 OH C H C OD H H Br 12) Solution: A CH3OCH2CH2Br Æ CH3OCH2CH2NH2 (nucleophilic substitution by ammonia) CH 3OCH2CN Æ CH3OCH2CH2NH2 (reduction of nitriles) 13) Solution: B Only ethanol undergoes oxidation with K2Cr2O7 in dilute H2SO4 whereas phenol does not. Orange K2Cr2O7 will turn green. Both ethanol and phenol will react with alkaline aqueous iodine and decolourised the brown iodine. Ethanol und ergoes iodoform test with alkaline aqueous iodine. Phenol undergoes easy electrophilic substitution with alkaline aqueous iodine. 14) Solution: D CH(CH COCH3 3)OH 2oROH which can be oxidised by acidified KMnO4 to ketone Chiral C resulting in optical isomers No aldehyde group present to react with Tollen’s reagent. 15) Solution: C C6H5CH2CN undergoes acidic hydrolysis to give C6H5CH2COOH C6H5CH2CH2OH undergoes oxidation to give C6H5CH2COOH

16) Solution: B C3H7CO2H + C2H5OH Æ C3H7CO OC2H5 OH-(aq) K2Cr2O7/H+