Solution - Career Point

1.[C]2.[B]3.[B]SOLUTION FOR MOCK TESTPAPER - IIIIT-JEE (PAPER - II)CH 2 =C'A''B'CHEMISTRYNH 2BrNOBrCOOHH 2 /PdPh+ EnantiomerNH 2CH 3 –CHBr –CH=CH–COOH H 2 /PdCH 2 –CH 2 –COOHOptically inacitve+COOH(resolvable)PhDistribution of electrons in the MO's in He 2 isσ 1s 2 σ∗ 1s 2 . He 2 is unstableDistribution of electrons in the MOS in H 2 is σ 1s2. H 2 is stable.Grignard reagent reacting with acyl halide usuallygives 3º alcohol.7.[B,C,D] The order of H-bond energiesF – H …….. F – > F – H …….. O > F – H …….. F >O – H …….. O > O – H …….. F > N – H …….. N8.[C,D]9.[C,D]RHE reaction : Hg 2 Cl 2 (s) + 2e → 2Hg(l) + 2Cl –LHE reaction : 2Ag(s) + 2Cl – → 2AgCl(s) + 2eNet reaction : Hg 2 Cl 2 (s) + 2Ag(s)⎯→ 2Hg (l) + 2AgCl(s)In case of same concentration of Cl – ions in thetwo half cells, E cell is independent on theconcentration of Cl – . Other substances are eitherpure solids or liquids, which have unit activitiesirrespective of their amounts.Column Matching :10. [A] → p, r, s ; [B] → q;[C] → p, q,,s ;[D] → q11. [A] → p, q,t ; [B] → p, r, s;[C] → p, q, r;[D] → p, q, r4.[B] After mixing total moles of A –= 100 × 0.2 × 10 –3 + 100 × 0.3 × 10 –3= 100 × 10 –3 × 0.5 molesAfter mixing total moles of HA= 100 × 0.1 × 10 –3 + 100 × 0.2 × 10 –3= 100 × 0.3 × 10 –3 moles5After mixing resulting pH = 5 + log 3NH 2HO CH 2 –CH*5.[A,B,C]HOCOOHIt contains 2 phenolic hydrogens and a carboxylicacidic hydrogens+NHCH 2 –HC 2HOCOO –HO Zwitter ion6.[A,B,C]Numerical Response type questions :12. [4]W Q 2.977 3 × 1×60×60= or =E 96500 106.4 / n 96500∴ n = 413. [8]O3/ H2O⎯⎯ ⎯⎯ →(Ozonolysis)HOHONH NH22O ⎯⎯ ⎯⎯→OOKOH / glycol(Wolff − Kishnerreduction)(X)(W)XtraEdge for IIT-JEE 95APRIL 2010