Solution - Career Point

⎛ dH ⎞For ideal gas, ⎜ ⎟⎠⎝ dPT= 0⎛ dE ⎞For the real gas, if ⎜ ⎟⎠⎝ dV⎛ dH ⎞then ⎜ ⎟⎠⎝ dPTT= 0,⎛ dV ⎞= P ⎜ ⎟⎠⎝ dPT+ V ≠ 013.[A] F(Monoester) Molecular weight = 186No Br 2 reaction ⇒ SaturatedTwo oxygen ⇒ 2 × 16 = 32No. of CH 2 = (186 – 32)/14 = 11Hence, molecular formula of saturated monoesterF, is C 11 H 22 O 2HydrolysisGOptically active,soluble inNaOH ⇒ AcidWe have,Ag + Brsalt ⎯ → 2 ± JHunsdiecker reaction;radical intermediate,so racemic mixture(J contains onecarbon less than G)HOptically active,(Alcohols are notsoluble in NaOH)+ve iodoform test,it suggestsHO– CH – R On warning| with H 2 SO 4CH 3 (dehydration)I (no diastereomers meansno geometrical isomers)It suggests same alkyl groupon one of the doubly bondedcarbon atomsI. TsClH optically active ⇒ JII. NaBrIt suggests G contains one morecarbon atom than H.Hence, molecular formula of ester F, isC 5 H 11 – C || O–O–C 5 H 11OC 5 H 11 – C || –OH(G) C 5 H 11 OH(H)CH 3*|H CH 3⇒ HO – CH – CH – CH 3 ⎯H 2⎯SO 4⎯, → ∆ C=C (I)HCH 3 C CH 33No cis-trans isomersHTsCl⎯ →NaBrO||So, G is CH 3 – CH − CH − C − OH| | |CH3 CH3CH3⎯ CH 3 – CH – C H – BrCH | CH|*3 3(J)⎛O ⎞⎜* || ⎟⎜Br2CH3− CH − CH − C−OAg ⎯⎯→CH3− CH − CH − Br⎟⎜ | || | ⎟⎜ CH3CH3CH3CH ⎟3⎜⎟⎝( ± J)⎠O||F = CH 3 – CH − CH − C − O − CH − CH − CH3| || |CH CH CH CH3314.[C] I exists as diastereomers and H is optically active.So, H is HO– CH − CH 2 CH 2 –CH 3|CH 3H 2 SO 4∆HHC = C, andH 3 CCH 2 CH 315.[D]16.[A]G is CH 3 CH 2 CH 2 – CHCO−|| OH|CH 3Since H is optically active and gives negativeiodoform test, so H isHO–CH 2 – CH – CH 2 –CH 2 and|CH 3OG is CH 3 CH 2 CHCH2C || –OH|CH317.[D] Final pH = 1.7 or – log [H + ] = 1.7or log [H + ] = – 2 + 0.3 or [H + ] = 2 × 10 –2Let v ml of 0.1 (M) HA solution is mixed with100 ml of 10 –2 (M) HCl. In the mixed solution,[HA] =4v× 10− × 10 3 =(100 + v)−3330.1v(1003(M) and+ v)10 × 10 1[HCl]= =(100 + v) (100 + v)HA H + + A –0.1v(1 – α)(100 + v)∴ [H + ] =0.1v α(100 + v)1+(100 + v)0.1v α(100 + v)(0.1v α + 1)= 2 × 10 –2(100 + v)XtraEdge for IIT-JEE 88APRIL 2010