Chapter_3

Organic Chemistry

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-03 JWDD052-Solomons-v2 May 23, 2007 14:57 3 AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR MECHANISMS: ACIDS AND BASES SOLUTIONS TO PROBLEMS 3.1 CH 3 OF F F HB + F F CH 3 O H B + F F F B CH 3 CH 3 + F F CH 3 CH 3 O B + F Cl Cl CH 3 Cl Al + Cl Cl Cl CH 3 Cl Cl Al + (a) (b) (c) 3.2 (a) Lewis base (b) Lewis acid (c) Lewis base (d) Lewis base (e) Lewis acid (f) Lewis base 3.3 F F Lewis base Lewis acid CH 3 CH 3 N H CH 3 CH 3 N B + + F H F F F B 3.4 (a) K a = [H 3 O + ][HCO 2 ] [HCO 2 H] = 1 . 77 × 10 4 Let x = [H 3 O + ] = [HCO 2 ] at equilibrium then, 0 . 1 x = [HCO 2 H] at equilibrium but, since the K a is very small, x will be very small and 0 . 1 x ' 0 . 1 32
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-03 JWDD052-Solomons-v2 May 23, 2007 14:57 ACIDS AND BASES 33 Therefore, (x)(x) 0 . 1 = 1 . 77 × 10 4 x 2 = 1 . 77 × 10 5 x = 0 . 0042 = [H 3 O + ] = [HCO 2 ] (b) % Ionized = [H 3 O + ] 0 . 1 × 100 or [HCO 2 ] 0 . 1 × 100 = . 0042 0 . 1 × 100 = 4 . 2% 3.5 (a) p K a =− log 10 7 ( 7) = 7 (b) p K a log 5 . 0 0 . 669 (c) Since the acid with a K a = 5 has a larger K a , it is the stronger acid. 3.6 When H 3 O + acts as an acid in aqueous solution, the equation is H 3 O + + H 2 O ® H 2 O + H 3 O + and K a is K a = [H 2 O][H 3 O + ] [H 3 O + ] = [H 2 O] The molar concentration of H 2 O in pure H 2 O, that is [H 2 O] = 55 . 5; therefore, K a = 55 . 5 The p K a is p K a log 55 . 5 1 . 74 3.7 The p K a of the methylaminium ion is equal to 10.6 (Section 3.5C). Since the p K a of the anilinium ion is equal to 4.6, the anilinium ion is a stronger acid than the methylaminium ion, and aniline (C 6 H 5 NH 2 ) is a weaker base than methylamine (CH 3 NH 2 ). 3.8 (a) Negative. Because the atoms are constrained to one molecule in the product, they have to become more ordered. (b) Approximately zero. (c) Positive. Because the atoms are in two separate product molecules, they become more disordered. 3.9 (a) If K eq = 1 then, log K eq = 0 = 1 G 2 . 303 RT 1 G = 0 (b) If K eq = 10 then, log K eq = 1 = 1 G 2 . 303
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-03 JWDD052-Solomons-v2 May 23, 2007 14:57 34 AN INTRODUCTION TO ORGANIC REACTIONS AND THEIR MECHANISMS 1 G =− (2 . 303)(0 . 008314 kJ mol 1 K 1 )(298 K) 5 . 71 kJ mol 1 (c) 1 G = 1 H T 1 S 1 G = 1 H 5 . 71 kJ mol 1 if 1 S = 0 3.10 Structures A and B make equal contributions to the overall hybrid. This means that the carbon-oxygen bonds should be the same length and that the oxygens should bear equal positive charges. A B C CH 3 O O CH 3 C O O 3.11 (a) CHCl 2 CO 2 H would be the stronger acid because the electron-withdrawing inductive effect of two chlorine atoms would make its hydroxyl proton more positive. The electron- withdrawing effect of the two chlorine atoms would also stabilize the dichloroacetate ion more effectively by dispersing its negative charge more extensively. (b) CCl 3 CO 2 H would be the stronger acid for reasons similar to those given in (a), except here there are three versus two electron-withdrawing chlorine atoms involved.
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Chapter_3 - P1: PBU/OVY JWDD052-03 P2: PBU/OVY QC: PBU/OVY...

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