Unformatted text preview: 1-15 (a) HCOOH stronger acid pKa 3.76 (b) CH3COOweaker base (c) CH30H stronger acid pKa 1 5.5 + -CN stronger base
+ -- .- HCOOweaker base + H CN FAVORS weaker PRODUCTS acid pK a 9.22
+ C H30H weaker acid pK a 1 5 .5 NaNH2 stronger base .. -- CH 3 COOH stronger acid pKa 4 . 74 CH30- Na+ weaker base C H3O- FAVORS stronger R EA CTANTS base FAVORS NH3 weaker PRODUCTS acid pK a 33 NaCN weaker base
PRODUCTS + -- .- + (d) Na+ -OCH3 stronger base (e)
(f) + H CN stronger acid pK a 9.22
-- -- .- HOCH 3 weaker acid pK a 15.5 + FAVORS H30+ + C H3OH2O + C H30H FAVORS P RODUCTS stronger weaker stronger weaker acid base base acid pK a -1 .7 pK a 1 5.5 The seventh reaction in Table 1-5 shows the Keq for the reverse of this reaction is 3 .2 x 1 0-16. Therefore, Keq for this reaction as written must be the inverse, or 3. 1 x 1 0 15, strongly favoring products.
-.- FAVORS HCl + H 2O H3O+ + CIPRODUCTS stronger stronger weaker weaker acid base base acid The first reaction in Table 1-5 shows the Keq for this reaction is 160, favoring products.
.- :0: II CH 3 - C - O-H +\ H Protonation of the double-bonded oxygen gives three resonance forms (as shown in Solved Problem 1-5(c)); protonation of the single-bonded oxygen gives only one. In general, the more resonance forms a species has, the more stable it is, so the proton would bond to the oxygen that gives a more stable species, that is, the double-bonded oxygen.
•• 1 -16 :0: II CH3 - C - O - H 9 ...
View Full Document