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2. Fill in the boxes with the missing reagents on both side of the arrow. Then, based on the pka table on the next page, determine which side of the equilibrium arrow the reaction would lie: pK, information R-s-H = 10.5 Cl-H =-7 R-N-H = 10,5 R CH, H =-20 H E3 RH = 10 RH X-NH or O RO `CH, H =-23 H =3-5 R. `N CH;-H -25 He OEt HE ~15 O H E 4.6 H E12 R-CEC-HE 25 E 6.4 R,N-H = -~35 EtO HC OEt H2 :13 H E 41 H E 15.7 H-o H = 43 R-o-H =15-19 10 CH2-H =-17 H,CH,CH,CH,C-H = 48 10.2

2. Fill in the boxes with the missing reagents on both side of the arrow. Then, based on the pka table on the next page, determine which side of the equilibrium arrow the reaction would lie: pK, information R-s-H = 10.5 Cl-H =-7 R-N-H = 10,5 R CH, H =-20 H E3 RH = 10 RH X-NH or O RO `CH, H =-23 H =3-5 R. `N CH;-H -25 He OEt HE ~15 O H E 4.6 H E12 R-CEC-HE 25 E 6.4 R,N-H = -~35 EtO HC OEt H2 :13 H E 41 H E 15.7 H-o H = 43 R-o-H =15-19 10 CH2-H =-17 H,CH,CH,CH,C-H = 48 10.2

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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2. Fill in the boxes with the missing reagents on both side of the arrow. Then, based on the pka table on the next page, determine which side of the equilibrium arrow the reaction would lie: pK, information R-s-H = 10.5 Cl-H E -7 R-N-H = 10.5 H2 X. -H E-20 -3 R.H = 10 RH X=NH or O RO `CH,-H -23 =3-5 OEt R. `CH, H =-25 He H2 HE 11 ~15 OH = 4.6 H E12 R-CEC-HE 25 6.4 OEt HE R,N-H = ~35 H2 `H = 41 Eto 13 H = 15.7. H-o 9. = 43 R-o-H =15-19 10 `CH,-H -17 H,CH,CH,CH,C-H = 48 E10.2
Transcribed Image Text:2. Fill in the boxes with the missing reagents on both side of the arrow. Then, based on the pka table on the next page, determine which side of the equilibrium arrow the reaction would lie: pK, information R-s-H = 10.5 Cl-H E -7 R-N-H = 10.5 H2 X. -H E-20 -3 R.H = 10 RH X=NH or O RO `CH,-H -23 =3-5 OEt R. `CH, H =-25 He H2 HE 11 ~15 OH = 4.6 H E12 R-CEC-HE 25 6.4 OEt HE R,N-H = ~35 H2 `H = 41 Eto 13 H = 15.7. H-o 9. = 43 R-o-H =15-19 10 `CH,-H -17 H,CH,CH,CH,C-H = 48 E10.2
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