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
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Chapter1: Chemical Foundations
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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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