2. Each of the circled hydrogens in the compound shown below could potentially be removed by reaction with an appropriate base (pK₂ values of each acidic hydrogen are provided below). pK₂ = 20 H₂ pKa = 38 (H₂ Hc pka = 10 pka = 16 In six separate reactions, the compound above is treated one at a time with a large excess of each of the six bases listed below. Reaction 1, base = NaOH Reaction 2, base = NaHCO3 Reaction 3, base = NaOCH3 Reaction 4, base = CH3MgBr Reaction 5, base = NaOPh Reaction 6, base = NaNH₂ +He pka = 5 Ha For each of the six reactions using a different base, indicate which, if any, of these hydrogens could be 100% (fully) deprotonated (i.e., reaction equilibrium is shifted far to the right) by the base indicated in that reaction. Some useful information: H₂O, pKa = 16; CH3OH, pKa = 16; H₂CO3, PK₂ = 7; NH3, PK₂ = 38; phenol (PhOH), pK₂ = 10; CH4, pK₂ = 50.

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2. Each of the circled hydrogens in the compound shown below could potentially be removed by reaction with
an appropriate base (pKa values of each acidic hydrogen are provided below).
pKa = 38 Ha
o+He pK₂ = 5
Hd pK₂ = 16
In six separate reactions, the compound above is treated one at a time with a large excess of each of the
six bases listed below.
Reaction 1, base
NaOH
Reaction 2, base = NaHCO3
Reaction 3, base = NaOCH3
Reaction 4, base = CH3MgBr
Reaction 5, base = NaOPh
Reaction 6, base = NaNH₂
pK₂ = 20
a
НЬ
=
Hc
pK₂ = 10
pka
For each of the six reactions using a different base, indicate which, if any, of these hydrogens could be
100% (fully) deprotonated (i.e., reaction equilibrium is shifted far to the right) by the base indicated in that
reaction.
Some useful information:
H₂O, PK₂ = 16; CH3OH, pK₂ = 16; H₂CO3, pK₂ = 7; NH3, pK₂ = 38; phenol (PhOH), pK₂ = 10; CH4, pka = 50.
pKa
Transcribed Image Text:2. Each of the circled hydrogens in the compound shown below could potentially be removed by reaction with an appropriate base (pKa values of each acidic hydrogen are provided below). pKa = 38 Ha o+He pK₂ = 5 Hd pK₂ = 16 In six separate reactions, the compound above is treated one at a time with a large excess of each of the six bases listed below. Reaction 1, base NaOH Reaction 2, base = NaHCO3 Reaction 3, base = NaOCH3 Reaction 4, base = CH3MgBr Reaction 5, base = NaOPh Reaction 6, base = NaNH₂ pK₂ = 20 a НЬ = Hc pK₂ = 10 pka For each of the six reactions using a different base, indicate which, if any, of these hydrogens could be 100% (fully) deprotonated (i.e., reaction equilibrium is shifted far to the right) by the base indicated in that reaction. Some useful information: H₂O, PK₂ = 16; CH3OH, pK₂ = 16; H₂CO3, pK₂ = 7; NH3, pK₂ = 38; phenol (PhOH), pK₂ = 10; CH4, pka = 50. pKa
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