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.

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.

Organic Chemistry

9th Edition

ISBN:9781305080485

Author:John E. McMurry

Publisher:John E. McMurry

Chapter24: Amines And Heterocycles

Section24.SE: Something Extra

Problem 61AP: Show the products from reaction of p-bromoaniline with the following reagents: (a) CH3I (excess) (b)...

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