Question

Which of the following bases are strong enough to deprotonate CH3CH2CH2C≡CH (pKa = 25), so that equilibrium favors the products? a. NaNH2 b. NaOH c. NaC≡N d. NaCH2(CO)N(CH3)2 e. H2O f. CH3CH2Li

Answer 1

Answer:

a, and f.

Explanation:

To be deprotonated, the conjugate acid of the base must be weaker than the acid that will react, because the reactions favor the formation of the weakest acid. The pKa value measures the strength of the acid. As higher is the pKa value, as weak is the acid. So, let's identify the conjugate acid and their pKas:

a. NaNH2 will dissociate, and NH2 will gain the proton and forms NH3 as conjugate acid. pKa = 38.0, so it happens.

b. NaOH will dissociate, and OH will gain the proton and forms H2O as conjugate acid. pKa = 14.0, so it doesn't happen.

c. NaC≡N will dissociate, and CN will gain a proton and forms HCN as conjugate acid. pKa = 9.40, so it doesn't happen.

d. NaCH2(CO)N(CH3)2 will dissociate and forms CH3(CO)N(CH3)2 as conjugate acid. pKa = -0.19, so it doesn't happen.

e. H2O must gain one proton and forms H3O+. pKa = -1.7, so it doesn't happen.

f. CH3CH2Li will dissociate, and the acid will be CH3CH3. pKa = 50, so it happens.