Question: Arrange each group of compounds in order of increasing basicity.

1. CH₃COO^-,CICH₂C00^- , and PhO^-
2. sodium acetylide, sodium amide, and sodium acetate
3. sodium benzoate, sodium ethoxide, and sodium phenoxide
4. pyridine, sodium ethoxide, and sodium acetate

a. Stronger conjugate acids are those from which weak bases come, and the pKa values of the given conjugate acids are, PhO^- = 10.0,CH₃COO^- = 4.7 and CICH₂COO^-= 2.8. Basicity increases with an increase of value. So, has the highest basicity and CICH₂COO^-is having the lowest basicity. Thus, the order of increasing basicity is, CICH₂C00^- < CH₃COO^- < PhO^- .

b. The pKa values of the given compounds are, NaNH₂= 36, HC=C^-= 25 and .CH₃COO^- Na⁺ = 4.7. Higher the pka value, the higher the basicity. So, has the highest basicity and is .CH₃COO^- Na⁺ having the lowest basicity. Thus, the order of increasing basicity is, b.CH₃COO^- Na⁺< HC $\equiv$C^- Na⁺ < NaNH₂.

C. The values of the given conjugate acids are,PhO^-Na+ = 10.0, CH₃COO^- Na⁺ = 15.9 and = 4.20. The lowest value of means, lowest basicity. So, CH₃COO^- Na⁺ has the highest basicity and is having the lowest basicity. Thus, the order of increasing basicity is, PhCOO^- Na+ <PhO^-Na+ < CH₃CH₂O^- Na+ .

d. The pKa values of pyridine, sodium ethoxide, and sodium acetate are, Pyridine = 5.2, CH₃CH₂O^- Na+= 15.9 and CH₃COO^- Na⁺= 4.7. Basicity increases with an increase of value. So, CH₃COO^- Na⁺ has the highest basicity and is having the lowest basicity. Thus, the order of increasing basicity is,CH₃COO^- Na⁺ <Pyridine <CH₃CH₂O^- Na+ .