Question

a. What is the product of the reaction of acetyl chloride with${\mathbf{HO}}^{-}$? The${\mathbf{pK}}_a$of HCl is -7; the${\mathbf{pK}}_a$of${\mathbf{H}}_2\mathbf{O}$is 15.7.

b. What is the product of the reaction of acetamide with${\mathbf{HO}}^{-}$? The${\mathbf{pK}}_a$of${\mathbf{NH}}_3$is 36; the${\mathbf{pK}}_a$of${\mathbf{H}}_2\mathbf{O}$is 15.7.

Short answer

a.

Formation of acetic acid

b.

No products formed

Step-by-step solution

Step-by-Step Solution Step 1: Tetrahedral intermediate

When a nucleophile adds to the carbonyl carbon of a carboxylic acid derivative, the weakest bond in the molecule—the p bond—breaks, and an intermediate is formed. It is called a tetrahedral intermediate because the ${\mathbf{sp}}^2$carbon in the reactant becomes a${\mathbf{sp}}^3$carbon (that is, a tetrahedral carbon) in the intermediate.

The product  of reaction with acetyl chloride

a.

HCl is a stronger acid than ${\mathrm{H}}_2\mathrm{O}$. So, ${\mathrm{Cl}}^{-}$is a weaker base than ${\mathrm{OH}}^{-}$. Therefore, ${\mathrm{Cl}}^{-}$will be eliminated from the tetrahedral intermediate. So, the product will be acetic acid represented as follows:

Formation of acetic acid

b.

${\mathrm{H}}_2\mathrm{O}$is stronger acid than ${\mathrm{NH}}_3$. So, ${\mathrm{OH}}^{-}$is weaker base than ${\mathrm{NH}}_2^{-}$. Therefore, ${\mathrm{OH}}^{-}$cannot eliminate the$-{\mathrm{NH}}_2$ group, so no reaction will occur which is represented as follows:

No products formed