Chapter 20, Problem 20.9P

Interpretation Introduction

(a)

Interpretation:

The complete and detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted.

Concept introduction:

A haloform reaction converts a methyl ketone (${\text{RCOCH}}_{\text{3}}$) into a carboxylic acid (${\text{RCO}}_{\text{2}}\text{H}$). The methyl group, ${\text{CH}}_3$, in methyl ketone is not a good leaving group, but under the conditions of the reaction, it is first converted into ${\text{CX}}_3$, where $\text{X=Cl, Br, I}$, which is a suitable leaving group. In steps 1-6 of this mechanism, the three alpha hydrogen atoms on the methyl group are replaced by bromine atoms via the alpha halogenation mechanism. The resulting tribromomethyl ketone then undergoes attack by ${\text{HO}}^{\text{-}}$ in step 7 to produce a tetrahedral intermediate, which subsequently eliminates ${\text{Br}}_3{\text{C}}^{-}$ in step 8 to produce an initial carboxylic acid. In step 9, that carboxylic acid is deprotonated by ${\text{Br}}_3{\text{C}}^{-}$, yielding a relatively stable carboxylate anion. In the acid workup in step 10, the proton is replenished on the carboxylate anion to produce the final carboxylic acid.

Interpretation Introduction

(b)

Interpretation:

The complete and detailed mechanism for the given reaction is to be drawn, and the major product is to be predicted.

Concept introduction:

A haloform reaction converts a methyl ketone (${\text{RCOCH}}_{\text{3}}$) into a carboxylic acid (${\text{RCO}}_{\text{2}}\text{H}$). The methyl group, ${\text{CH}}_3$, in methyl ketone is not a good leaving group, but under the conditions of the reaction, it is first converted into ${\text{CX}}_3$, where $\text{X=Cl, Br, I}$, which is a suitable leaving group. In steps 1-6 of this mechanism, the three alpha hydrogen atoms on the methyl group are replaced by bromine atoms via the alpha halogenation mechanism. The resulting tribromomethyl ketone then undergoes attack by ${\text{HO}}^{\text{-}}$ in step 7 to produce a tetrahedral intermediate, which subsequently eliminates ${\text{Br}}_3{\text{C}}^{-}$ in step 8 to produce an initial carboxylic acid. In step 9, that carboxylic acid is deprotonated by ${\text{Br}}_3{\text{C}}^{-}$, yielding a relatively stable carboxylate anion. In the acid workup in step 10, the proton is replenished on the carboxylate anion to produce the final carboxylic acid.