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Chapter 17: Problem 43

In Section 17.7B, we suggested that the mechanism of Fischer esterification of carboxylic acids is a model for the reactions of functional derivatives of carboxylic acids. One of these reactions is that of an acid chloride with water (Section 18.4A). Suggest a mechanism for this reaction.

Short Answer

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Question: Describe the mechanism for the reaction of an acid chloride with water in three steps. Answer: The mechanism for the reaction of an acid chloride with water involves the following three steps: (1) Formation of the tetrahedral intermediate, where water acts as a nucleophile and attacks the carbonyl carbon of the acid chloride. (2) Proton transfer within the tetrahedral intermediate, transforming the negatively charged oxygen into a hydroxyl group and the neutral oxygen into a hydroxide ion. (3) Formation of the carboxylic acid and chloride ion, where the hydroxide ion acts as a leaving group and forms a chloride ion by taking up a proton from the carboxylic acid. This mechanism is analogous to the Fischer esterification of carboxylic acids.

Step by step solution

01

Formation of the tetrahedral intermediate

In the first step, water acts as a nucleophile and attacks the carbonyl carbon of the acid chloride. This results in the formation of a tetrahedral intermediate. The pi bond of the carbonyl group breaks and the electrons shift to the oxygen. $$\text{RC(O)Cl} + \text{H}_2\text{O} \longrightarrow \text{RC(OH)(O^{-})} + \text{Cl}^{-}$$
02

Proton transfer

In the second step, a proton transfer occurs within the tetrahedral intermediate formed in step 1. This transforms the negatively charged oxygen into a hydroxyl group and the neutral oxygen becomes a hydroxide ion. $$\text{RC(OH)(O^{-})} \longrightarrow \text{RC(OH)}_2 + \text{OH}^{-}$$
03

Formation of carboxylic acid and chloride ion

In the third step, the carboxylic acid product is formed through the reformation of the carbonyl pi bond. The hydroxide ion acts as a leaving group and forms a chloride ion by taking up a proton from the carboxylic acid. $$\text{RC(OH)}_2 + \text{OH}^{-} \longrightarrow \text{RC(O)OH} + \text{Cl}^{-}$$ In conclusion, the mechanism for the reaction of an acid chloride with water involves three steps: formation of the tetrahedral intermediate, proton transfer, and the formation of the carboxylic acid and chloride ion. This mechanism is analogous to the Fischer esterification of carboxylic acids.

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