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Chapter 19: Problem 49

Consider the following reaction to produce methyl acetate: When this reaction is carried out with \(\mathrm{CH}_{3} \mathrm{OH}\) containing oxygen18, the water produced does not contain oxygen-18. Explain.

Short Answer

Expert verified
In the esterification reaction between acetic acid \((CH_3COOH)\) and oxygen-18 labeled methanol \((CH_3O^{18}H)\), the oxygen-18 remains bonded to the methyl carbon (CH3) throughout the reaction. The mechanism proceeds in two steps: nucleophilic attack, forming a tetrahedral intermediate, and elimination of water. The oxygen-18 atom becomes part of the ester product (methyl acetate), and the water molecule produced does not contain the oxygen-18 isotope. This explains why the water produced during the reaction does not contain oxygen-18.

Step by step solution

01

Examining the reaction in question

The reaction to produce methyl acetate is an esterification reaction between acetic acid (CH3COOH) and methanol (CH3OH). The general form of the reaction can be written as: \[CH_3COOH + CH_3OH \rightarrow CH_3COOCH_3 + H_2O\] In this particular case, we are given that the methanol molecule contains oxygen-18, which we can represent as \(CH_3O^{18}H\). So, the reaction becomes: \[CH_3COOH + CH_3O^{18}H \rightarrow CH_3COOCH_3 + H_2O\]
02

Understanding Esterification

Esterification is the reaction between a carboxylic acid (in this case, acetic acid) and an alcohol (in this case, methanol) to produce an ester (methyl acetate) and water. The main functional groups involved in the reaction are the hydroxyl group (-OH) of the carboxylic acid and the oxygen atom that links the alkyl group of the alcohol to its corresponding hydrogen atom. The mechanism of esterification involves the nucleophilic addition of the alcohol to the carbonyl carbon of the carboxylic acid, followed by the elimination of a water molecule.
03

Reaction Mechanism

To explain the observed outcome of the reaction, we need to understand the detailed reaction mechanism. For the esterification reaction, the mechanism proceeds in two steps: 1. Nucleophilic attack: The lone pair of electrons on the oxygen atom of the alcohol attacks the electrophilic carbonyl carbon of the carboxylic acid, forming a tetrahedral intermediate: \[CH_3C(=O)OH + CH_3O^{18}H \rightarrow CH_3C(OH)(O^{18}CH_3)OH\] 2. Elimination of water: The tetrahedral intermediate collapses, reforming the carbonyl double bond and simultaneously removing the water molecule: \[CH_3C(OH)(O^{18}CH_3)OH \rightarrow CH_3COO^{18}CH_3 + H_2O\]
04

Conclusion

From the reaction mechanism, we can see that the oxygen-18 atom remains bonded to the methyl carbon (CH3) throughout the entire reaction. In other words, the oxygen atom originally present in the methanol molecule becomes part of the ester product, and the water molecule produced does not contain the oxygen-18 isotope. This explains the observation that the water produced during the reaction does not contain oxygen-18.

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