Methyl ethyl ketone can undergo Claisen condensation with a given ester, for example, ethyl benzoate, to yield either of two products, depending upon experimental conditions. What are these two products? How can one tell which product one has obtained?

Short Answer

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Methyl ethyl ketone can undergo Claisen condensation with ethyl benzoate to yield either Product A (\(CH_3-CO-CH_2-CO-Ph\)) or Product B (\(C_2H_5-O-CO-CH_2-CO-CH_3\)), depending on the reaction conditions. To determine which product has been obtained, spectroscopic techniques such as infrared (IR) spectroscopy, nuclear magnetic resonance (NMR) spectroscopy, or mass spectrometry (MS) can be employed.

Step by step solution

01

Write the structure of reactants

First, let's write the structure of methyl ethyl ketone (MEK) and ethyl benzoate. Methyl ethyl ketone: \(CH_3-CO-CH_2-CH_3\) Ethyl benzoate: \(C_2H_5-O-CO-Ph\), where 'Ph' represents the phenyl group (\(C_6H_5\)) Now, let's proceed with the Claisen condensation process.
02

Claisen condensation reaction

In the Claisen condensation reaction, an ester reacts with an enolizable ketone or another enolizable ester, in the presence of a base, to form a β-keto ester or a β-diketone (depending on the reaction conditions). Considering the reactants, we have two possible reaction pathways, leading to two products. Reaction pathway A: MEK acts as an ester, and ethyl benzoate acts as a nucleophile (enol form). Reaction pathway B: Ethyl benzoate acts as an ester, and MEK acts as a nucleophile (enol form).
03

Write the structure of products

Now, let's write the structure of the products obtained through the two reaction pathways. Product A: Obtained when MEK acts as the ester and ethyl benzoate as a nucleophile (enol form). Product A structure: \(CH_3-CO-CH_2-CO-Ph\) Product B: Obtained when ethyl benzoate acts as the ester and MEK as a nucleophile (enol form). Product B structure: \(C_2H_5-O-CO-CH_2-CO-CH_3\)
04

Identifying the obtained product

To determine which product has been obtained, we can use spectroscopic techniques, such as infrared (IR) spectroscopy, nuclear magnetic resonance (NMR) spectroscopy, or mass spectrometry (MS). - Infrared (IR) spectroscopy: Product A will show a strong absorption band for the carbonyl group (\(C=O\)) in the range of 1690-1730 cm\(^{-1}\), while product B will show an ester absorption band in the range of 1735-1750 cm\(^{-1}\). - Nuclear magnetic resonance (NMR) spectroscopy: The chemical shifts (δ) of the protons in products A and B will show significant differences, specifically for the methylene group (\(CH_2\)) and the ester-related protons. Comparing these proton chemical shifts with reference values can help to identify the obtained product. - Mass spectrometry (MS): The mass spectra of the products will exhibit distinct molecular ion peaks, which can be used to differentiate between them. In conclusion, methyl ethyl ketone can undergo Claisen condensation with ethyl benzoate to yield either Product A (\(CH_3-CO-CH_2-CO-Ph\)) or Product B (\(C_2H_5-O-CO-CH_2-CO-CH_3\)), depending on the reaction conditions. Spectroscopic techniques, such as IR, NMR, and MS, can be employed to determine which product has been obtained.

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