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An enantiomerically pure is treated with...

An enantiomerically pure is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be :

A

Optically active mixture

B

Pure enantiomer

C

Mesi-compound

D

Racemic mixture

Text Solution

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To solve the problem, we need to analyze the reaction between an enantiomerically pure acid and a racemic mixture of an alcohol that has one chiral carbon. Here’s a step-by-step breakdown of the solution: ### Step 1: Identify the Reactants We have two reactants: - An enantiomerically pure acid (let's denote it as \( R \)-acid). - A racemic mixture of an alcohol with one chiral carbon (let's denote it as \( S \)-alcohol and \( R \)-alcohol). ### Step 2: Understand the Reaction The reaction between the acid and the alcohol is known as Fischer esterification. In this reaction, the acid reacts with the alcohol in the presence of an acid catalyst (often \( H^+ \)), resulting in the formation of an ester and water. ### Step 3: Determine the Products When the enantiomerically pure acid reacts with the racemic alcohol, we can form two different esters: 1. The ester formed with \( R \)-alcohol. 2. The ester formed with \( S \)-alcohol. ### Step 4: Analyze the Stereochemistry Since the alcohol is racemic, it contains both \( R \) and \( S \) forms. The resulting esters will also have different configurations based on the alcohol used: - The ester from \( R \)-acid and \( R \)-alcohol will have a specific configuration. - The ester from \( R \)-acid and \( S \)-alcohol will have a different configuration. ### Step 5: Determine Optical Activity Since the acid is enantiomerically pure, it will not introduce any additional chirality. However, the esters formed from the racemic alcohol will be optically active because they will have one chiral center from the alcohol. Thus, both esters will be optically active. ### Conclusion The final conclusion is that the esters formed from the reaction will be a mixture of optically active compounds. ### Answer The ester formed will be an optically active mixture. ---

To solve the problem, we need to analyze the reaction between an enantiomerically pure acid and a racemic mixture of an alcohol that has one chiral carbon. Here’s a step-by-step breakdown of the solution: ### Step 1: Identify the Reactants We have two reactants: - An enantiomerically pure acid (let's denote it as \( R \)-acid). - A racemic mixture of an alcohol with one chiral carbon (let's denote it as \( S \)-alcohol and \( R \)-alcohol). ### Step 2: Understand the Reaction ...
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