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Which of the following gives an opticall...

Which of the following gives an optically inactive aldaric acid on oxidation with dilute nitric acid ?

A

B

C

D

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The correct Answer is:
To solve the problem of identifying which aldaric acid becomes optically inactive upon oxidation with dilute nitric acid, we will analyze each of the given options step by step. ### Step 1: Understanding Aldaric Acids Aldaric acids are a type of sugar acid that contain both an aldehyde group (CHO) and a hydroxyl group (OH) on adjacent carbon atoms. When oxidized, the aldehyde group is converted to a carboxylic acid (COOH), and the hydroxymethyl group (CH2OH) is also oxidized to a carboxylic acid. ### Step 2: Oxidation with Dilute Nitric Acid When an aldose sugar is treated with dilute nitric acid, the following transformations occur: - The aldehyde group (CHO) at the first carbon oxidizes to a carboxylic acid (COOH). - The hydroxymethyl group (CH2OH) at the last carbon also oxidizes to a carboxylic acid (COOH). This results in the formation of a dicarboxylic acid, specifically an aldaric acid. ### Step 3: Checking for Optical Activity A compound is optically inactive if it possesses a plane of symmetry. To determine if the resulting aldaric acid is optically inactive, we need to analyze the structure of each product after oxidation and check for symmetry. ### Step 4: Analyzing Each Option 1. **Option 1**: - After oxidation, the structure does not have a plane of symmetry. Therefore, it is optically active. 2. **Option 2**: - Similar to Option 1, this structure also lacks a plane of symmetry, making it optically active. 3. **Option 3**: - Upon oxidation, this structure has a plane of symmetry. The arrangement of groups allows for mirror images, indicating that it is optically inactive. 4. **Option 4**: - This structure also does not have a plane of symmetry, thus it is optically active. ### Conclusion After analyzing all options, we find that **Option 3** produces an optically inactive aldaric acid upon oxidation with dilute nitric acid due to the presence of a plane of symmetry. ### Final Answer The correct option is **Option 3**. ---
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