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QJEEAdv21Ch03cf: JEE Advanced 2021 Chemistry - Class 12 JEEAdv21Ch03cf Solution: Given : The compound(s), which on reaction with HNO 3 will give …

Given :

JEE Advanced 2021 Paper 1 Online Chemistry - Biomolecules Question 15 English
The compound(s), which on reaction with HNO3 will give the product having degree of rotation, [ α\alpha ]D = −- 52.7 ∘^\circ is (are)

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/72dba50af242904d30acd710d14e170f49b3a926d0214ae862071c4b9716ebe2.webp
  • B https://ik.imagekit.io/novatra/photos/82ee9997dbe4bf0647cbf8115c15a32ff34da50cd14acd3f9a978fbd57f19211.webp
  • C https://ik.imagekit.io/novatra/photos/183b60642e938a59e372fc8334e8ef2e45cf87b47fd926db4b6eea526f5d75a6.webp
  • D https://ik.imagekit.io/novatra/photos/f06044722d52525e721893a27ecd9316759044c2aee9cf806d23535040ede6a4.webp

Step-by-Step Verified Solution

Monosaccharide oxidises on reacting with HNO3 to form carboxylic acids. Primary alcohols also oxidise to carboxylic acid alongwith aldehyde.

Complete reaction is as follows :

JEE Advanced 2021 Paper 1 Online Chemistry - Biomolecules Question 15 English Explanation 1
The enantiomer of product (P) has rotation −52.7o . The two enantiomers (c and d) with their reactions are as follows : JEE Advanced 2021 Paper 1 Online Chemistry - Biomolecules Question 15 English Explanation 2 JEE Advanced 2021 Paper 1 Online Chemistry - Biomolecules Question 15 English Explanation 3

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