
Define peptide linkage. What is the action of the following on glucose?
a) Dilute nitric acid.
b) Hydroxylamine.
Answer
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Hint:Write the structure of glucose and find the reactive centers. Find reactive centers in both dilute nitric acid and hydroxylamine that can attack glucose molecules. Dilute nitric acid is a strong oxidizing agent and hydroxylamine contains an amine group that can react with the carbonyl group of glucose.
Complete answer:
Peptide linkage: Peptide linkage which is also known as peptide bond is an amide formed between an acid and an amino group with the elimination of water molecules. In this linkage $ - COOH$ and $ - N{H_2}$ bond together with elimination of water to form $ - \{ CO - NH\} - $ , this linkage is known as peptide linkage.
a) We have to find the reaction between glucose and dilute nitric acid. Since dilute nitric acid is a strong oxidizing agent, we can understand that glucose will be oxidized in this reaction. The terminal groups in glucose, $ - CHO$ and $ - C{H_2}OH$ are both oxidized to form two carboxylic acid groups $ - COOH$ and hence glucose becomes gluconic acid or saccharic acid.
b) We know that glucose contains one terminal carbonyl group $ - CHO$ which is an aldehyde. This carbonyl group can react actively with Hydroxylamine to from an Oxime $ - C = NOH$ at the terminal end. It also adds up a molecule of hydrogen cyanide $HCN$ , to form cyanohydrin.
Note:
Nitric acid is a strong oxidizing agent and hence it oxidizes the terminal groups of glucose to form two carboxylic acid groups. Glucose molecules react with hydroxylamine to form an oxime and it also adds a molecule of hydrogen cyanide to give cyanohydrin.
Complete answer:
Peptide linkage: Peptide linkage which is also known as peptide bond is an amide formed between an acid and an amino group with the elimination of water molecules. In this linkage $ - COOH$ and $ - N{H_2}$ bond together with elimination of water to form $ - \{ CO - NH\} - $ , this linkage is known as peptide linkage.
a) We have to find the reaction between glucose and dilute nitric acid. Since dilute nitric acid is a strong oxidizing agent, we can understand that glucose will be oxidized in this reaction. The terminal groups in glucose, $ - CHO$ and $ - C{H_2}OH$ are both oxidized to form two carboxylic acid groups $ - COOH$ and hence glucose becomes gluconic acid or saccharic acid.
b) We know that glucose contains one terminal carbonyl group $ - CHO$ which is an aldehyde. This carbonyl group can react actively with Hydroxylamine to from an Oxime $ - C = NOH$ at the terminal end. It also adds up a molecule of hydrogen cyanide $HCN$ , to form cyanohydrin.
Note:
Nitric acid is a strong oxidizing agent and hence it oxidizes the terminal groups of glucose to form two carboxylic acid groups. Glucose molecules react with hydroxylamine to form an oxime and it also adds a molecule of hydrogen cyanide to give cyanohydrin.
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