ABC26OR0796 · Natural Products
Subject: Organic Chemistry · Chapter: Natural Products · Topic: Natural Products & – General · Exam: JAM 2005-2022 · Marks: 2 · Difficulty: Medium
5(a) For the following scheme of transformations, draw the structures of A, B, C and D. 
(b) Complete hydrolysis of a pentapeptide with 6N HCl at 110°C in a sealed tube gave 2 equivalents of glycine, one equivalent each of tyrosine, leucine and phenylalanine. Reaction of the pentapeptide with Sanger's reagent (2,4-dinitrofluorobenzene, DNFB) and subsequent hydrolysis gave the DNFB derivative of tyrosine. Chymotryspin cleavages of this peptide yielded tyrosine, leucine and a tripeptide. Deduce the sequence of the pentapeptide.
(b) Complete hydrolysis of a pentapeptide with 6N HCl at 110°C in a sealed tube gave 2 equivalents of glycine, one equivalent each of tyrosine, leucine and phenylalanine. Reaction of the pentapeptide with Sanger's reagent (2,4-dinitrofluorobenzene, DNFB) and subsequent hydrolysis gave the DNFB derivative of tyrosine. Chymotryspin cleavages of this peptide yielded tyrosine, leucine and a tripeptide. Deduce the sequence of the pentapeptide.Answer
Answer (as printed): A
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Explanation
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