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Besides asymmetric hydrogenation (Section 29.4), several other methods are now available for the synthesis of optically active amino acids. How might a reaction like the Strecker synthesis be adapted to the preparation of chiral amino acids?

Short Answer

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Asymmetric hydrogenation constitutes a reaction that incorporates hydrogen atoms to a target molecule comprising three-dimensional selectivity. This selectivity arises from the reagents or catalysts.

Step by step solution

01

Asymmetric hydrogenation

Asymmetric hydrogenation constitutes a reaction that incorporates hydrogen atoms to a target molecule comprising three-dimensional selectivity. This selectivity arises from the reagents or catalysts.

02

Synthesis of amino acids

Amino acids can be generated using several ways, and one among them is the Strecker synthesis. Aldehydes are transformed into amino acids in this synthesis. The aldehyde combines with ammonia using a reagent like potassium cyanide.

An α-amino acid is generated by condensation reaction. This gets subjected to hydrolysis to generate amino acid.

03

Preparation of chiral amino acids by Strecker synthesis

The chiral amine like R*NH2 or other chiral nitrogen-containing compound can create a chiral imine. This will favor the generation of one of the amino nitriles in Strecker synthesis.

The hydrolysis of the CN group and the removal of R*would then lead to the formation of the resulting amino acid.

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Most popular questions from this chapter

What types of stabilizing interactions exist between each of the following pairs of amino acids?

a. Ser and Tyr

b. Val and Leu

c. two Phe residues

An octapeptide contains the following amino acids: Arg, Glu, His, Ile, Leu, Phe, Tyr, and Val. Carboxypeptidase treatment of the octapeptide forms Phe and a heptapeptide. Treatment of the octapeptide with chymotrypsin forms two tetrapeptides, A and B. Treatment of A with trypsin yields two dipeptides, C and D. Edman degradation cleaves the following amino acids from each peptide: Glu (octapeptide), Glu (A), Ile (B), Glue (C), and Val (D). Partial hydrolysis of tetrapeptide B forms Ile-Leu in addition to other products. Deduce the structure of the octapeptide and fragments A-D.

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