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Solutions to Problems 491 36. This problem is related to the last, except that the starting material for a Strecker synthesis is an alde- hyde, and the Strecker sequence adds a carbon atom (as a CN, which becomes the carboxy group). So each starting material must be one carbon shorter than the intended target: 1. 2. HCN H⁺,H₂O (a) For Gly, H₂NCH₂CN 1. NH3 2. HCN (b) For Ile, CH₃CH₂CH(CH₃)CHO 1. NH₃ 2. HCN (c) For Ala, CH₃CHO CH₃CH(NH₂)CN 0 N-C(CO₂CH₂CH₃)₂ 37. After steps 1 and 2 you have o Acidic hydrolysis cleaves not only the phthalimido group from the ring nitrogen but also the acetyl group from the second nitrogen. giving Lys. 38. (a) Because the R group is secondary, alkylation routes should be avoided. Use the Strecker synthesis. NH₂ 1. NH3 2. HCN (CH₃)₂CHCHO (CH₃)₂CHCHCN (b) The R group is primary; now you have a choice. Either the Strecker synthesis starting with or a Gabriel-based method will do. 2. N-CH(CO₂CH₂CH₃)₂ Even the sequence works just fine here. Br