Asymmetric Stetter Catalyzed by Triazolium Salts

三唑盐催化的不对称Stetter

基本信息

批准号：
6948472
负责人：
JAVIER READ DE ALANIZ
金额：
$ 2.33万
依托单位：
COLORADO STATE UNIVERSITY
依托单位国家：
美国
项目类别：
财政年份：
2004
资助国家：
美国
起止时间：
2004-09-01 至 2006-06-30
项目状态：
已结题

来源：
https://reporter.nih.gov/project-details/6948472
关键词：
aldehydes angiogenesis inhibitors carbene catalyst chemical reaction drug design /synthesis /production infrared spectrometry ligands neoplasm /cancer chemotherapy nuclear magnetic resonance spectroscopy predoctoral investigator salts stereochemistry stereoisomer triazoles

项目摘要

DESCRIPTION (provided by applicant): The discovery of new reactions will facilitate the development of new drug entities and promising new therapeutics. This proposal will seek to apply novel chiral organic salts to catalyze the formation of carbon-carbon bonds possessing stereochemistry at both ends of the bond. Stereochemically dense functionality is associated with improved specificity for one target. The research plan involves the use of tri- and tetra-substituted Michael acceptors with tethered aldehydes in an asymmetric Stetter reaction. Reversing the normal reactivity of aldehydes as electrophiles to make them function as nucleophiles is one of the tenets of this research. Successful realization of these goals will allow us to use the chemistry developed to facilitate a rapid synthesis of azaspirene, a potent angiogenesis inhibitor, a strategy of significant interest in cancer chemotherapy. It is envisioned, however, that the broader impact of this science may be in revealing a new reaction manifold and a novel strategy to solving problems of molecular complexity particularly relevant to synthetic approaches to pharmaceutical targets.

描述（由申请人提供）：新反应的发现将促进新药物实体和有前途的新疗法的开发。该提案将寻求应用新型手性有机盐来催化在键的两端具有立体化学的碳-碳键的形成。立体化学致密的功能与提高某一靶标的特异性相关。该研究计划涉及在不对称Stetter反应中使用三取代和四取代的迈克尔受体与束缚醛。逆转醛作为亲电子试剂的正常反应性，使其发挥亲核试剂的作用是这项研究的宗旨之一。成功实现这些目标将使我们能够利用开发的化学方法来促进氮杂螺环的快速合成，氮杂螺环是一种有效的血管生成抑制剂，这是癌症化疗中具有重要意义的策略。然而，预计这门科学的更广泛影响可能在于揭示一种新的反应流形和一种解决分子复杂性问题的新策略，特别是与药物靶点的合成方法相关的问题。