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hanjing011

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注册: 2010-05-05
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专业: 色谱分析

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2,3,4,5-tetrahydro-7-methoxy-1H-2-benzazepin-1-one is a very important intermediate in our synthesis. But mixtuer of two isomers was obtained when 6-methoxy-3,4-dihydronaphthalen-1(2H)-one oxime was used as the starting material to carry out the Bakemann reaction. This is due to the two groups at the both ends of the C=N double bond are close, which resulted in less selectivity. We guess that trans-6-methoxy-3,4-dihydronaphthalen-1(2H)-one oxime would give 1,3,4,5-tetrahydro-7-methoxy -2H-1-benzazepin-2-one ane cis-6-methoxy-3,4-dihydronaphthalen-1(2H)-one oxime gives 2,3,4,5-tetrahydro-7- methoxy -1H-2-benzazepin-1-one. Therefore, we synthesized and trained the title compound so as to improve the selectivity of Backmann rearrangement.
2,3,4,5-四氢-7-甲氧基-1H-2-苯并氮卓-1-酮是我们合成研究中重要的中间体。但是，用6-甲氧基-3，4-二氢萘-1（2H）酮肟为原料进行贝克曼重排时得到了两种同分异构体的混合物。这是由于C=N双键两端的基团大小相近而导致重排的选择性较差。我们推测，用反式6-甲氧基-3，4-二氢萘-1(2H)酮肟会得到1,3,4,5-四氢-7-甲氧基-2H-1-苯并氮卓-2-酮，顺式酮肟会得到2,3,4,5-四氢-7-甲氧基-1H-2-苯并氮卓-1-酮。因而，我们合成并培养了题目物质以便于提高贝克曼重排的选择性。

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