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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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charleygan 
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| 2,3,4,5-tetrahydro-7-methoxy-1H-2-benzazepin-1-one is a very important intermediate in our research. But mixture of two isomers is obtained when 6-methoxy-3,4-dihydronaphthalen-1(2H)-one oxime is used as the starting material to carry out the Bakemann reaction, which resulted from the less selectivity of rearrangement because of the close size of two groups at the both ends of the C=N double bond. We propose 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 and cis-6-methoxy-3,4-dihydronaphthalen-1(2H)-one oxime would give 2,3,4,5-tetrahydro-7- methoxy -1H-2-benzazepin-1-one. Therefore, the goal compound is synthesized to improve the selectivity of Backmann rearrangement. |
3楼2010-07-28 11:55:39
cxsunkun
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2楼2010-07-28 11:32:23
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