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Author: Subject: cyclisation
bharathi
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[*] posted on 30-1-2006 at 01:57
cyclisation


Hi all,
I am new to this forum and hope to find some response with respect to my query.
I have been working with a cyclisation reaction wherein chloroacetonitrile and methylcarbazate are involved. The reaction is carried out with traces of sodium methoxide in methanol at 0-5°C.
Can someone help me know how this reaction mechanism looks like ? What is the role of sodium methoxide ? Is the methyl carbazate involved in reaction likely to undergo/degrade to side products?

Thnx in advance,
Regds/Bharathi
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sparkgap
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[*] posted on 31-1-2006 at 00:09


First, for the uninitiated:

carbazic acid is hydrazinecarboxylic acid, H<sub>2</sub>N-NH-COOH.

OK... you're making pyrazolones? Could you please give more specific details?

In any case, the N-N single bond really isn't one of the most stable bonds there is. With the carbazate ester, I see the possibility of a decomposition into CO<sub>2</sub> and N<sub>2</sub> in strongly acidic or basic conditions. Although, seeing that you're just adding a touch of methoxide, I guess decomposition would be negligible. The role of methoxide is probably catalytic. :)

As to mechanism I'll give a hint: check the polarities of the atoms in the starting materials, and compare with the structure of the expected product. ;)

sparky (~_~)




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bharathi
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[*] posted on 5-2-2006 at 22:31


Thnx for the reply. Since this reaction is a cyclisation with no by products, all the atoms of the reagents remain intact in the expected product with none of them being knocked out as byproducts. So where do we find the difference in polarities between the reactants and products? Plz guide me.

Regds,
Bharathi
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