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According to my class notes, acetoacetic ester doesn't give haloform test, but I feel that it should.

I know that the methyl group on acetoacetic ester is not responsible for the haloform test. First, the active methylene group is attacked by the base and halogen ion is substituted there. After the attack of $\ce{OH-}$ on the carbonyl carbon we will also get an ethanol molecule (from the ester part), which apparently should give positive haloform test.

I am confused. Which is correct?

granthium
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1 Answers1

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As @user55119 says it is unlikely the di-iodide is formed under iodoform conditions.

According to this paper1, acetoacetate gives only mono-iodination at the methylene centre under more powerful iodinating conditions than the standard iodoform $\ce{I2/OH-}$. The mono-iodo product in the presence of excess base will exist as the iodo-enolate effectively stopping iodination at the methyl ketone centre. Thus no iodoform product will be formed.

References

  1. Goswami, Papori & Ali, Shahzad & Khan, Dr. M. Musawwer & Khan, Abu. (2007). Selective and effective oxone-catalysed α-iodination of ketones and 1, 3-dicarbonyl compounds in the solid state. General Papers ARKIVOC. 2007. 82-89. (Link)
Nilay Ghosh
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Waylander
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