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?

  • $\begingroup$ With just dil. $\ce {NaOH}$ at r.t.p., the hydrolysis may be rather slow? Only with heating may sufficient ethanol be produced to give a positive iodoform test. I would agree with you that under heating, with $\ce {NaOH}$ (aq), $\ce {I2}$ (aq), iodoform test would give a positive result. $\endgroup$ Commented Jun 24, 2020 at 8:17
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    $\begingroup$ Does this answer your question? Does acetic acid give a positive result with the iodoform test? $\endgroup$ Commented Jun 24, 2020 at 9:22
  • $\begingroup$ This has an accepted answer. $\endgroup$ Commented Jun 24, 2020 at 9:23
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    $\begingroup$ @MathewMahindaratne granthium is not asking about acetic acid. granthium is asking if iodoform test reagents would be able to hydrolyse the acetoacetic ester and produce suffiicient ethanol to give a positive iodoform test result. $\endgroup$ Commented Jun 24, 2020 at 10:42
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    $\begingroup$ Look at this: chemistry.stackexchange.com/questions/135626/… ---- Search "acetoacetic" in reference i PDF. NaOCl gives dichloroacetic acid. It is unlikely that the diiodide, under iodoform conditions, is ever formed although the diiodide of AAE is known. The methyl group, which is less acidic than the methylene group, is not involved. $\endgroup$
    – user55119
    Commented Jun 24, 2020 at 12:38

1 Answer 1


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.


  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)
  • $\begingroup$ what about any other haloform test? $\endgroup$
    – granthium
    Commented Jun 30, 2020 at 18:27
  • $\begingroup$ i wanted to know for any halogen ( i have now edited the question) $\endgroup$
    – granthium
    Commented Jun 30, 2020 at 18:29
  • $\begingroup$ See the comment from @user55119 above. The reaction works with NaOCl and Brominating agents. $\endgroup$
    – Waylander
    Commented Jun 30, 2020 at 19:21

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