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We know that in haloform reaction alpha acidic hydrogen are substituted by halogen atom. But in the above case we find two acidic hydrogens in between the two keto groups (due to -R effect) which are more acidic than the hydrogens on the terminal methyl groups.

So what should be the products of this reaction?

  • $\begingroup$ That would depend on concentrations - how far reaction will go. $\endgroup$
    – Mithoron
    Jul 18 '18 at 12:58
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    $\begingroup$ @Mithoron do you have a pdf or any file regarding this. That'll help me. I've googled it but found nothing. $\endgroup$
    – user187604
    Jul 18 '18 at 13:07

I assume that there is sufficient base and chlorine and that the conditions are vigorous enough to get beyond dichloride 4, which, being a non-enolic β-diketone at C3, is labile toward Haller-Bauer-type cleavage. Acetylacetone 1 is highly enolic and readily deprotonated. The first two chlorinations lead to dichloride 4. Cleavage of dichloride 4 with base gives enolate 6 which can be protonated by water, or to save space, chlorinated to provide the trichloroketone 8, which is an intermediate in the haloform reaction of acetone. Cleavage with base leads to chloroform (10) and a second equivalent of sodium acetate (7).

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