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The iodoform test involves an acid base reaction with NaOH in which the alpha hydrogen atoms (hydrogen atoms on carbon 3) react with hydroxide ions followed by substitution of iodine. This is expected in case of ketones like acetone, alcohols like propanol and an aldehyde like methanal. However, my question is whether this compound (1-phenylpropan-2-one) also undergo a similar reaction.

My thoughts:

What I think is that instead of the hydroxide ions attacking the hydrogen atoms on carbon 3, they should attack those on carbon 1 as the carbon 1 has more acidic hydrogen atoms.Thus iodoform wouldn't form. Is my logic correct?

Edit:After the iodination of carbon 1, the attack,I believe, would happen on the carbonyl carbon just like a di keto compound, say acetylacetone would react after all the hydrogen atoms attached to the active methylene are iodinated. For reference, see https://chemistry.stackexchange.com/a/84448/144210

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    $\begingroup$ You are correct that the protons in position 1 are the most acidic so will iodinate first. However, once that has ocurred, the protons in position 3 will iodination resulting in a positive test. $\endgroup$
    – Waylander
    Commented Jun 4 at 6:35
  • $\begingroup$ The major problem I have is that I believe that this reaction must proceed in the same way as a di-keto compound would react with halogen.See chemistry.stackexchange.com/a/84448/144210 $\endgroup$
    – The Fright Of The Night
    Commented Jun 4 at 15:42
  • $\begingroup$ Acetylacetone forms the dibromide under haloform conditions. However, it is unlikely that the diiodide forms under these conditions. Look here. $\endgroup$
    – user55119
    Commented Jun 4 at 17:23
  • $\begingroup$ With benzyl methyl ketone after di-halogenation the attack of OH- on the carbonyl cannot go as drawn in the diketo example as there is no di-haloenolate that can be formed $\endgroup$
    – Waylander
    Commented Jun 4 at 17:41
  • $\begingroup$ @Waylander why is that required. I mean even when acetone reacts with the given reactants, after the iodination on one side is over , the carbonyl carbon is attacked. $\endgroup$
    – The Fright Of The Night
    Commented Jun 6 at 15:02

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