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I am currently learning about the iodoform reaction, i.e. when CH3COR compounds react with iodine and sodium hydroxide to form iodoform.

My course book and Chemguide both says that the reaction is as follows:

$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow CHI_3 + RCOO^- +3I^- +3H_2O}$$

However, this picture from Wikipedia suggests that hydrogen iodide is formed instead of water:

enter image description here

Which one is correct?

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Your balanced equation is correct, both water and iodide (and iodide in water will be in equilibrium with hydrogen iodide) are formed in the iodoform reaction.

$$ \mathrm{CH_3COR + 3I_2 +4OH^- \rightarrow CHI_3 + RCOO^- +3I^- +3H_2O}$$

The Wikipedia article that you mention, could have been written a bit clearer. Specifically, there are two key steps in the iodoform reaction. First, there is formation of the tri-iodo ketone and water is generated in this step.

enter image description here

In a subsequent step, the carbonyl undergoes an SN2 reaction to generate iodoform.

enter image description here

The Wikipedia drawing in your post seems to focus on the mechanism of the second step (3 to 4b). The first step (2 to 3) is just illustrated in a cursory fashion and the formation of water is not included in the figure.

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  • $\begingroup$ Do you mean that the reaction does not directly produce hydrogen iodide (as Wikipedia suggested), but produces iodide ion which forms an equilibrium with water and hydrogen iodide? So, is it wrong to say $ \mathrm{CH_3COR + 3I_2 +OH^- \rightarrow CHI_3 + RCOO^- +3HI}$? $\endgroup$ – ace_HongKongIndependence Sep 17 '14 at 15:04
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    $\begingroup$ @ace "So, is it wrong to say..." Well technically yes because 1) you don't show the water that is formed and 2) you show HI which is really formed indirectly - from the iodide that is formed first. $\endgroup$ – ron Sep 17 '14 at 15:31
  • $\begingroup$ Also remember that the reaction takes place in strongly basic solution. Any HI, a very strong acid, that may be formed will not stay in the form very long. $\endgroup$ – Ian Bush Mar 16 '18 at 8:55

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