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“What is the product of the reaction between propanone, sodium hyroxide and Iodine? Is the carboxylic acid form or does it break down to the carbonate ion?
8 years ago
Chemistry Question asked by
Add a Commenttriiodomethane (iodoform) reaction with alcohols
http://www.chemguide.co.uk/organicprops/alcohols/iodoform.html
Please see this link for more information. All best, Alan
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3 Answers
The reaction between Propanone (a methyl ketone, Iodine and Sodium hydroxide is called as the IODOFORM REACTION. The oxidation product here is always a Carboxylate ion (generally RCOO-, CH3COO- in this case), NOT Carbonate ion (which is an inorganic ion). For details, please go through the Chemistry of Iodoform reaction.
Answered by [Deleted Member]
Equations for the triiodomethane (iodoform) reaction
We will take the reagents as being iodine and sodium hydroxide solution.
The first stage involves substitution of all three hydrogens in the methyl group by iodine atoms. The presence of hydroxide ions is important for the reaction to happen - they take part in the mechanism for the reaction (not required for UK A level).
In the second stage, the bond between the C I3 and the rest of the molecule is broken to produce triiodomethane (iodoform) and the salt of an acid.
Putting all this together gives the overall equation for the reaction:
you can see the equations in link below:
www.chemguide/organicprops/carbonyls/iodoform.html
We will take the reagents as being iodine and sodium hydroxide solution.
The first stage involves substitution of all three hydrogens in the methyl group by iodine atoms. The presence of hydroxide ions is important for the reaction to happen - they take part in the mechanism for the reaction (not required for UK A level).
In the second stage, the bond between the C I3 and the rest of the molecule is broken to produce triiodomethane (iodoform) and the salt of an acid.
Putting all this together gives the overall equation for the reaction:
you can see the equations in link below:
www.chemguide/organicprops/carbonyls/iodoform.html
Answered by [Deleted Member]
Iodine is a very good oxidising agent in alkaline solution but it can not oxidise Propanone further to Propanoic acid because it is a Ketone and not an aldehyde and there is no H attached to the Carbon in the Carbonyl group which could have led to formation of a COOH group. However, this reaction of Propanone with I2 in NaOH is a usual iodoform reaction in which CHI3 alongwith Sodium Ethanoate and Iodide ions would result.
