Sign Up
Tutor Hunt on twitter Tutor Hunt on facebook

Tutor HuntQuestions

Tutor Hunt Questions

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?
2 years ago

Chemistry Question asked by Vijaylaxmi

triiodomethane (iodoform) reaction with alcohols http://www.chemguide.co.uk/organicprops/alcohols/iodoform.html Please see this link for more information. All best, Alan
21/11/2015 20:08:02 | comment by Alan
+ 8     Rate Up  Rate Down
Know the Answer?

Please enter your response to the question below. The student will get a notification as soon your response has been approved by our moderation team.

Submit Answer

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 Monika | 2 years ago
    Rate Up  Rate Down     
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


Answered by Abdulahad | 24 months ago
    Rate Up  Rate Down     
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.
Answered by Deepa | 2 years ago
    Rate Up  Rate Down     
Ask an Academic Question
Popular Links
© 2005-2017 Tutor Hunt - All Rights Reserved

Privacy | Copyright | Terms of Service
loaded in 0.031 seconds
twitter    facebook