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Topic: Preparation of Iodoform  (Read 11428 times)

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Offline Chump0

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Preparation of Iodoform
« on: April 23, 2010, 04:09:16 PM »
Hello again,

I am now trying to resolve the following question:

Treatment of pentan-2-one with sodium hydroxide and iodine in aqueous solution causes a reaction that results in a yellow precipitate with an "antiseptic" odor. (hmm what would that be :P) However, treatment of pentan-3-one does not give the same precipitate. Show the mechanisms of the reactions and suggest what the precipitate is.

Obviously (for me) the precipitate is Iodoform CHI3 and it's know as an disinfectant with an "antiseptic" odor.
So let's move on, I know the mechanism of pentan-2-one, and can be found underneath:


The point where I am stuck is that I'm not sure what the mechanism of pentan-3-one is.
I have deduced the mechanism of it from the mechanism of pentan-2-one. In the question is says that pentan-3-one does not form Iodoform so obviously the end product has to be something slightly different. Here's what I came up with:


My question is; Is my previous deduction correct? Or is my mechanism wrong?

Thank you :D

Offline Doc Oc

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Re: Preparation of Iodoform
« Reply #1 on: April 23, 2010, 05:25:41 PM »
You won't see hydrolysis of the 3-pentanone, even if you iodinate both alpha carbons.  What is the key difference and what explanation can you make?

Offline Chump0

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Re: Preparation of Iodoform
« Reply #2 on: April 24, 2010, 12:16:08 PM »
You won't see hydrolysis of the 3-pentanone, even if you iodinate both alpha carbons.  What is the key difference and what explanation can you make?

From what I understood is that you say the mechanism ends with eventually this product, because 3-pentanone does not hydrolyse.


The key difference that I can make is the fact that both alpha carbons are connected to a methyl group. While in 2-pentanone the alpha carbon is not connected to a methyl group.
Are you suggesting that the methyl group is hindering the hydroylisis?
« Last Edit: April 24, 2010, 12:37:13 PM by Chump0 »

Offline Doc Oc

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Re: Preparation of Iodoform
« Reply #3 on: April 24, 2010, 11:06:47 PM »
Yes, the alkyl group is the key.  What properties are different between the methyl group and an iodide?

Offline Chump0

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Re: Preparation of Iodoform
« Reply #4 on: April 25, 2010, 08:58:34 AM »
Yes, the alkyl group is the key.  What properties are different between the methyl group and an iodide?

I don't want to say it, but I really don't know.... :(

Offline Doc Oc

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Re: Preparation of Iodoform
« Reply #5 on: April 25, 2010, 09:35:35 AM »
That's okay, this is how you learn things.

Think in terms of sterics or electronics.  In this situation, is it more likely that sterics are affecting the hydrolysis or the electronics?

Offline Chump0

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Re: Preparation of Iodoform
« Reply #6 on: April 25, 2010, 09:43:47 AM »
That's okay, this is how you learn things.

Think in terms of sterics or electronics.  In this situation, is it more likely that sterics are affecting the hydrolysis or the electronics?

I don't think sterics have any effect, I thinks it's due the electronics. If both groups left and right to the middle carbon atom break the bond you will get a 2+ charge on the carbon.
And that's not favorable.

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