Iodine propanone mechanism

Which step is slow?

Composition and transformation of substance.

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Iodine propanone mechanism

#1  Postby Corke » Dec 26, 2012 12:02 pm

I'm revising for my exams, which are coming up in January. One of the things that's come up is this mechanism:

Image

Iodine reacts with propanone to form an iodoketone. We're looking at the rate of reaction, which is given by this:

rate = k[CH3COCH3][H+]

So is the first step the rate-determining step? Or is it the third step?

The textbook says it's the third step, but I came across a forum full of teachers saying that it's much easier to teach that it's the first step. Also, my chemistry teacher said that that mechanism is wrong and gave me another version, but I lost it.

So now I'm really confused. The mechanism looks okay to me, and I can understand why it's the third step that's slowest (because you have to break a C-H bond) but is it?
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Re: Iodine propanone mechanism

#2  Postby TopCat » Jan 15, 2015 1:51 pm

There are a few things to say about this. Firstly, your forum full of 'teachers' should be shot, or at least, whacked hard with a big heavy implement until the urge to talk bollocks goes away. Protonation of an oxygen lone pair is almost never the rate determining step, it's always fast. And whether they think it's easier to teach that way should be irrelevant - they shouldn't be teaching it if it's wrong.

Secondly, no one with any sense writes organic chemical structures like in your diagram any more (not for the last thirty-plus years at least).

You'd be much better off drawing the structures and the mechanism like this:

Image

Hope you'll agree that's clearer! All the literature does it this way (ie without including an explicit 'C' for carbons or 'H' for their attached hydrogens, unless there's some particular reason for drawing attention to it), so best get used to it unless you have teachers from the early 20th century insisting on the older convention.

Now, you're welcome to draw H+ as H3O+ if you like, I've omitted all that for clarity.

The rate-determining step is indeed loss of the C1 proton to form the enol tautomer (your step 3).

But why, I hear you ask, in that case, is the concentration of H+ included in the rate equation?

Well, the rate of the reaction is the rate of the rate-determining step, so:

Rate = k1P, where P is the concentration of the protonated keto-form of the starting material.

We don't know what P is, but we do know that it's in an equilibrium reaction with the unprotonated form, with another rate constant I've labelled k2.

k2 = P/AH, where A and H are the concentrations of acetone, and H+ respectively.

Rearranging and substituting for P, we get:

Rate = k1k2AH, and since if you multiply a constant by another constant, you get a constant,

Rate = kAH, as given originally.

Hope that helps! Sorry I didn't see your question earlier, I'm fairly new here.


By the way, if you want a comprehensive undergraduate-level organic chemistry textbook, I can wholeheartedly recommend this one, by Stuart Warren (the principal author), Jonathan Clayden and Nick Greeves, both of whom were in Warren's research group in Cambridge in the 80s and are very well respected in their own right, now.

Stuart Warren is without a doubt the most lucid explainer of organic chemistry concepts I have ever encountered, he was a fantastic lecturer and tutor, and his clear style comes across very strongly in that textbook (which I'm actually in the process of reading cover to cover right now!).

I have no financial connection with any of the authors.
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Re: Iodine propanone mechanism

#3  Postby Calilasseia » Mar 08, 2015 5:57 pm

I now really want to add that textbook to my own library. :)
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