[djiang87]

I just did a prac at uni which was titled 'kinetics of propanone iodination by titration'. I had to perform 5 different 'runs'. One was a reference and the other 4 i varied the value of propanone and sulfuric acid. Then i added some iodine in it. After i left the flask in a waterbath for 10 minutes i added more iodine into it. Then after adding some acetate i titrated the solution against sodium thiosulphate. Now i have to interpret my data that i got from this experiment which is a problem for me. I then calculated the conc of iodine in each flask from the titrations. However i need help with some of the questions:

For the eqn's [A]~[A]^0 - k't for a=0 and 1/[A]~1/[A]o+k't for a=2 i have to relate them to the relevant graph. Problem is wot types of graphs do these eqns go to? Im thinking the first one goes to a linear plot with negative slope? and im not too sure bout the 2nd one

wot species are involved in the rate determining step? This i have no clue at all

And finally given that there is an equilibrium between propanone and its enol and that halogens rapidly attatch carbon to carbon double bonds devise a plausible mechanism for the reaction? this im not too sure either

There are other questions but i think i got them. Any help is much appreciated. Thanks

[Johnny010]

I am going to assume k' is a pseudo first order rate constant:

First ln[A]_t=ln[A]₀ - k't

Plot ln[A]_t vs. t
This will give a negative slope with a gradient of -k'

Then using this value for the rxn rate (assuming this is a first order rxn):

ln k' = ln A - Ea/R * 1/T

Plot ln k' vs 1/T to get another negative sloped graph, with -Ea/R as the gradient.

From that you get Ea.


Look up Enolates for the mechanism:

The acidic H causes enolisation (the formation of a dbl bond adjacent to a hydroxyl carbon) C=C(OH).

The enolisation is highly in favour of the ketone so the enolate is the rate determining step/species of the reaction.

Although in a very low conc. (approx to the order of 1 in a million) it is a very reactive species: The iodine will add to the enol:

The OH group has a lone pair, which falls into the dbl bond, which then attacks one of the Iodine atoms of I₂.
The resultant compound has a C=O⁺H group, which naturally falls to form a keto group again + H⁺.

Hope this helped.



An attempt at a mechanism (RDS=rate determining step):

C-C-C=O:  H⁺_H2SO4 < - >R.D.S  C-C=C-O:H   I-I  (electrons from O go to dbl bond, electrons from dbl bond attack an I) -> C-C(Br)-C=O+H -> Product