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The iodination of propanone is a reaction with interesting kinetics. The reaction is acid-catalysed.

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Introduction

A DETAILED KINETIC STUDY The iodination of propanone CH3COCH3 (aq) + I2 (aq) ?? CH3COCH2I (aq) + HI (aq) is a reaction with interesting kinetics. The reaction is acid-catalysed. From the equation, one might postulate that the rate equation would be 123 A study of the reaction will involve finding a and b. First: Find the order with respect to iodine It is arranged that the propanone concentration is much greater than the iodine concentration, e.g., [CH3COCH3] = 1.00 mol dm?3 and [I2] = 0.00500 rnol dm?3 so that, at the end of run, [CH3COCH3] = 0.995 mol dm?3, a decrease of 0.5 %. ...read more.

Middle

This stops the reaction instantly by neutralizing the acid. The time at which the reaction stops is recorded. The iodine that remains is determined by titration against a standard solution of sodium thiosulphate. The analysis is repeated at intervals of a few minutes. The volume of thiosulphate required is plotted against the time elapsed since the start of the reaction. The figure attached shows the plot obtained. It is a straight line: the gradient of the graph does not change as the concentration of iodine decreases. This shows that the rate of reaction remains constant as the iodine concentration decreases: The reaction is zero-order with respect to iodine, and the rate equation becomes Second: Find the order with respect to propanone The procedure is repeated with different concentrations of propanone. ...read more.

Conclusion

The rate equation is Interpretation of the rate equation Iodine must be present for iodination to occur, but its concentration does not appear in the rate equation. To explain this, it is suggested that the reaction takes place in steps. It is the rate of the slowest step that determines the rate of the overall reaction. If iodine is involved in a step which is too fast to be rate-determining, it will not appear in the rate equation. The suggested mechanism for the reaction is This is the slow, rate-determining step. An enol is formed. Since halogens are electrophiles, iodine reacts rapidly with the C=C bond through its ?-electrons: The intermediate formed has a positive charge on a carbon atom, and fast loses a proton to form the iodoketone: ?? ?? ?? ?? ...read more.

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