What is the mechanism for the iodine substitution reaction of ketones?

Propanone readily forms 1-iodopropanone on reaction with acidified iodine solution, as do all 2–ones ('methyl ketones') I assume?

• propanone + iodine ==> iodopropanone + hydrogen iodide

  • Note: (a) Strictly speaking the 1-iodo isn't needed in the name

  • (b) The hydrogen iodide is in aqueous solution and therefore strictly speaking is hydriodic acid HI(aq).

• CH₃COCH_3(aq) + I_2(aq) ==> CH₃COCH₂I_(aq) + H⁺_(aq) + I^–_(aq) 

However, from experimental observations, the rate expression is:

• rate = k [ketone/propane] [acid hydrogen ion]

• rate = k₂[CH₃COCH_3(aq)][H⁺_(aq)]

• and iodine is not in the rate expression but one of the products is!

• Therefore the reaction is zero order for iodine and it also zero order for bromine in the similar bromination reaction.

This suggests there is a slow rate determining step involving the ketone and the hydrogen ion in the mechanism and what ever happens next e.g. involving the iodine, is much faster. A proposed mechanism is shown below, where R = CH₃ if it was propanone. Note that very little can be absolutely proved in mechanistic detail and you will find variations of this diagram on the web.

Other examples of iodine reacting with ketones (specifically '2-ones')

• (ii) + I₂ ==> + HI

• butanone + iodine ==> 1-iodobutanone + hydrogen iodide

• (ii) + I₂ ==>   + HI

• pentan-2-one + iodine ==> 1-iodopentan-2-one + hydrogen iodide

So, what is the reaction mechanism for propanone reacting with iodine under acidic conditions.

The mechanism for propanone, with only one slow step suggested is ...

• Step (1) (CH₃)₂C=O + H₃O⁺ (CH₃)₂C=O⁺H + H₂O

  • The ketone is reversibly protonated on the oxygen (+) by the acid in an acid–base reaction (proton transfer).

• Step (2) (CH₃)₂C=O⁺H   (CH₃)₂C⁺–OH

  • The electrons 'between' the C–O partly shift to form a carbocation i.e. the positive charge is transferred from the oxygen to the carbon.

• Step (3) (CH₃)₂C⁺–OH + H₂O CH₃C(OH)=CH₂ + H₃O⁺ 

  • The carbocation, derived from the protonated ketone, loses a proton and slowly changes into the 'enol'* form.

  • This involves breaking a strong C–H bond, hence a high activation energy and slow speed. The positive charge on the adjacent carbon of the carbocation facilitates in 'pulling' the C–H bond pair to form the C=C bond and release the proton to a water molecule.

  • The rate of formation of the enol thus depends on the concentrations of the ketone and the acid, explaining the rate equation experimentally found. Also note that it is an example of autocatalysis because one of the reaction products is the oxonium ion!

  • * An 'enol' has both alkene and alcohol functional groups and is isomeric with the original ketone. This is an example of functional group isomerism involving a dynamic equilibrium of the two isomers (the original ketone and enol formed) and is sometimes called an example of tautomerism.

• Step (4)  CH₃C(OH)=CH₂ + I₂ CH₃C(=O⁺H)–CH₂I + I^– 

  • Half of the iodine molecule (an electrophile) then quickly adds to the 'enol' (just like any reactive alkene), and the oxygen then carries the positive charge (not on the carbon as in electrophilic addition to alkenes), and the protonated iodoketone is formed.

• Step (5) CH₃C(=O⁺H)–CH₂I + H₂O    CH₃COCH₂I + H₃O⁺ 

  • Then a water molecule rapidly, and reversibly, removes the proton in another acid–base reaction to leave the iodoketone.

• For the slow step (3), the rate depends on the isomerisation of protonated ketone/carbocation, which in turn will depend on the concentration of the ketone AND the acid providing the H₃O⁺ ion.

Note:

• The same reaction is catalysed by bases and proceeds by a different mechanism and gives different products ultimately. Multiple substitution takes place, initially forming 1–iodopropanone, then 1,1–diidopropanone, and then 1,1,1–triiodopropanone. Finally, the carbon chain splits to give triiodomethane, CHI₃, i.e. its the 'iodoform' reaction given by ethanol, ethanal, and all 2–ones ('methyl ketones').

  • Two typical bases that catalyse this reaction are the  hydroxide ion, OH^– from aqueous sodium hydroxide, and the ethanoate ion CH₃COO^– from aqueous sodium ethanoate.

  • e.g. from kinetic studies of the base catalysed iodination/halogenation of ketones, the mechanism of the halogenation (iodination) of propanone catalysed by the hydroxide ion is ...

    • step (1) CH₃COCH₃ + OH^– [CH₃COCH₂]^– + H₂O  

      • (rds, rate determining step)

    • step (2) [CH₃COCH₂]^– + I₂ ===> CH₃COCH₂I + I^–

    • Note:

    • (a) The mechanism would be similar for the ethanoate ion.

    • (b) [CH₃COCH₂]^– is an example of a carbanion, a negatively charged carbon based anion.

    • (c) overall reaction: CH₃COCH₃ + OH^– +  I₂ ===> CH₃COCH₂I + H₂O + I^–

    • (d) The rate expression is: rate = k [ketone/propanone] [base/hydroxide ion]

    • i.e. rate = k₂ [CH₃COCH₃] [OH^–]

• The individual products are almost impossible to isolate in the base catalysed reaction, but in the acid catalysed reaction, the rate of halogenation decreases with successive halogen atom substitution, so it is possible to isolate e.g. 1–iodopropanone, and 1,1–diiodopropanone and 1,1,1–triiodopropanone presumably? and molecules such as 1,3–diiodopropanone may be formed, but I'm not sure on this one?

• The acid catalysed reaction mechanism is often presented, and not unreasonably at UK pre-university A level, as a simpler three step mechanism, with Step (1) as the rds and clearly showing the proton's role in this acid catalysed reaction.

  1. Step (1)  (CH₃)₂C=O + H⁺ (CH₃)₂C=O⁺H

     • SLOW protonation of the ketone by the hydrogen ion is the rate determining step.

  2. Step (2)  (CH₃)₂C=O⁺H CH₃C(OH)=CH₂ + H⁺

     • FAST deprotonation and rearrangement to give the enol form

  3. Step (3)  CH₃C(OH)=CH₂ + I₂ CH₃COCH₂I + HI

     • FAST equivalent of adding I–I across the carbon–carbon double bond >C=C< and then elimination of HI.