There is a question in my organic chemistry textbook with an answer. What it doesn't explain (and neither does my theory textbook, I checked) is the mechanism involved in the reaction. The reaction is as follows:

synthesis of 5‐(iodomethyl)oxolan‐2‐one

Can someone help me?

  • $\begingroup$ If the reaction was simply iodine adding to a double bond do you know how it works? $\endgroup$
    – Waylander
    Jul 13, 2019 at 17:21
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    $\begingroup$ @ClaudioLancia Well I- is not a great nucleophile, and internal nucleophilic attack goes better than external nucleophilic attack. Carboxylate may not be a great nucleophile but it is internal, and the formation of 5-membered rings is favoured. See where this is going? $\endgroup$
    – Waylander
    Jul 13, 2019 at 17:33
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    $\begingroup$ en.wikipedia.org/wiki/Iodolactonization $\endgroup$ Jul 13, 2019 at 17:43
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    $\begingroup$ Oh, I get it now! Thanks a lot for the help. I didn't know the name of the reaction (especially since the exercise was in the alkene section, and my book didn't mention anything about Halolactonization), so this is a great help! $\endgroup$ Jul 13, 2019 at 18:05
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    $\begingroup$ @ClaudioLancia Now that you understand, could you write up your own answer to summarize what you learned? $\endgroup$
    – Karsten
    Jul 13, 2019 at 19:23

1 Answer 1


https://en.wikipedia.org/wiki/Iodolactonization I+ reacts with the CC-π electron donor. A cyclic iodonium three-membered ring intermediate is formed. The carboxylate group opens the electrophilic intermediate to form the product lactone. 5-membered rings (gamma-lactones) https://en.wikipedia.org/wiki/Lactone are prefered due to optimum entropic and enthalpic parameters of activation.

  • $\begingroup$ Uh ... Why the -1? $\endgroup$ Jul 13, 2019 at 19:17
  • $\begingroup$ @OscarLanzi Well, it's not my downvote, but there are many bad details - links to improve, weird wording/terminology, e.g. it's simply cyclic iodonium not "epi-iodium" etc. $\endgroup$
    – Mithoron
    Jul 13, 2019 at 21:58
  • $\begingroup$ In this case, the regioselectivity (to form the 5-membered instead of the 6-membered ring) also has a stereoelectronic factor: chemistry.stackexchange.com/q/106846/16683 $\endgroup$ Jul 14, 2019 at 6:48

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