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I am having some trouble figuring out how this reaction proceeds:

Reaction of (S)-cyclohex-3-ene-1-carboxylic acid to form (1S,5S)-6-oxabicyclo[3.2.1]oct-3-en-7-one

Attempt 1: The most obvious thing to do is deprotonate the acid group. I tried using the carboxylate group to attack the alkene, leaving the molecule negatively charged and not really knowing what to do next! If this happened, water would certainly protonate it and then it wouldn't be unsaturated, so it can't be that simple.

Attempt 2: Using the alkene to deprotonate the acid, there will be a positive charge at (one of) the alkene carbons. Then I tried using hydroxide to add in an alcohol group but to no avail.

Is there something special going on because of the cyclic structure?

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    $\begingroup$ This doesn't make any sense. How did you lose an equivalent of $\ce{H2}$ from the reactant? $\endgroup$ – Zhe Oct 26 '17 at 2:43
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The reaction as you are given turns up no results in SciFinder, meaning it has not been reported in any scientific journal. It is thus the imagination of whoever set the test.

However, closely related reactions are reported and these may come as no surprise to the advanced chemist. They mainly differ in that iodine $\ce{I2}$ is added to the reaction mixture. The mechanism is called iodolactonisation.

Basically, iodine is attacked by the double bond to form an iodonium ion (whether or not this is actually a three-membered ring shall be deemed irrelevant in the context of this answer). This is then rear-side attacked in an intramolecular $\mathrm{S_N2}$ reaction by the deprotonated carboxyl group to form the five-membered lactone kinetically. In a subsequent step, the iodo substituent is removed in an $\mathrm{E2}$ type reaction with a base to regenerate the double bond; due to steric constraints relating to the required anti-periplanar transition state, this newly-formed double bond is shifted along the six-membered ring.

Mechanism of the iodolactonisation

This is described for your compound multiple times in the literature using different bases, for example by Khanjin et al.[1] They use DBU as a milder alternative.

Reference

[1]: N. A. Khanjin, J.P. Snyder, F. M. Menger, J. Am. Chem. Soc. 1999, 121, 11831–11846. DOI: 10.1021/ja992453d.

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  • $\begingroup$ Makes a lot more sense. I wonder if steps 1 and 2 are under Curtin-Hammett conditions or if there is good selectivity in the first step to put the iodinium anti from the carboxylate... $\endgroup$ – Zhe Oct 26 '17 at 13:27
  • $\begingroup$ my question sheet had an error! thank god, otherwise I would have been very confused! thanks for clarifying the mechanism with halogen. $\endgroup$ – gamma1 Oct 26 '17 at 15:46
  • $\begingroup$ @Zhe I guess we can argue with steric hindrance of the carboxy group to direct the attack of iodine to the back side. Or, we get equilibrium conditions. $\endgroup$ – Jan Oct 27 '17 at 3:40

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