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The problem is such: enter image description here

So far my working has been:

enter image description here

I'm concerned about the last step where the ester forms a ring. The methyl group doesn't look like a plausible leaving group, but it does not appear in the product so how can I get rid of it?

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  • $\begingroup$ @A.K. Sorry! I was having difficulty uploading the image.. Please have a look now! $\endgroup$ – justbehappy Dec 29 '15 at 23:12
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    $\begingroup$ I think your top drawing is incomplete. Specifically, hydrolysis of the ester is required in order to first convert it to the carboxylate anion. It is the oxygen in this anion that reacts with the bromonium ion to form the intramolecular lactone. To achieve this the bromination takes place in the presence of sodium bicarbonate which hydrolyzes the ester. You can google "bromolactonization" or see this link for the analogous iodolactonization. $\endgroup$ – ron Dec 29 '15 at 23:19
  • $\begingroup$ @ron yes thanks for pointing it out. My working contains the next few steps of the reaction mechanism, but i had the most difficulty with the first steps as pictured $\endgroup$ – justbehappy Dec 29 '15 at 23:20
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A methyl cation definitely won’t leave by itself. However, you can consider bromide ions floating around in solution. These bromide, being nucleophilic can attack the methyl group in an $\mathrm{S_N2}$ manner, because any positively charged oxygen is a good leaving group. Therefore, the side product would be bromomethane.

Other than that, your mechanism is looking good.

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  • $\begingroup$ i was wondering what CHCl3 does in the reaction? $\endgroup$ – justbehappy Dec 29 '15 at 23:17
  • $\begingroup$ @justbehappy Be a solvent, I guess. (There shouldn’t be anything it does to participate.) $\endgroup$ – Jan Dec 29 '15 at 23:17
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I propose:

  1. Hydrolysis of ester to acid through basic hydrolysis
  2. Halogenation of the double bond
  3. Having in mind the leaving groups, you get the desired cycle.
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    $\begingroup$ The first set of conditions provided by the o/p don't permit the ester to be 'reduced' (though the link provided is for basic hydrolysis). @ Jan's solution seems more plausible, though I agree with @Ron that it is much more common to hydrolyse the ester first before closing the ring. $\endgroup$ – NotEvans. Dec 30 '15 at 0:08

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