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enter image description here

In the above reaction, I remember reading that Br2 in CCl4 develops a partial polarity and adds on to the double bond, resulting in addition of Br on 2 adjacent carbons.

But, as per the text, the end product is a ester. (As given below)

enter image description here

Why is this the final product, and how to judge the products in cases like this?

Note - If I look at the answer and work backwards I would say that first Br+ attacks, and a carbocation is formed, on which hydroxy group attacks to form the final product. Is this right? And of it is then, why is this happening?

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  • $\begingroup$ Under basic conditions I would definitely say that this is the product. Under neutral conditions I'm not so sure. $\endgroup$ – bon Jul 19 '16 at 9:08
  • $\begingroup$ Thanks, @bon, and is the reaction mechanism that I proposed correct? $\endgroup$ – Brahmnoor Singh Jul 19 '16 at 9:10
  • $\begingroup$ Yes it is essentially right. The intermediate will be a cyclic bromonium ion. $\endgroup$ – bon Jul 19 '16 at 9:12
  • $\begingroup$ Just google iodolactonisation. It's basically that but with bromine rather than iodine. Ideally you'd want base involved to deprotonate the carboxylic acid $\endgroup$ – RobChem Jul 19 '16 at 17:22
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The first intermediate is indeed Br+ atom attached to the double bond forming a cycle (see step one):
enter image description here
On the next step, however, oxygen of carboxylic acid attacks the intermediate and forms the product you show on second figure. This happens because carboxylic acid group is in the right position for intramolecular reaction.

First step follows the diagram. In second step the attacking nucleopline is not Br$^-$, but R-COOH or RCOO$^-$ depending on the reaction condition. To obtain high yield of the cyclic product on your second figure you need to slowly add dilute solution of Br$_2$ to a solution of organic reagent. If you add Organic reagent to neat (pure) Br2, then you get a significant amount of the product you predicted on figure 1. Don't do it in real life because a reaction of neat bromine with alkenes is violent and dangerous.

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    $\begingroup$ Reaction of bromine with olefins is in no wah dangerous— its a routine reaction. Careful addition and control of temperatire is all that's required $\endgroup$ – NotEvans. Jul 19 '16 at 13:45
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You also need to think about the frontier orbitals or Baldwin's rules and why you'll get the five membered lactone ring rather than a six membered one - bearing in mind that the nucleophilic oxygen on the carboxy group can attack two possible carbon atoms on the bromonium ion. 5 exo tet vs 6 endo tet. Or if you draw a chair confirmation you should be able to see that the anti-bonding sigma orbital on the carbon atom that would make a six membered ring is pointing at a tricky/inaccesible angle compared to the one on the 5 membered ring.

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