# Addition of Br2 to Double Bond and Carboxylic Acid

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)

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?

• Under basic conditions I would definitely say that this is the product. Under neutral conditions I'm not so sure.
– bon
Jul 19 '16 at 9:08
• Thanks, @bon, and is the reaction mechanism that I proposed correct? Jul 19 '16 at 9:10
• Yes it is essentially right. The intermediate will be a cyclic bromonium ion.
– bon
Jul 19 '16 at 9:12
• Just google iodolactonisation. It's basically that but with bromine rather than iodine. Ideally you'd want base involved to deprotonate the carboxylic acid Jul 19 '16 at 17:22

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.