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What products will be formed when $\ce{Br2/KOH}$ reacts with the following compound in water at room temperature?

Succinic anhydride

Here is what I think is possible:

  1. $\ce{OH-}$ will take up the acidic hydrogen.
  2. $\ce{OH-}$ can attack the carbonyl group.
  3. $\ce{OBr-}$ can attack the carbonyl group.
  4. $\ce{Br+}$ can attack the lone pair of the doubly bonded oxygen and break the carbonyl bond.
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