Question

The reaction below is called ___.
$ C{{H}_{3}}COOAg+B{{r}_{2}}\to C{{H}_{3}}Br+C{{O}_{2}}+AgBr. $

Answer 1

Hint: We know that the Hunsdiecker Reaction is defined as a chemical reaction which involves in the carboxylic acid silver salts reacting to halogens to create an unstable intermediate that further undergoes decarboxylation thermally leading to the formation of the final product referred to as alkyl halides.

Complete answer:
In the Hunsdiecker reaction, silver salts of carboxylic acid react with a halogen to form an organic halide. Therefore, the reaction mechanism of Hunsdiecker reaction involves organic radical intermediates instead of carbocation or carbanion. $ C{{H}_{3}}COOAg+B{{r}_{2}}\to C{{H}_{3}}Br+C{{O}_{2}}+AgBr. $
Now we can see from the above reaction, The Hunsdiecker reaction is the reaction of silver carboxylate with a halogen to form an alkyl halide. The Hunsdiecker reaction which is also known as Borodin-reaction or Hunsdiecker-Borodin-reaction in which carboxylation and halogenation both can be seen.
 $ C{{H}_{3}}COOAg+B{{r}_{2}}\to C{{H}_{3}}Br+C{{O}_{2}}+AgBr, $ is known as Hunsdiecker reaction. In this reaction, silver salt of a carboxylic acid is heated with bromine to form alkyl halide.

Additional Information:
The Hunsdiecker reaction also contains several variations. There exist cases where silver carboxylate is exchanged with thallium carboxylate, and then the carboxylic acid is either reacted or treated with halide ions of chloride and iodide bromine, and lead to tetra acetate majorly to effect decarboxylation and halogenation.

Note:
Remember; don’t confuse the Hunsdiecker reaction with the Kochi reaction because as we know Kochi reaction is an organic reaction in which decarboxylation of carboxylic acids to form alkyl halide with lead $ \left( IV \right) $ acetate and a lithium halide.