Question

silver acetate + br2 in presence of ccl4 gives​

Answer 1

Explanation:

In this second propagation step, an alkyl bromide - the product - is formed and a carboxyl radical is recovered. ➡️The Hunsdiecker reaction takes place when a silver carboxylate is heated in CCl4 in the presence of bromine. ... This yields a carboxyl radical and a bromine atom.

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Answer 2

Answer:

In this reaction silver salt of carboxylic acid is treated with bromine in presence of CCl₄ to form alkyl bromides.

It is the reaction which proceeds through free radical mechanism.