Hunsdiecker reaction involve free radical in intermediate steps.

Answer

**Step-by-Step Solution:**

1. **Understanding the Hunsdiecker Reaction:**
The Hunsdiecker reaction is a chemical reaction that involves the conversion of carboxylic acids into alkyl halides. The general reaction can be represented as:
\[ RCOOAg + Br_2 \xrightarrow{CCl_4, \text{reflux}} RBr + CO_2 + AgBr \]

2. **Identifying the Role of Free Radicals:**
In this reaction, we need to determine if free radicals are involved in the mechanism. Free radicals are species that have unpaired electrons and are highly reactive.

3. **Initiation Step:**
The reaction begins with the formation of a silver carboxylate complex. The carboxylic acid reacts with silver(I) ions to form the silver carboxylate:
\[ RCOO^- + Ag^+ \rightarrow RCOOAg \]

4. **Formation of Free Radicals:**
The next step involves the reaction of the silver carboxylate with bromine (Br2). This step leads to the formation of a radical intermediate:
\[ RCOOAg + Br_2 \rightarrow RCOOBr + AgBr \]
Here, RCOOBr can undergo homolytic cleavage to form free radicals:
\[ RCOOBr \rightarrow RCOO^\cdot + Br^\cdot \]

5. **Propagation Steps:**
The free radical RCOO^\cdot can then decompose to form carbon dioxide and another radical:
\[ RCOO^\cdot \rightarrow R^\cdot + CO_2 \]
The alkyl radical (R^\cdot) can then react with bromine to form the final product:
\[ R^\cdot + Br_2 \rightarrow RBr + Br^\cdot \]

6. **Conclusion:**
The Hunsdiecker reaction indeed involves free radicals as intermediates in its mechanism. The presence of radical species during the reaction confirms that the statement about the involvement of free radicals is true.

---