In a set of the given reactions, acetic acid yielded a product C: $CH_3COOH + PCl_5 \rightarrow A \xrightarrow[Anh. AlCl_3]{C_6H_6} B \xrightarrow[ether]{C_2H_5MgBr} C$ Product C would be:

The reaction sequence proceeds as follows:

1. Formation of A: Acetic acid ($CH_3COOH$) reacts with phosphorus pentachloride ($PCl_5$) to form acetyl chloride ($CH_3COCl$). This is a standard method for preparing acyl chlorides. $CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$
2. Formation of B: Acetyl chloride (A) reacts with benzene ($C_6H_6$) in the presence of anhydrous aluminium chloride ($AlCl_3$) to form acetophenone ($C_6H_5COCH_3$). This is the Friedel-Crafts Acylation reaction. $CH_3COCl + C_6H_6 \xrightarrow{AlCl_3} C_6H_5COCH_3$
3. Formation of C: Acetophenone (B), a ketone, reacts with ethyl magnesium bromide ($C_2H_5MgBr$, a Grignard reagent) followed by hydrolysis to form a tertiary alcohol. The alkyl group from the Grignard reagent adds to the carbonyl carbon. $C_6H_5COCH_3 + C_2H_5MgBr \xrightarrow{ether} C_6H_5-C(OMgBr)(CH_3)-C_2H_5 \xrightarrow{H_3O^+} C_6H_5-C(OH)(CH_3)-C_2H_5$

The final product C is 2-phenylbutan-2-ol.