1. Benzoic acid (C$_6$H$_5$COOH) to n-propylbenzene (C$_6$H$_5$CH$_2$CH$_2$CH$_3$).
2. Step 1: SOCl$_2$ converts COOH to COCl (benzoyl chloride).
3. Step 2: (C$_2$H$_5$)$_2$Cd reacts with COCl to form ketone C$_6$H$_5$COCH$_2$CH$_3$.
4. Step 3: Zn-Hg/HCl (Clemmensen reduction) reduces ketone to C$_6$H$_5$CH$_2$CH$_2$CH$_3$.
5. Other options: (1) LiAlH$_4$ reduces to alcohol; (3) Grignard gives tertiary alcohol; (4) HCl is ineffective.
6. Thus, the answer is (2) SOCl$_2$, (C$_2$H$_5$)$_2$Cd, Zn-Hg, HCl.