The compound shown is P. Q is 1-chlorobutane, R is chlorobenzene and S is 2-chloro-2-methylpropane. Arrange them in decreasing order of reactivity towards $\mathrm{NaI}$ in dry acetone (SN2). Treat S as very slow but not zero, and R as essentially inert.

P is benzyl chloride, a benzylic halide whose SN2 transition state is stabilised by the ring, so it is faster than the unactivated primary halide Q. The tertiary halide S is hindered to backside attack and reacts only very slowly by SN2. Chlorobenzene R has an sp2 carbon with a stronger C-Cl bond and the ring blocks backside attack, so it is inert. Order: P > Q > S > R.