Step-by-step Derivations

1. Identify the Reaction and Mechanism: The reaction involves alkyl chlorides (P, Q, R, S) reacting with potassium iodide (KI) in acetone. This is a classic Finkelstein reaction. KI provides the nucleophile, iodide ion ($I^{-}$), and acetone is a polar aprotic solvent. These conditions strongly favor an $S_N2$ (bimolecular nucleophilic substitution) mechanism. The question explicitly states that the reaction is $S_N2$.

2. Governing Factor for $S_N2$ Rate: The rate of an $S_N2$ reaction is primarily determined by steric hindrance around the α-carbon (the carbon atom bonded to the leaving group, which is Chlorine in this case). The nucleophile ($I^{-}$) must attack the α-carbon from the side opposite to the leaving group (backside attack). Any bulky groups on or near the α-carbon will hinder this attack and slow down the reaction rate. The general reactivity order for $S_N2$ reactions based on the substrate structure is: Methyl > Primary (1°) > Secondary (2°) >> Tertiary (3°) (practically no reaction). Furthermore, branching at the β-carbon (the carbon adjacent to the α-carbon) also increases steric hindrance and decreases the reaction rate.

3. Analyze the Structures of Substrates P, Q, R, and S: Let's analyze the structure of each reactant to assess the steric hindrance.

   • P: $CH_3-CH_2-CH_2-CH_2-Cl$ (1-chlorobutane or n-butyl chloride) This is a primary (1°) alkyl halide. The carbon chain is unbranched. Steric hindrance is minimal.

   • Q: $CH_3-CH_2-CH(Cl)-CH_3$ (2-chlorobutane or sec-butyl chloride) This is a secondary (2°) alkyl halide. The α-carbon is bonded to a methyl group, an ethyl group, and a hydrogen atom. The presence of two alkyl groups on the α-carbon creates significant steric hindrance compared to a primary halide.

   • R: $(CH_3)_2CH-CH_2-Cl$ (1-chloro-2-methylpropane or isobutyl chloride) This is a primary (1°) alkyl halide. However, there is branching at the β-carbon (an isopropyl group is attached to the $-CH_2Cl$ group). This β-branching hinders the nucleophile's approach, making it less reactive than an unbranched primary halide like P.

   • S: $(CH_3)_3C-CH_2-Cl$ (1-chloro-2,2-dimethylpropane or neopentyl chloride) This is also a primary (1°) alkyl halide. However, it has a very bulky tertiary-butyl group at the β-position. This creates extreme steric hindrance, effectively blocking the backside attack of the nucleophile. Neopentyl halides are famously known for being extremely unreactive in $S_N2$ reactions.

4. Compare the Reaction Rates:

   • Comparing Primary Halides (P, R, S): All are 1° halides. The rate will decrease as steric hindrance from β-branching increases.

     • P has no β-branching.
     • R has one additional methyl group on the β-carbon compared to a straight chain.
     • S has a t-butyl group on the β-carbon, which is extremely bulky. The order of increasing steric hindrance is P < R < S. Therefore, the order of reaction rates is P > R > S.

   • Comparing with the Secondary Halide (Q): Q is a 2° halide. It is more hindered at the reaction site than unbranched (P) or moderately branched (R) primary halides. Thus, Q will be slower than P and R. P > Q and R > Q.

   • Comparing Q and S: Q is a secondary halide. S is a neopentyl halide. The steric hindrance in neopentyl systems is so severe that they are even less reactive than secondary halides in $S_N2$ reactions. Therefore, Q > S.

5. Determine the Overall Order: Combining the comparisons from step 4:

   • P is the most reactive (unbranched 1°).
   • R is next (β-branched 1°).
   • Q is next (2°).
   • S is the least reactive (extremely hindered neopentyl system). The overall order of reaction rates is: P > R > Q > S.

6. Conclusion: The derived order of rates is P > R > Q > S. This corresponds to option C. The stored answer is B (S > P > R > Q), which claims that the most sterically hindered compound (S) is the most reactive. This contradicts the fundamental principles of the $S_N2$ mechanism. Therefore, the stored answer appears to be incorrect.