Halogen Exchange and Properties of Haloalkanes

To determine the correct statement(s), let us analyze the reaction sequence step by step:

1. Identification of Compounds $\mathbf{P}$, $\mathbf{Q}$, and $\mathbf{R}$:

   • Formation of $\mathbf{P}$: Addition of $\text{HBr}$ across the double bond of cyclohexene yields bromocyclohexane. $\text{Cyclohexene} + \text{HBr} \longrightarrow \mathbf{P} \quad (\text{Bromocyclohexane, } \text{C}_6\text{H}_{11}\text{Br})$ Here, $\text{X} = \text{Br}$.

   • Formation of $\mathbf{Q}$ (Finkelstein Reaction): Treatment of an alkyl bromide with $\text{NaI}$ in dry acetone replaces bromine with iodine. $\mathbf{P} \xrightarrow{\text{NaI / acetone}} \mathbf{Q} \quad (\text{Iodocyclohexane, } \text{C}_6\text{H}_{11}\text{I})$ Here, $\text{X} = \text{I}$.

   • Formation of $\mathbf{R}$ (Swarts Reaction): Treatment of an alkyl bromide with metallic fluorides (e.g., $\text{AgF}$, $\text{Hg}_2\text{F}_2$, $\text{CoF}_2$, or $\text{SbF}_3$) replaces bromine with fluorine. $\mathbf{P} \xrightarrow{\text{Metallic Fluoride}} \mathbf{R} \quad (\text{Fluorocyclohexane, } \text{C}_6\text{H}_{11}\text{F})$ Here, $\text{X} = \text{F}$.

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1. Evaluation of the Options:

   • Option (A): $\text{C}-\text{X}$ Bond Length As the size of the halogen atom increases down the group ($\text{F} < \text{Br} < \text{I}$), the $\text{C}-\text{X}$ bond length increases: $\text{C}-\text{I} > \text{C}-\text{Br} > \text{C}-\text{F} \implies \mathbf{Q} > \mathbf{P} > \mathbf{R}$ Therefore, option (A) is incorrect.

   • Option (B): $\text{C}-\text{X}$ Bond Enthalpy Smaller halogen atoms form shorter and stronger bonds with carbon. The order of bond dissociation enthalpy is: $\text{C}-\text{F} > \text{C}-\text{Br} > \text{C}-\text{I} \implies \mathbf{R} > \mathbf{P} > \mathbf{Q}$ Therefore, option (B) is correct.

   • Option (C): Relative Reactivity Toward $\text{S}_\text{N}2$ Reaction The reactivity in nucleophilic substitution ($\text{S}_\text{N}2$) strongly depends on the leaving group ability of the halide ion ($\text{I}^- > \text{Br}^- > \text{F}^-$): $\text{Reactivity: } \mathbf{Q} > \mathbf{P} > \mathbf{R}$ Therefore, option (C) is incorrect.

   • Option (D): $\text{p}K_\text{a}$ Values of Conjugate Acids The leaving groups are $\text{Br}^-$ (for $\mathbf{P}$), $\text{I}^-$ (for $\mathbf{Q}$), and $\text{F}^-$ (for $\mathbf{R}$). Their conjugate acids are $\text{HBr}$, $\text{HI}$, and $\text{HF}$, respectively. The acidic strength order is: $\text{HI} > \text{HBr} > \text{HF}$ Since $\text{p}K_\text{a} = -\log_{10} K_\text{a}$, the $\text{p}K_\text{a}$ order is the reverse of acidity: $\text{p}K_\text{a}(\text{HF}) > \text{p}K_\text{a}(\text{HBr}) > \text{p}K_\text{a}(\text{HI}) \implies \mathbf{R} > \mathbf{P} > \mathbf{Q}$ Therefore, option (D) is incorrect.

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Correct Answer: B