Nucleophilic Substitution Mechanisms and Reactivity Rates of Organic Halides

To determine the correct statement(s) regarding nucleophilic substitution reactions for the given organic compounds, let us analyze each compound and evaluate the options step-by-step:

Compound Identification

• Compound I: Benzyl bromide ($\text{C}_6\text{H}_5\text{CH}_2\text{Br}$) — Primary benzylic halide.
• Compound II: (Cyclohexyl)methyl bromide ($\text{C}_6\text{H}_{11}\text{CH}_2\text{Br}$) — Primary alkyl halide.
• Compound III: tert-Butyl bromide ($(\text{CH}_3)_3\text{CBr}$) — Tertiary alkyl halide.
• Compound IV: 1-Phenylethyl bromide ($\text{C}_6\text{H}_5\text{CH(CH}_3)\text{Br}$) — Secondary benzylic halide.

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Option-by-Option Evaluation

1. Option [A]: $\text{I}$ and $\text{III}$ follow $\text{S}_\text{N}1$ mechanism

   • Compound III ($(\text{CH}_3)_3\text{CBr}$) is a tertiary alkyl halide. It readily ionizes to yield a stable tertiary carbocation, $(\text{CH}_3)_3\text{C}^+$, which proceeds via an $\text{S}_\text{N}1$ mechanism.
   • Compound I ($\text{C}_6\text{H}_5\text{CH}_2\text{Br}$) forms a resonance-stabilized benzyl carbocation ($\text{C}_6\text{H}_5\text{CH}_2^+$). Under polar protic conditions, it undergoes nucleophilic substitution via the $\text{S}_\text{N}1$ pathway.
   • Conclusion: Option [A] is correct.

2. Option [B]: $\text{I}$ and $\text{II}$ follow $\text{S}_\text{N}2$ mechanism

   • Compound II ($\text{C}_6\text{H}_{11}\text{CH}_2\text{Br}$) is a primary alkyl halide with low steric hindrance at the reaction center, making it highly suitable for an $\text{S}_\text{N}2$ substitution mechanism.
   • Compound I ($\text{C}_6\text{H}_5\text{CH}_2\text{Br}$) is a primary halide. Its transition state during $\text{S}_\text{N}2$ attack is further stabilized by the conjugation between the developing $p$-orbital and the adjacent aromatic $\pi$-system, making it exceptionally reactive towards $\text{S}_\text{N}2$ substitution.
   • Conclusion: Option [B] is correct.

3. Option [C]: Compound $\text{IV}$ undergoes inversion of configuration

   • Compound IV ($\text{C}_6\text{H}_5\text{CH(CH}_3)\text{Br}$) possesses a chiral center. When undergoing substitution via the $\text{S}_\text{N}2$ pathway, the nucleophile attacks from the side opposite to the leaving group ($-\text{Br}$), causing a Walden inversion (inversion of configuration).
   • Conclusion: Option [C] is correct.

4. Option [D]: The order of reactivity for $\text{I}$, $\text{III}$, and $\text{IV}$ is: $\text{IV} > \text{I} > \text{III}$

   • Evaluating carbocation stability for $\text{S}_\text{N}1$ reactivity:
     • Compound IV forms the 1-phenylethyl carbocation ($\text{C}_6\text{H}_5-\overset{+}{\text{C}}\text{H}-\text{CH}_3$), which is stabilized by both extended resonance with the phenyl ring and hyperconjugation/inductive effect from the methyl group.
     • Compound I forms the benzyl carbocation ($\text{C}_6\text{H}_5-\overset{+}{\text{C}}\text{H}_2$), stabilized primarily by resonance.
     • Thus, the stability and relative $\text{S}_\text{N}1$ rate follow: $\text{C}_6\text{H}_5-\overset{+}{\text{C}}\text{H}-\text{CH}_3 \ (\text{IV}) > \text{C}_6\text{H}_5-\overset{+}{\text{C}}\text{H}_2 \ (\text{I}) > (\text{CH}_3)_3\text{C}^+ \ (\text{III})$
   • Under specific polar solvolytic conditions favoring benzylic stabilization, the overall reactivity order follows: $\text{IV} > \text{I} > \text{III}$.
   • Conclusion: Option [D] is correct.

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Final Conclusion

Statements A, B, C, and D are all correct.