Question

The S\(_{\text{N}}\)1 reactivity of the following halides will be in the order
I) C\(_{\text{6}}\)H\(_{\text{5}}\)CH\(_{\text{2}}\)Br
II) (C\(_{\text{6}}\)H\(_{\text{5}}\))\(_2\)CHBr
III) (C\(_{\text{6}}\)H\(_{\text{5}}\))\(_2\)C(CH\(_{\text{3}}\))Br
IV) (CH\(_{\text{3}}\))\(_2\)CHBr

• A. \(III>II>I>IV\)
• B. \(III>I>II>IV\)
• C. \(IV>I>II>III\)
• D. \(II>III>I>IV\)

Hint:

For S\(_{\text{N}}\)1 reactions, the key is carbocation stability. Remember these general trends: \begin{itemize} \item Tertiary>Secondary>Primary for simple alkyl carbocations. \item Resonance stabilization (e.g., benzylic, allylic) is highly significant and often outweighs simple alkyl inductive stabilization. More resonance structures mean more stability. \end{itemize}

Answer 1

The Correct Option is A

Step 1: Understand S\(_{\text{N}}\)1 Reactivity.
S\(_{\text{N}}\)1 (Substitution Nucleophilic Unimolecular) reactions are two-step processes. The rate-determining step is the dissociation of the leaving group to form a carbocation intermediate. Therefore, the stability of this carbocation intermediate is the most critical factor determining the rate and thus the reactivity in an S\(_{\text{N}}\)1 reaction. A more stable carbocation leads to a faster S\(_{\text{N}}\)1 reaction. Carbocation stability is generally enhanced by: \begin{itemize} \item Alkyl substituents: Alkyl groups are electron-donating (via inductive effect and hyperconjugation), which helps stabilize the positive charge. The order of stability due to alkyl groups is Tertiary (3°)>Secondary (2°)>Primary (1°)>Methyl. \item Resonance stabilization: Delocalization of the positive charge through resonance structures (e.g., with adjacent phenyl groups or double/triple bonds) significantly increases stability. This is generally a much stronger stabilizing effect than inductive stabilization from alkyl groups. Benzylic carbocations (where the positive charge is adjacent to a benzene ring) are prime examples. \end{itemize} Step 2: Determine the structure and carbocation stability for each given halide. \begin{itemize} \item I) C\(_{\text{6}}\)H\(_{\text{5}}\)CH\(_{\text{2}}\)Br (Benzyl bromide): When the bromide ion (\( \text{Br}^- \)) leaves, it forms the primary benzylic carbocation (C\(_{\text{6}}\)H\(_{\text{5}}\)-CH\(_{\text{2}}^+\)). This carbocation is stabilized by resonance with the single adjacent phenyl group. \item II) (C\(_{\text{6}}\)H\(_{\text{5}}\))\(_2\)CHBr (Diphenylmethyl bromide): Upon the departure of \( \text{Br}^- \), a secondary benzylic carbocation ((C\(_{\text{6}}\)H\(_{\text{5}}\))\(_2\)CH\(^+\)) is formed. This carbocation benefits from resonance stabilization by two adjacent phenyl groups, providing greater stability than a single phenyl group. \item III) (C\(_{\text{6}}\)H\(_{\text{5}}\))\(_2\)C(CH\(_{\text{3}}\))Br (1-bromo-1,1-diphenyl-1-methylmethane): Loss of \( \text{Br}^- \) leads to the formation of a tertiary carbocation ((C\(_{\text{6}}\)H\(_{\text{5}}\))\(_2\)C\(^+\)(CH\(_{\text{3}}\))). This carbocation is highly stabilized by resonance from two phenyl groups and also by the inductive effect of the methyl group. This combination makes it exceptionally stable. \item IV) (CH\(_{\text{3}}\))\(_2\)CHBr (Isopropyl bromide): When \( \text{Br}^- \) leaves, a secondary carbocation ((CH\(_{\text{3}}\))\(_2\)CH\(^+\)) is formed. This carbocation is stabilized only by the inductive effect and hyperconjugation from the two methyl groups. It lacks any resonance stabilization. \end{itemize} Step 3: Compare carbocation stabilities and deduce S\(_{\text{N}}\)1 reactivity order.
The order of carbocation stability (and thus S\(_{\text{N}}\)1 reactivity) from highest to lowest is: % Option (A) III>The tertiary carbocation with two phenyl groups provides the most extensive resonance stabilization and inductive stabilization, making it the most stable. % Option (B) II>The secondary carbocation with two phenyl groups is highly stabilized by resonance, but slightly less than the tertiary one in III. % Option (C) I>The primary benzylic carbocation with one phenyl group has good resonance stabilization, but less than those with two phenyl groups. % Option (D) IV The secondary carbocation, stabilized only by alkyl groups, is the least stable among these as it lacks resonance stabilization. Therefore, the order of S\(_{\text{N}}\)1 reactivity is: \[ \text{III}>\text{II}>\text{I}>\text{IV} \] Step 4: Match the derived order with the given options.
The determined order, III>II>I>IV, precisely matches option (1).