Question

Which one of the following compounds is the most reactive in \( S_N1 \) reaction?

• A. \( \text{C}_6\text{H}_5 \text{-CH}_2\text{Br} \)
• B. \( \text{C}_6\text{H}_5 \text{-CH}_2\text{Br} \)
• C. \( \text{C}_6\text{H}_5 \text{-CH}_2\text{Br} \)
• D. \( \text{C}_6\text{H}_5 \text{-CH}_2\text{-C}_6\text{H}_5 \)

Hint:

In \( S_N1 \) reactions, the stability of the carbocation intermediate is key. Tertiary carbocations are the most stable, followed by secondary, then primary carbocations.

Answer 1

The Correct Option is C

In the \( S_N1 \) reaction, the rate-determining step involves the formation of a carbocation intermediate. The reactivity of the compound is largely dependent on the stability of the carbocation formed. The more stable the carbocation, the more reactive the compound will be in an \( S_N1 \) reaction.
The general order of reactivity for \( S_N1 \) reactions is: \[ 3^\circ>2^\circ>1^\circ \] Where \( 3^\circ \) represents a tertiary carbon, \( 2^\circ \) represents a secondary carbon, and \( 1^\circ \) represents a primary carbon. - Compound A (\( \text{C}_6\text{H}_5 \text{-CH}_2\text{Br} \)): This is a primary alkyl halide, which forms a less stable primary carbocation, making it less reactive in \( S_N1 \). - Compound B (\( \text{C}_6\text{H}_5 \text{-CH}_2\text{Br} \)): This is a primary alkyl halide, similar to compound A. - Compound C (\( \text{C}_6\text{H}_5 \text{-CH}_2\text{Br} \)): Again a primary halide, yet it would still undergo \( S_N1 \) in the presence of stabilizing effects from a group like \( C_6H_5 \), which increases its reactivity. - Compound D (\( \text{C}_6\text{H}_5 \text{-CH}_2\text{-C}_6\text{H}_5 \)): This is a tertiary halide, and tertiary carbocations are more stable, so compound D is most reactive in \( S_N1 \) reactions. However, compound C is more reactive. Thus, the compound in option \( C \) is most reactive in an \( S_N1 \) reaction.