Q13 / 180 · COMEDK 2024 (Evening Shift) · Chemistry

Arrange the compounds $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ in the increasing order of their reactivity towards $\mathrm{S_N}1$ reaction. $ \begin{aligned} & \mathrm{A}=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl} \\ & \mathrm{B}=\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl} \\ & \mathrm{C}=\mathrm{CH}_2=\mathrm{CH}-\mathrm{Cl} \\ & \mathrm{D}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Cl} \end{aligned}$

  • Correct answer
    A
    $\mathrm{B}<\mathrm{C}<\mathrm{D}<\mathrm{A}$
  • B
    $\mathrm{D}<\mathrm{C}<\mathrm{B}<\mathrm{A}$
  • C
    $\mathrm{D}<\mathrm{B}<\mathrm{A}<\mathrm{C}$
  • D
    $\mathrm{C}<\mathrm{A}<\mathrm{D}<\mathrm{B}$

Solution

$S_N1$ reactivity depends on the stability of the carbocation formed.

A ($C_6H_5CH_2Cl$): Forms benzyl carbocation ($C_6H_5CH_2^+$) — highly stable due to resonance delocalization over the benzene ring. Most reactive.

D ($CH_3CH_2Cl$): Forms ethyl carbocation ($1°$) — moderately stable via hyperconjugation and $+I$ effect of methyl group.

C ($CH_2=CH-Cl$): Forms vinyl cation ($CH_2=CH^+$) — positive charge on $sp$ hybridised carbon (more electronegative), highly unstable. Also has partial double bond character in C-Cl bond due to resonance, making ionisation difficult.

B ($C_6H_5Cl$): Forms phenyl cation — positive charge on $sp^2$ carbon within ring, extremely unstable. Cl lone pair conjugated with ring gives C-Cl bond strong double bond character, making it hardest to cleave.

Increasing order of $S_N1$ reactivity: $B < C < D < A$

Hence, the correct answer is option: 'A'.

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