Q32 / 125 · BITSAT 2009 · Chemistry

Which of the following order is not correct?

  • A
    $\mathrm{MeBr}>\mathrm{Me}_{2} \mathrm{CHBr}>\mathrm{Me}_{3} \mathrm{CBr}>$ $\mathrm{Et}_{3} \mathrm{CBr}\left(\mathrm{S}_{\mathrm{N}} 2\right)$
  • B
    $\mathrm{Me}_{3} \mathrm{CBr}>\mathrm{Me}_{2} \mathrm{CHBr}>\mathrm{Me}_{2} \mathrm{CH.CH}_{2} \mathrm{Br}>$ $\mathrm{MeCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} .\left(\mathrm{E}_{2}\right)$
  • Correct answer
    C
    $\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{Ph} \mathrm{CHBrMe}>\mathrm{PhCBrMe}_{2}>$ $\operatorname{PhCBrMePh}\left(\mathrm{S}_{\mathrm{N}} 1\right)$
  • D
    $\mathrm{MeI}>\mathrm{MeBr}>\mathrm{MeCl}>\mathrm{MeF}\left(\mathrm{S}_{\mathrm{N}} 2\right)$

Solution

The more is the stability of intermediate carbonium ion, the more is the chance of $\mathrm{S}_{\mathrm{N}} 1$ mechanism. The intermediates obtained will be $\mathrm{Ph}^{+} \mathrm{C} \mathrm{H}_{2}(\mathrm{i}), \mathrm{Ph} \mathrm{C} \mathrm{H}-\mathrm{Me}(\mathrm{ii})$ $\mathrm{Ph} \mathrm{C}-\mathrm{Me}_{2}\left(\right.$ iii) $, \mathrm{Ph} \mathrm{C}^{+} \mathrm{MePh}$ (iv) The stabilty is of the order iv $>\mathrm{iii}>\mathrm{ii}>\mathrm{i}$

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

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