$S_N1$ પ્રક્રિયા માટે નીચેના ભાગોની પ્રતિક્રિયાશીલતાનો વધતો ક્રમ કયો છે ?

$(I)\,\begin{array}{*{20}{c}}
  {C{H_3}CHC{H_2}C{H_3}} \\ 
  {|\,\,\,\,\,\,\,\,\,\,\,} \\ 
  {Cl\,\,\,\,\,\,\,\,\,} 
\end{array}$

$(II)\,CH_3CH_2CH_2Cl$

$(III)\,H_3CO-C_6H_4-CH_2Cl$

  • A$(III) < (II) < (I)$
  • B$(II) < (I) < (III)$
  • C$(I) < (III) < (II)$
  • D$(II) < (III) < (I)$
JEE MAIN 2017, Diffcult
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b
Secondary, tertiary and benzylic halides appear to react by a mechanism that involves the formation of a carbocation and proceeds through \(\mathrm{S}_{\mathrm{N}} 1\) mechanism which takes place in two steps. The first step is the slow step \(-\) it is the rate-determining step. In this step, a molecule of alkyI halide ionizes and becomes an alkyl cation. Therefore, greater the stability of carbocation greater will be its tendency to react via \(S_{N} 1\) mechanism.

\(\mathrm{R}-\mathrm{X} \rightleftharpoons \mathrm{R}^{+}+\mathrm{X}^{-}\)

In the second step, the intermediate alkyl cation reacts rapidly with water to produce alcohol.

\(\mathrm{R}^{+}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{ROH}\)

Hence, the order of reactivity of halides for the \(\mathrm{S}_{\mathrm{N}} 1\) mechanism is \(\mathrm{II}<\mathrm{I}<\mathrm{III}\).

art

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