question_answer

Arrange the following compounds in increasing order of \[{{S}_{{{N}^{1}}}}\] reactivity. (a) \[\underset{\begin{smallmatrix}   \\  (I) \end{smallmatrix}}{\mathop{ClC{{H}_{2}}CH=CHC{{H}_{2}}C{{H}_{3}},}}\,\] \[\underset{\begin{smallmatrix}   \\  (II) \end{smallmatrix}}{\mathop{C{{H}_{3}}C(Cl)=CHC{{H}_{2}}C{{H}_{3}},}}\,\] \[\underset{\begin{smallmatrix}   \\  (III) \end{smallmatrix}}{\mathop{C{{H}_{3}}CH=CHC{{H}_{2}}C{{H}_{2}}Cl}}\,\] (b) \[\underset{\begin{smallmatrix}   \\  (I) \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}Br,}}\,\]   \[\underset{\begin{smallmatrix}   \\  (II) \end{smallmatrix}}{\mathop{C{{H}_{2}}=CHCH(Br)C{{H}_{3}},}}\,\]  \[\underset{(III)}{\mathop{C{{H}_{2}}=CHBr,}}\,\]                                \[\underset{\begin{smallmatrix}   \\  (IV) \end{smallmatrix}}{\mathop{C{{H}_{3}}CH(Br)C{{H}_{3}}}}\,\] (c) \[\underset{(I)}{\mathop{{{(C{{H}_{3}})}_{3}}CBr,}}\,\]  \[\underset{\begin{smallmatrix}   \\  (II) \end{smallmatrix}}{\mathop{{{(C{{H}_{3}})}_{2}}CHBr,}}\,\]                                \[\underset{\begin{smallmatrix}   \\  (III) \end{smallmatrix}}{\mathop{C{{H}_{3}}\,C{{H}_{2}}C{{H}_{2}}Br}}\,\]

Answer:

                The order of reactivity of alkyl halides towards \[{{S}_{{{N}^{1}}}}\] reactions is : Allylic > Tertiary > Secondary > Primary > Vinylic Based on this, the increasing order of reactivity in different alkyl halides listed above is : (a) (II) < (III) < (I) (b) (III) < (I) < (IV) < (II) (c) (III) < (II) < (I).