Answer:
All are primary alkyl halides and their structural formulae are:
\[\underset{(I)}{\mathop{{{C}_{2}}{{H}_{5}}-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}Br}}\,\]\[\underset{(II)}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}Br}}\,\]\[\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(III)}{\mathop{{{C}_{3}}{{H}_{7}}-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-Br}}\,\]
The order of reactivity is: (III) > (I) > (II)
Explanation: The\[{{S}_{{{N}^{2}}}}\] reactions are sensitive to steric hindrance. Greater the steric hindrance to the attacking nucleophile, lesser will be the reactivity. In the light of this, the reactivity order is justified.
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