A) \[C{{H}_{3}}Cl>C{{H}_{3}}Br>C{{H}_{3}}I\]
B) \[C{{H}_{3}}I>C{{H}_{3}}Br>C{{H}_{3}}Cl\]
C) \[C{{H}_{3}}Br>C{{H}_{3}}I>C{{H}_{3}}Cl\]
D) \[C{{H}_{3}}Br>C{{H}_{3}}Cl>C{{H}_{3}}I\]
Correct Answer: B
Solution :
Key Idea When an alkyl halide reacts with Mg metal in dry ether, as a solvent, in an atmosphere of nitrogen then alkyl magnesium halide is obtained. This is called Grignard reagent. \[RX+Mg\xrightarrow{Dry\,\,ether}R-Mg-X\] The reactivity with respect to halide RI > RBr > RCl For a given halide the order is Allyl > alkyl \[({{3}^{o}}>{{2}^{o}}>{{1}^{o}})>aryl\]. The order of reactivity of methyl halide in the formation of Grignard reagent is as \[C{{H}_{3}}I>C{{H}_{3}}Br>C{{H}_{3}}Cl\] Here the alkyl group is same, therefore the reactivity depend on halides. The ease of the leaving group in alkyl halide is as \[-I>-Br>-Cl-F\] and the bond strength being in order C ? F > C ? Cl > C ? Br > C ? I Therefore, the reactivity of methyl halide in the formation of Grignard reagent is as \[C{{H}_{3}}I>C{{H}_{3}}Br>C{{H}_{3}}Cl\]
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