A) \[{{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CCl}\]
B) \[{{(C{{H}_{3}})}_{2}}CHCl\]
C) \[{{\text{(}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{)}}_{\text{2}}}\text{CHCl}\]
D) \[\text{C}{{\text{H}}_{\text{3}}}\text{Cl}\]
Correct Answer: D
Solution :
Since, we know that primary halides show inversion during \[{{S}_{N}}2\] reaction more than secondary while secondary show more than tartiary. Therefore, \[\text{C}{{\text{H}}_{\text{3}}}\text{C1 }\left( {{\text{l}}^{\circ }}\text{halide} \right)\] show sterechemical inversion more in comparison to \[{{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{3}}}\text{CCl (}{{\text{3}}^{\text{o}}}\text{halide), }\] \[{{\text{(C}{{\text{H}}_{\text{3}}}\text{)}}_{\text{2}}}\text{CHCl }\] \[\text{(}{{\text{2}}^{\text{o}}}\text{halide)}\] and\[{{\text{(}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}\text{)}}_{\text{2}}}\text{CHCl (}{{\text{2}}^{\text{o}}}\text{halide)}\]You need to login to perform this action.
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