A) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br\]
B) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}Br\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}Br\]
D) \[C{{H}_{3}}C{{H}_{2}}Br\]
Correct Answer: D
Solution :
| [d] Key Idea: The relative reactivity of alkyl halides towards \[{{S}_{N}}2\] reactions is as follows: |
| Primary > Secondary > Tertiary |
| However, if the primary alkyl halide or the nucleophile/base is sterically hindered the nucleophile will have difficulty to getting the back side of the a-carbon as a result of this, the elimination product will be predominant. Here \[C{{H}_{3}}C{{H}_{2}}Br\] is the least hindered, hence it has the highest relative rate towards \[{{S}_{N}}2\] reaction. |
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