A) \[{{S}_{N}}1\] mechanism
B) \[{{S}_{N}}2\]mechanism
C) Any of (a) and (b)
D) None of the above
Correct Answer: A
Solution :
Tertiary halide preferentially undergo \[{{S}_{N}}1\] substitution as they can give stable carbocation. \[\underset{t-butyl\,\,chloride}{\mathop{{{H}_{3}}C-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-Cl}}\,\xrightarrow[-Cl]{slow}\underset{\begin{smallmatrix} {{3}^{o}}carbocation \\ (most\,\,stable) \end{smallmatrix}}{\mathop{{{({{H}_{3}}C)}_{3}}{{C}^{+}}}}\,\xrightarrow[fast]{+O{{H}^{-}}}\] \[\underset{t-butyl\,\,alcohol}{\mathop{{{({{H}_{3}}C)}_{3}}COH}}\,\]
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