A) \[{{(C{{H}_{3}})}_{2}}C-O{{C}_{2}}{{H}_{5}}\]
B) \[{{(C{{H}_{3}})}_{3}}C-{{C}_{2}}{{H}_{5}}\]
C) \[{{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}\]
D) \[C{{H}_{3}}-CH=CH-{{C}_{2}}{{H}_{5}}\]
Correct Answer: C
Solution :
Tertiary alkyl halides when treated with a strong nucleophile\[({{C}_{2}}{{H}_{5}}{{O}^{-}}N{{a}^{+}})\], preferentially undergo elimination reactions. The reason is due to steric reasons, nucleophile will prefer to act as a base and abstract a proton (which is smaller in size) to give the elimination product rather than attacking a tetrahedron carbon (which is bigger in size) to form the substitution product. \[\underset{tert-butylchloride}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-Cl}}\,\xrightarrow[\text{Elimination}]{{{C}_{2}}{{H}_{5}}{{O}^{-}}N{{a}^{+}}}\underset{2-methylpropene(97%)}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}}}\,\] \[\underset{tert-butylethyl\,\,ether\,\,(3%)}{\mathop{+{{(C{{H}_{3}})}_{3}}-C-O{{C}_{2}}{{H}_{5}}}}\,\]
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