A) \[{{C}_{2}}{{H}_{5}}ONa+{{C}_{2}}{{H}_{5}}I\]
B) \[{{C}_{2}}{{H}_{5}}ONa+{{(C{{H}_{3}})}_{3}}CBr\]
C) \[{{C}_{2}}{{H}_{5}}I+dry\,A{{g}_{2}}O\]
D) \[{{C}_{2}}{{H}_{5}}PH+{{H}_{2}}S{{O}_{4}}(140{}^\circ C)\]
Correct Answer: B
Solution :
\[{{C}_{2}}{{H}_{5}}ONa+{{(C{{H}_{3}})}_{3}}C-Br\xrightarrow{{}}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\] \[+NaBr+{{C}_{2}}{{H}_{5}}OH\] With tertiary alkyl halide, alkene is formed instead of ether.
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