A) sodium iso-propoxide
B) sodium n-propoxide
C) 150-propyl bromide
D) n-propyl bromide
Correct Answer: A
Solution :
\[\underset{\begin{smallmatrix} ethyl \\ bromide \end{smallmatrix}}{\mathop{{{C}_{2}}{{H}_{5}}Br}}\,+\underset{sodium\,iso-propoxide}{\mathop{C{{H}_{3}}-\underset{\underline{O}\,{{N}^{a}}^{^{+}}}{\mathop{\underset{|}{\mathop{CH}}\,}}\,-C{{H}_{3}}\xrightarrow{\Delta }}}\,\] \[\underset{2-ethoxy\,propane}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{\underset{O-{{C}_{2}}{{H}_{5}}}{\mathop{\underset{|}{\mathop{C{{H}_{3}}-C-C{{H}_{3}}+NaBr}}\,}}\,}}\,}}\,}}\,\]
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