A) \[C{{H}_{3}}-C{{H}_{2}}-OH\xrightarrow{Na}\]
B) \[C{{H}_{3}}-C{{H}_{2}}-OH\xrightarrow{K}\]
C) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-OH\xrightarrow{Al}\]
D) \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-OH\xrightarrow{C{{H}_{3}}MgBr}\]
Correct Answer: D
Solution :
[d] \[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-\overset{-}{\mathop{O}}\,-\overset{+}{\mathop{H}}\,\xrightarrow{\overset{-}{\mathop{C}}\,{{H}_{3}}\overset{+}{\mathop{M}}\,gBr}C{{H}_{4}}\](Major product) Acid-base reaction takes place.
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