A compound `(C_(5)H_(8))` reacts with ammoniacal `AgNO_(3)` to give a white precipitate and reacts with excess of `KMnO_(4)` solution to give `(CH_(3))_(2)CH-COOH`. The compound is

Options

A. `CH_(2) = CH - CH = CH -CH_(3)`
B. `CH_(3)-CH= CH - CH_(2) - CH_(3)`
C. `(CH_(3))_(2)CH- C -= CH`
D. `(CH_(3))_(2) CH - C =CH_(2)`

Answer

c

Explanation

Compound X - reacts with ammonical `AgNO_(3)` solution , so it must be a terminal alkyne .
Formation of `(CH_(3))_(2)CHCOOH` on oxidation of X with hot alkaline `KMnO_(4)` further confirms that X is `(CH_(3))_(2)CHC -= CH`