MCQ
$Ph - C \equiv C - C{H_3}\xrightarrow{{H{g^{2 + }}/{H^ + }}}A$, $A$ is
- ✓

- B

- C

- D







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$2MnO_4^ - + 5{C_2}O_4^ - + 16{H^ + } \to 2M{n^{ + + }} + 10C{O_2} + 8{H_2}O$
Here $20\, mL$ of $0.1\, M\, KMnO_4$ is equivalent to