$CH_3-C(CH_3)=CH-CH_3 \xrightarrow[KMnO_4/\Delta]{NaIO_4}$ Product will be

  • A
    $CH_3-COOH + CH_3-CH_2-COOH$
  • B
    $CH_3-CHO + CH_3-CH_2-COOH$
  • C
    $CH_3-CHO + CH_3-C(=O)-CH_3$
  • D
    $CH_3-COOH + CH_3-C(=O)-CH_3$

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