$CH_3-CH_2-CH(Cl)_2$ $\xrightarrow[{[Excess]}]{{NaNH_2}} P$ $\xrightarrow{{CH_3Cl}} Q$,$Q$ will be

  • A
    $CH_3-CH_2-C \equiv CH$
  • B
    $CH_3-CH_2-C \equiv C-CH_3$
  • C
    $CH_3-CH_2-CH_2-CH_2-Cl$
  • D
    $CH_2=CH-CH=CH_2$

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