$Ph-CH=CH-Ph$ $\xrightarrow[CCl_4]{Cl_2} X$ $\xrightarrow{2NaNH_2} Y$ $\xrightarrow[BaSO_4]{H_2+Pd} Z$. Product $Z$ is

  • A
    Option A
  • B
    Option B
  • C
    $Ph-C(Ph)=CH_2$
  • D
    $Ph-C \equiv C-Ph$

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