In the reaction $Me-C \equiv C-Et$ $\xrightarrow{Na/liq. NH_3} P$ $\xrightarrow[CCl_4]{Br_2} (Q)$; then $Q$ is

  • A
    $A$ pure compound which is optically inactive due to internal compensation
  • B
    $A$ binary mixture which is optically inactive due to external compensation
  • C
    $A$ binary mixture which is optically active
  • D
    $A$ pure compound which is optically inactive due to absence of chiral centre

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