$CH_{3}C \equiv CMgBr$ can be prepared by the reaction of:

  • A
    $CH_{3}C=CBr$ with $MgBr_{2}$
  • B
    $CH_{3}C \equiv CH$ with $MgBr_{2}$
  • C
    $CH_{3}C \equiv CH$ with $KBr$ and $Mg$ metal
  • D
    $CH_{3}C \equiv CH$ with $CH_{3}MgBr$

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