What is the final product A in this two-step reaction? H-Cequiv C-H ([1]NaNH_(2))/([2]CH_(3)Br) A CH_(2)Br-Cequiv CH CH_(3)-Cequiv CBr CH_(3)-Cequiv CH CH_(3)-CH=CH_(2)

The final product A in this two-step reaction is $CH_{3}-C\equiv CBr$.<br /><br />In the first step, the acetylene molecule ($H-C\equiv C-H$) reacts with NaNH2 (sodium amide) to form the acetylide ion ($C_{2}H^{-}$). In the second step, the acetylide ion reacts with CH3Br (methyl bromide) to form the product A, which is $CH_{3}-C\equiv CBr$.