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(C) Step $1$: Ethyne $(HC \equiv CH)$ reacts with excess $NaNH_2$ (a strong base) to remove both acidic protons,forming the disodium acetylide interme

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$5$-decyne

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(C) Step $1$: Ethyne $(HC \equiv CH)$ reacts with excess $NaNH_2$ (a strong base) to remove both acidic protons,forming the disodium acetylide intermediate,$Na^{\oplus} [C \equiv C]^{2-} Na^{\oplus}$.Step $2$: The disodium acetylide acts as a nucleophile and undergoes an $S_N2$ reaction with excess $I-CH_2-CH_2-CH_2-CH_3$ (butyl iodide). Each carbon of the acetylide attacks a butyl group,displacing the iodide ion.Step $3$: The final product formed is $CH_3-CH_2-CH_2-CH_2-C \equiv C-CH_2-CH_2-CH_2-CH_3$,which is $5$-decyne.

Predict the product of the following reaction sequence:
$HC \equiv CH$ $\xrightarrow[\begin{smallmatrix} (2) \text{ excess } I-CH_2-(CH_2)_2-CH_3 \\ (3) \text{ } H^{\oplus} \end{smallmatrix}]{(1) \text{ excess } NaNH_2}$

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Predict the product of the following reaction sequence:$HC \equiv CH$ $\xrightarrow[\begin{smallmatrix} (2) \text{ excess } I-CH_2-(CH_2)_2-CH_3 \\ (3) \text{ } H^{\oplus} \end{smallmatrix}]{(1) \text{ excess } NaNH_2}$