CH_3C equiv C^-Na^+ + (CH_3)_2CH-Cl to What w | vedclass

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Question

(C) The reaction involves a strong nucleophile,the acetylide ion $(CH_3C \equiv C^-)$,and a secondary alkyl halide,isopropyl chloride $((CH_3)_2CH-Cl)

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Propyne and propene

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(C) The reaction involves a strong nucleophile,the acetylide ion $(CH_3C \equiv C^-)$,and a secondary alkyl halide,isopropyl chloride $((CH_3)_2CH-Cl)$.Since the nucleophile is a strong base and the substrate is a secondary alkyl halide,the reaction will primarily undergo an $E2$ elimination reaction rather than an $S_N2$ substitution.The acetylide ion acts as a base,abstracting a proton from the $\beta$-carbon of the isopropyl chloride,leading to the formation of propene $(CH_3-CH=CH_2)$,propyne $(CH_3-C \equiv CH)$,and sodium chloride $(NaCl)$.Therefore,the major products are propene and propyne.

$CH_3C \equiv C^-Na^+ + (CH_3)_2CH-Cl \to$ What will be the product of this reaction?

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