Ph-C equiv CH xrightarrow[MeOH]{MeO^{-}} The | vedclass

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Question

(B) The reaction involves the nucleophilic addition of the methoxide ion $(MeO^{-})$ to the alkyne. Since the terminal alkyne $(Ph-C \equiv CH)$ is no

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$Ph-CH=CH-OMe$

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(B) The reaction involves the nucleophilic addition of the methoxide ion $(MeO^{-})$ to the alkyne. Since the terminal alkyne $(Ph-C \equiv CH)$ is not acidic enough to be deprotonated by $MeO^{-}$ to a significant extent,the methoxide ion acts as a nucleophile. The $MeO^{-}$ attacks the electrophilic carbon of the triple bond,followed by protonation by the solvent $(MeOH)$ to yield the enol ether. The reaction proceeds as follows: $Ph-C \equiv CH + MeO^{-} \rightarrow Ph-C^{-}=CH-OMe$ $Ph-C^{-}=CH-OMe + MeOH \rightarrow Ph-CH=CH-OMe + MeO^{-}$ Thus,the major product is $Ph-CH=CH-OMe$.

$Ph-C \equiv CH \xrightarrow[MeOH]{MeO^{-}}$ The major product of the reaction is:

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