CH_3-CH_2-CH_2-Cequiv CH + LiNH_2 to (A) xrig | vedclass

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(B) $1$. The reaction of $CH_3-CH_2-CH_2-C\equiv CH$ with $LiNH_2$ is an acid-base reaction where the terminal alkyne proton is removed to form the li

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$CH_3-(CH_2)_2-C\equiv C-CH_3$

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(B) $1$. The reaction of $CH_3-CH_2-CH_2-C\equiv CH$ with $LiNH_2$ is an acid-base reaction where the terminal alkyne proton is removed to form the lithium acetylide intermediate $(A)$,which is $CH_3-CH_2-CH_2-C\equiv C^-Li^+$.$2$. The intermediate $(A)$ then reacts with dimethyl sulfate $(CH_3)_2SO_4$ in a nucleophilic substitution reaction $(S_N2)$.$3$. The acetylide ion acts as a nucleophile and attacks the methyl group of $(CH_3)_2SO_4$,resulting in the methylation of the alkyne.$4$. The final product $(B)$ is $CH_3-CH_2-CH_2-C\equiv C-CH_3$ (pent$-2-$yne).

$CH_3-CH_2-CH_2-C\equiv CH + LiNH_2 \to (A) \xrightarrow{(CH_3)_2SO_4} (B)$
Give the structural formula of compound $(B)$.

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$CH_3-CH_2-CH_2-C\equiv CH + LiNH_2 \to (A) \xrightarrow{(CH_3)_2SO_4} (B)$Give the structural formula of compound $(B)$.