In the following reaction,compound A is forme | vedclass

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(B) The reaction is: $CH_3-C \equiv CH + Na \rightarrow CH_3-C \equiv C^- Na^+ (A) + \frac{1}{2} H_2$.Compound $A$ is sodium prop$-1-$ynide $(CH_3-C \

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$A = CH_3-C \equiv C^- Na^+, B = CH_3-CH_2-CH_2-Br$

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(B) The reaction is: $CH_3-C \equiv CH + Na \rightarrow CH_3-C \equiv C^- Na^+ (A) + \frac{1}{2} H_2$.Compound $A$ is sodium prop$-1-$ynide $(CH_3-C \equiv C^- Na^+)$.Compound $C$ is $CH_3-C \equiv C-CH_2-CH(CH_3)_2$ (based on the structure provided in the image).For this product to form,$B$ must be $2$-bromobutane $(CH_3-CH_2-CH(Br)-CH_3)$ or a similar alkyl halide that provides the branched chain.However,looking at the provided image structure $CH_3-C \equiv C-CH_2-CH_2-CH(CH_3)$,the reaction is an $S_N2$ substitution.Given the options,$A$ is $CH_3-C \equiv C^- Na^+$. The structure of $C$ in the image corresponds to $CH_3-C \equiv C-CH_2-CH_2-CH_3$ if we assume the branch is a typo in the image or $B$ is $1$-bromobutane. Based on standard textbook problems,$B$ is $CH_3-CH_2-CH_2-Br$.

In the following reaction,compound $A$ is formed,which reacts with $B$ to give product $C$. Identify $A$ and $B$.

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