In the reaction sequence:CH_3-Cequiv CH xrigh | vedclass

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(C) $1$-Propyne reacts with methyl magnesium bromide to form prop-$1$-ynylmagnesium bromide,which is compound $(A)$. Methane is the byproduct.The prop

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$CH_3-C\equiv C-COOH$

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(C) $1$-Propyne reacts with methyl magnesium bromide to form prop-$1$-ynylmagnesium bromide,which is compound $(A)$. Methane is the byproduct.The prop-$1$-ynyl anion acts as a nucleophile and attacks carbon dioxide. This is followed by hydrolysis to form but-$2$-ynoic acid,which is compound $(B)$.$CH_3-C\equiv CH$ $\xrightarrow{CH_3MgBr} CH_4 + CH_3-C\equiv C-MgBr$ $\xrightarrow[ (ii) H_2O/H^{\oplus} ]{(i) CO_2} CH_3-C\equiv C-COOH 	ext{ (Compound } B)$

In the reaction sequence:
$CH_3-C\equiv CH$ $\xrightarrow{CH_3MgBr}$ $CH_4 + (A)$ $\xrightarrow[ (ii) H_2O/H^{\oplus} ]{(i) CO_2}$ $(B)$
$(B)$ will be:

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In the reaction sequence:$CH_3-C\equiv CH$ $\xrightarrow{CH_3MgBr}$ $CH_4 + (A)$ $\xrightarrow[ (ii) H_2O/H^{\oplus} ]{(i) CO_2}$ $(B)$$(B)$ will be: