In the reaction sequence CH_3-C equiv CH xrig | vedclass

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(C) The reaction of propyne $(CH_3-C \equiv CH)$ with Grignard reagent $(CH_3MgBr)$ is an acid-base reaction.Terminal alkynes have acidic hydrogen,whi

Vedclass · Accepted Answer

$CH_3-C \equiv C-COOH$

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(C) The reaction of propyne $(CH_3-C \equiv CH)$ with Grignard reagent $(CH_3MgBr)$ is an acid-base reaction.Terminal alkynes have acidic hydrogen,which reacts with the Grignard reagent to form an alkynyl magnesium bromide and methane $(CH_4)$.$CH_3-C \equiv CH + CH_3MgBr \rightarrow CH_3-C \equiv C-MgBr + CH_4$Here,$(A)$ is $CH_3-C \equiv C-MgBr$.Next,the reaction of $(A)$ with $CO_2$ followed by acidic hydrolysis $(H_2O/H^+)$ is a standard method to prepare carboxylic acids from Grignard reagents.$CH_3-C \equiv C-MgBr + CO_2 \rightarrow CH_3-C \equiv C-COOMgBr$$CH_3-C \equiv C-COOMgBr + H_2O/H^+ \rightarrow CH_3-C \equiv C-COOH + Mg(OH)Br$Thus,$(B)$ is but$-2-$ynoic acid $(CH_3-C \equiv C-COOH)$.

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

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