What will be the product A? | vedclass

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$(A)$ The acidic strength of the functional groups present in the molecule follows the order: $-COOH > 	ext{phenolic } -OH > 	ext{terminal alkyne }

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$(A)$ The acidic strength of the functional groups present in the molecule follows the order: $-COOH > 	ext{phenolic } -OH > 	ext{terminal alkyne } (-C \equiv CH)$.Given that $2 	ext{ moles}$ of $NaNH_2$ are used, $NaNH_2$ is a strong base that will deprotonate the two most acidic protons.The most acidic proton is the carboxylic acid proton $(-COOH)$, followed by the phenolic $-OH$ group (specifically the one activated by the electron-withdrawing $-NO_2$ group).Therefore, the $-COOH$ group becomes $-COO^-$ and the phenolic $-OH$ group (ortho to $-NO_2$) becomes $-O^-$. The terminal alkyne and the other phenolic $-OH$ remain unchanged. This corresponds to option $A$.

Column-$I$: Compound / Reagent	Column-$II$: Other name / Chemical / Process
$A$. Methanol	$I$. Lucas reagent
$B$. $ZnCl_2$ / Conc. $HCl$	$II$. Baeyer's reagent
$C$. Rectified spirit	$III$. Wood spirit
$D$. Dil. $KMnO_4$	$IV$. $95 \% C_2H_5OH$
-	$V$. $75 \% C_2H_5OH$
-	$VI$. $90 \% C_3H_7OH$

What will be the product $A$?

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