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-$I$	Column-$II$
$A$. Methanol	$p$. Conversion of phenol to salicylic acid
$B$. Kolbe's reaction	$q$. Wood spirit
$C$. Williamson's reaction	$r$. Vapours are passed over heated copper at $573 \ K$
$D$. Conversion of $2^\circ$ alcohol to ketone	$s$. Reaction of alkyl halide with sodium alkoxide

What will be the product $A$?

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