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 $A$.

Molecule	Label
$CH_3-O-CH_3$	$I$
$CH_3CHO$	$II$
$CH_3CH_2CH_3$	$III$
$CH_3CH_2OH$	$IV$

What will be the product $A$?

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The increasing order of boiling points of the following is:
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$CH_3-O-CH_3$ $I$
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