In the given pairs,identify the most acidic c | vedclass

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(B) For $(A)$: $o$-toluic acid $(II)$ is more acidic than benzoic acid $(I)$ due to the ortho-effect,which inhibits resonance and destabilizes the pla

Vedclass · Accepted Answer

$A-II, B-I, C-I, D-II$

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(B) For $(A)$: $o$-toluic acid $(II)$ is more acidic than benzoic acid $(I)$ due to the ortho-effect,which inhibits resonance and destabilizes the planar structure,making the carboxylate anion more stable relative to the acid.For $(B)$: Cyclohexanecarboxylic acid $(I)$ is more acidic than $2$-methylcyclohexanecarboxylic acid $(II)$ because the methyl group is electron-donating $(+I 	ext{ effect})$,which destabilizes the carboxylate anion.For $(C)$: Aniline $(I)$ is more basic than $o$-toluidine $(II)$ due to the ortho-effect,where the methyl group sterically hinders the protonation of the nitrogen atom.For $(D)$: $2$-methylcyclohexylamine $(II)$ is more basic than cyclohexylamine $(I)$ because the methyl group is electron-donating $(+I 	ext{ effect})$,which increases the electron density on the nitrogen atom.Thus,the most acidic in $(A)$ is $(II)$,in $(B)$ is $(I)$,the most basic in $(C)$ is $(I)$,and in $(D)$ is $(II)$. The correct option is $(B)$.

List-$I$ (Pair of Compounds)	List-$II$ (Isomerism)
$A$. $n$-propanol and Isopropanol	$(I)$ Metamerism
$B$. Methoxypropane and ethoxyethane	$(II)$ Chain Isomerism
$C$. Propanone and propanal	$(III)$ Position Isomerism
$D$. Neopentane and Isopentane	$(IV)$ Functional Isomerism

In the given pairs,identify the most acidic compound in $(A)$ and $(B)$,and the most basic compound in $(C)$ and $(D)$.

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