Match List-I with List-II:List-I (Pair of Com | vedclass

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(D) . $n$-propanol $(CH_3CH_2CH_2OH)$ and Isopropanol $(CH_3CH(OH)CH_3)$ differ in the position of the $-OH$ group,hence they are Position isomers $(I

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) . $n$-propanol $(CH_3CH_2CH_2OH)$ and Isopropanol $(CH_3CH(OH)CH_3)$ differ in the position of the $-OH$ group,hence they are Position isomers $(III)$.$B$. Methoxypropane $(CH_3OCH_2CH_2CH_3)$ and ethoxyethane $(CH_3CH_2OCH_2CH_3)$ have different alkyl groups attached to the same functional group $(-O-)$,hence they are Metamers $(I)$.$C$. Propanone $(CH_3COCH_3)$ and propanal $(CH_3CH_2CHO)$ have different functional groups (ketone vs aldehyde),hence they are Functional isomers $(IV)$.$D$. Neopentane $(C(CH_3)_4)$ and Isopentane $(CH_3CH(CH_3)CH_2CH_3)$ differ in the arrangement of the carbon chain,hence they are Chain isomers $(II)$.Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

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

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