Match List - I with List - II.List - I (Pair of Compounds) List - II (Isomerism) (A) n-propanol and Isopropanol (I) Metamerism (B) Methoxypropane and ethoxyethane (IV) Functional Isomerism (C) Propanone and propanal (III) Position Isomerism (D) Neopentane and Isopentane (II) Chain Isomerism
| List - I (Pair of Compounds) | List - II (Isomerism) |
|---|---|
| (A) n-propanol and Isopropanol | (I) Metamerism |
| (B) Methoxypropane and ethoxyethane | (IV) Functional Isomerism |
| (C) Propanone and propanal | (III) Position Isomerism |
| (D) Neopentane and Isopentane | (II) Chain Isomerism |
- (A)–(II), (B)–(I), (C)–(IV), (D)–(III)
- (A)–(III), (B)–(I), (C)–(II), (D)–(IV)
- (A)–(I), (B)–(III), (C)–(IV), (D)–(II)
- (A)–(III), (B)–(I), (C)–(IV), (D)–(II)
The Correct Option is D
Approach Solution - 1
To determine the correct matches, we analyze the isomeric relationships of the compounds in List-I:
Step 1: Analyze (A) \textit{n}-propanol and isopropanol
{n}-propanol (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\)) and isopropanol (\(\text{CH}_3\text{CHOHCH}_3\)) differ in their functional group positions.
This is an example of functional isomerism.
\[(A) \rightarrow \text{(IV)}.\]
Step 2: Analyze (B) Methoxypropane and ethoxyethane
Methoxypropane (\(\text{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3\)) and ethoxyethane (\(\text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3\)) differ in the arrangement of the ether groups.
This is an example of metamerism.
\[(B) \rightarrow \text{(I)}.\]
Step 3: Analyze (C) Propanone and propanal
Propanone (\(\text{CH}_3\text{COCH}_3\)) and propanal (\(\text{CH}_3\text{CH}_2\text{CHO}\)) have different functional groups (ketone vs. aldehyde).
This is an example of functional isomerism.
\[(C) \rightarrow \text{(III)}.\]
Step 4: Analyze (D) Neopentane and isopentane
Neopentane (\(\text{C}(\text{CH}_3)_4\)) and isopentane (\(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_3\)) differ in the arrangement of their carbon chains.
This is an example of chain isomerism.
\[(D) \rightarrow \text{(II)}.\]
Final Matches:
\[(A) \rightarrow \text{(IV)}, \, (B) \rightarrow \text{(I)}, \, (C) \rightarrow \text{(III)}, \, (D) \rightarrow \text{(II)}.\]
Correct Answer: (4).
Approach Solution -2
(A) n-Propanol and Isopropanol → Both have the same molecular formula C₃H₈O but differ in the position of the –OH group. In n-propanol, –OH is attached to the terminal carbon, while in isopropanol, it is attached to the middle carbon.
Hence, they exhibit Position Isomerism → (III).
(B) Methoxypropane and Ethoxyethane → Both have the same molecular formula C₄H₁₀O and are ethers with different alkyl groups attached to oxygen. This is a case of Metamerism → (I).
(C) Propanone and Propanal → Both have the same molecular formula C₃H₆O, but one is a ketone and the other an aldehyde. This is Functional Isomerism → (IV).
(D) Neopentane and Isopentane → Both have the molecular formula C₅H₁₂ but differ in the arrangement of carbon atoms. This is Chain Isomerism → (II).
Step 2: Final matching.
(A)–(III), (B)–(I), (C)–(IV), (D)–(II)
Final Answer: (A)–(III), (B)–(I), (C)–(IV), (D)–(II)
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