Question

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

• A. (A)–(II), (B)–(I), (C)–(IV), (D)–(III)
• B. (A)–(III), (B)–(I), (C)–(II), (D)–(IV)
• C. (A)–(I), (B)–(III), (C)–(IV), (D)–(II)
• D. (A)–(III), (B)–(I), (C)–(IV), (D)–(II)

Answer 1

Answer: (D)

(A)–(III), (B)–(I), (C)–(IV), (D)–(II)

Answer 2

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).