A) 1
B) 3
C) 8
D) 6
Correct Answer: B
Solution :
Since the compound has \[{{C}_{n}}C{{}_{2n}}O\]type general formula, it must be a ketone (or aldehyde). It has the following metamers (the isomers having same functional group but different alkyl groups attached with the same functional group): (a) \[\underset{2-\text{pentannone}}{\mathop{C{{H}_{3}}\overset{O}{\mathop{\overset{|\,|}{\mathop{C}}\,}}\,C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\] (b) \[\underset{3-\text{pentannone}}{\mathop{C{{H}_{3}}C{{H}_{2}}-\overset{O}{\mathop{\overset{|\,|}{\mathop{C}}\,-}}\,C{{H}_{2}}C{{H}_{3}}}}\,\] (c) \[\underset{3-\text{methy}\,1-2-\text{butanone}}{\mathop{C{{H}_{3}}-\overset{\text{O}}{\mathop{\overset{|\,|}{\mathop{C}}\,}}\,-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}}}\,\] Hence, it has three metamers.
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