A) 1
B) 3
C) 8
D) 6
Correct Answer: B
Solution :
Since the compound has\[{{C}_{n}}{{H}_{2n}}O\]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): (i) \[\underset{2-pentanone}{\mathop{C{{H}_{3}}\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\] (ii) \[\underset{3-pentanone}{\mathop{C{{H}_{3}}C{{H}_{2}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}C{{H}_{3}}}}\,\] (iii) \[\underset{3-\text{ }methyl-\text{ }2-\text{ }butanone}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{C}}\,-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}}}\,\] Hence, it has three metamers.
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