A) 1
B) 4
C) 5
D) 3
Correct Answer: B
Solution :
Four primary alcohols of the formula \[{{C}_{2}}{{H}_{12}}O\] are possible. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH,\] \[CH{{ & }_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}OH\] \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}-C{{H}_{2}}OH,\] \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}OH\]
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