A) \[\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,\]
B) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ HO \end{smallmatrix}}{\mathop{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-OH\]
D) \[\underset{\begin{smallmatrix} | \\ HO \end{smallmatrix}}{\mathop{C{{H}_{2}}}}\,-\underset{\begin{smallmatrix} | \\ HO \end{smallmatrix}}{\mathop{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}\,}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{3}}\]
Correct Answer: B
Solution :
: \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{2{}^\circ }{\mathop{CH}}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{2{}^\circ }{\mathop{CH}}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{2{}^\circ }{\mathop{CH}}}\,-C{{H}_{3}}\]
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