A) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}\]
B) \[C{{H}_{3}}-CHOH-C{{H}_{3}}\]
C) \[C{{H}_{3}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,-\underset{\begin{smallmatrix} | \\ Br \end{smallmatrix}}{\overset{\begin{smallmatrix} Br \\ | \end{smallmatrix}}{\mathop{C}}}\,-COOH\]
D) \[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{C}}\,=O\]
Correct Answer: A
Solution :
Optically active compounds have at least one asymmetric or chiral carbon atom. Compound given in option [a] contains two chiral carbon atoms as \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ *| \end{smallmatrix}}{\mathop{C}}}\,-\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ *| \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{3}}\] \[\therefore \]it is optically active.
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