A compound is optically active if it has at least one chiral center (a carbon atom attached to four different groups). Let us analyze the given compounds: CH$_3$-CH(OH)-CH(OH)-CH$_3$: This compound has two hydroxyl groups (OH) on adjacent carbon atoms. Both hydroxyl groups are equivalent, and the molecule has a plane of symmetry, making it optically inactive. CH$_3$-CH$_2$-CH$_2$-OH: This compound has no chiral centers, as all carbons are attached to at least two identical groups. Therefore, it is optically inactive. CH$_3$-CH$_2$-CH-CH$_3$ (with a Cl on the second carbon):} The second carbon atom is a chiral center, as it is attached to four different groups: CH$_3$, H, Cl, and CH$_2$CH$_3$. Hence, this compound is optically active. (CH$_3$)$_2$CH-CH$_2$-CH$_2$-Cl: The molecule does not have any chiral centers, as the carbon bonded to the chlorine atom is not attached to four different groups. Therefore, it is optically inactive. Conclusion: Among the given compounds, only CH$_3$-CH$_2$-CH-CH$_3$ (with Cl on the second carbon) is optically active.
