Step 1: Understanding the Concept:
A compound is optically inactive if it lacks a chiral center (a carbon atom bonded to four different groups).
Step 2: Formula Application:
Analyze the substituents on the second carbon of 2-chloro-2-methylbutane.
Step 3: Explanation:
In 2-chloro-2-methylbutane, the second carbon is bonded to: a chlorine atom, two methyl groups ($CH_3$), and one ethyl group ($C_2H_5$). Because two of the groups are identical (the two methyl groups), this carbon is not chiral. Therefore, the molecule is achiral and optically inactive. All other options have at least one carbon bonded to four different groups.
Step 4: Final Answer:
The optically inactive compound is 2-chloro-2-methylbutane.