Step 1: Understanding the Question:
Optical isomerism requires the presence of a chiral center (an asymmetric carbon atom). A carbon atom is chiral if it is covalently bonded to exactly four different atoms or functional groups. We must analyze the structure of each option to find the one lacking a chiral center.
Step 2: Detailed Explanation:
Let's draw and analyze each molecule:
(a) 3-Iodohexane: $\text{CH}_3\text{-CH}_2\text{-CH}(\text{I})\text{-CH}_2\text{-CH}_2\text{-CH}_3$.
Look at Carbon 3. It is bonded to: a hydrogen atom (-H), an iodine atom (-I), an ethyl group ($-\text{C}_2\text{H}_5$), and a propyl group ($-\text{C}_3\text{H}_7$). All four groups are different. Thus, it has a chiral center and exhibits optical isomerism.
(b) 2-Iodopentane: $\text{CH}_3\text{-CH}(\text{I})\text{-CH}_2\text{-CH}_2\text{-CH}_3$.
Look at Carbon 2. It is bonded to: a hydrogen atom (-H), an iodine atom (-I), a methyl group ($-\text{CH}_3$), and a propyl group ($-\text{C}_3\text{H}_7$). All four groups are different. Thus, it has a chiral center and exhibits optical isomerism.
(d) 2-Iodo-3-methylbutane: $\text{CH}_3\text{-CH}(\text{I})\text{-CH}(\text{CH}_3)_2$.
Look at Carbon 2. It is bonded to: a hydrogen atom (-H), an iodine atom (-I), a methyl group ($-\text{CH}_3$), and an isopropyl group ($-\text{CH}(\text{CH}_3)_2$). All four groups are different. Thus, it has a chiral center and exhibits optical isomerism.
(c) 2-Iodo-2-methylbutane: $\text{CH}_3\text{-C}(\text{I})(\text{CH}_3)\text{-CH}_2\text{-CH}_3$.
Look at Carbon 2. It is bonded to: an iodine atom (-I), an ethyl group ($-\text{C}_2\text{H}_5$), and two identical methyl groups ($-\text{CH}_3$). Because two of the four groups are exactly the same, this central carbon is superimposable on its mirror image. It is an achiral molecule and therefore does NOT exhibit optical isomerism.
Step 3: Final Answer:
2-Iodo-2-methylbutane does not exhibit optical isomerism, matching option (c).