See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Optical Isomerism in Acyclic Alkanes **Step 1: Understand optical isomerism requirement** Optical isomerism requires a **chiral center** — a carbon bonded to four different groups. Acyclic alkanes are saturated hydrocarbons with only single bonds. **Step 2: Check smaller alkanes (C₃, C₅)** - **Propane** ($C_3H_8$): Only one carbon type; no chiral center possible - **Pentane** ($C_5H_{12}$): Even with branching (isopentane, neopentane), every carbon still has at least two identical H atoms or identical alkyl groups — no chiral center **Step 3: Examine heptane ($C_7H_{16}$)** Consider **3-methylhexane**: $$CH_3-CH_2-CH(CH_3)-CH_2-CH_2-CH_3$$ The third carbon is bonded to: - $H$ (one H atom) - $CH_3$ (methyl group) - $CH_2CH_2CH_3$ (n-propyl group) - $CH_2CH_3$ (ethyl group) All four groups are **different** → **chiral center confirmed!** **Step 4: Why not C₅?** With only 5 carbons total, the branching patterns don't allow a carbon to have four distinctly different groups (at least two groups remain identical in structure). **Answer: (C) 7** The minimum carbon requirement for an acyclic alkane to exhibit optical isomerism is **7 carbons** (as exemplified by 3-methylhexane).