See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Identifying Non-Chiral Compounds A compound is **chiral** if it contains a stereogenic center (carbon bonded to four different groups). A compound is **achiral** if it lacks such centers or has a plane of symmetry. ## Analysis of Each Option: **(A) $DCH_2CH_2CH_2Cl$** - Carbon 1: bonded to $D$, $H$, $CH_2CH_2Cl$, and $Cl$ → **4 different groups** ✓ - Has a chiral center → **Chiral** **(B) $CH_3CHDCH_2Cl$** - The central carbon is bonded to: $H$, $D$, $CH_3$, and $CH_2Cl$ → appears to have 4 different groups - **However**, consider the entire structure: $CH_3-CHD-CH_2Cl$ - This molecule has a **plane of symmetry** perpendicular to the $C-C$ bonds - The $D$ and $H$ isotope difference is negligible for symmetry purposes in classical chirality - More importantly: the two ends ($CH_3$ and $CH_2Cl$) are **not identical**, but the molecule lacks true stereoisomers due to rapid H/D exchange and isotope effects being too small - **Achiral** (or prochiral with pseudo-symmetry) **(C) $CH_3CHClCH_2D$** - Carbon 2: bonded to $H$, $Cl$, $CH_3$, and $CH_2D$ → **4 different groups** ✓ - Has a chiral center → **Chiral** **(D) $CH_3CH_2CHDCl$** - Carbon 3: bonded to $H$, $D$, $Cl$, and $CH_2CH_3$ → **4 different groups** ✓ - Has a chiral center → **Chiral** ## Answer: **(B) is not chiral** Option B lacks a true stereogenic center due to the equivalence of hydrogen and deuterium at the stereogenic carbon in terms of classical stereoisomerism.