See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Identifying Pseudo-asymmetric Carbon A **pseudo-asymmetric carbon** (prochiral center) has four groups where two are identical, but those identical groups are themselves chiral centers with opposite configurations. ## Analysis of Each Option: **Option (A): $CH_3-CHBr-CHBr-CH_2Br$** - All three carbons bearing Br have different substituents - No pseudo-asymmetric carbon (carbons with identical substituents needed) - ✗ Incorrect **Option (B): $CH_3-CHBr-CH(OH)-CHBr-CH_3$** - The second carbon (from left) bears: $CH_3$, $H$, $Br$, and $CH(OH)CHBr-$ - The fourth carbon bears: $CH_3$, $H$, $Br$, and $-CHBr CH(OH)$ - The **second carbon** has two identical groups ($CH_3$ and $H$ present once each) BUT the two $Br$-containing arms have opposite stereochemistry (one goes to $CHBr$, other from $CHBr$) - More precisely: this carbon is bonded to groups where two substituents differ only in the stereochemical configuration at a nearby chiral center - ✓ **Correct** — Contains a pseudo-asymmetric center **Option (C): $CH_3-CHBr-CH(OH)-CHBr-CH_3$** - Similar to (B) but has only one $OH$ group - Still lacks the necessary condition for a clear pseudo-asymmetric center - ✗ Incorrect **Option (D): $CH_3-CH(Br)-CH(OH)-CH(OH)-CH_3$** - Both $OH$ groups make it symmetric or differently chiral - ✗ Incorrect **Answer: (B)** — The second carbon atom is pseudo-asymmetric because it bears two groups ($Br$-containing arms) that are identical in connectivity but opposite in stereochemical configuration at adjacent chiral centers.