See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

**Step-by-step Analysis:** **Identify the stereochemistry of each compound:** Compound (1): Central carbon has $COOc_2H_5$, $COOCH_3$, $H$, $Br$ with both $H$ and $Br$ on the same side (wedge/dash shown). This is a chiral center. Compound (2): Central carbon has $COOC_2H_5$, $COOCH_3$, $H$, $Br$ with $H$ and $Br$ on opposite sides. This is also chiral. Compound (3): Central carbon has $COOC_2H_5$, $COOCH_3$, $H$, $Br$ with $H$ and $Br$ positions swapped relative to compound (1). **Determine the relationship between (1) and (3):** Using priority rules (CIP system): - Both have the same four groups attached to the chiral center - In compound (1): groups are arranged in one stereochemical order - In compound (3): the positions of $H$ and $Br$ are swapped—this inverts the stereochemistry at that center When two compounds have the **same molecular formula and same connectivity but opposite configurations at all chiral centers**, they are **enantiomers** (non-superimposable mirror images). **Why not the other options?** - (A) False: (1) and (2) differ in ester groups arrangement - (B) False: (1) and (2) have opposite stereochemistry at one center but different ester groups - (D) False: (1) and (2) have different connectivity in ester groups; they're not enantiomers **Answer: (C) (1) and (3) are enantiomers** ✓