See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Stereoisomers in Oxime Formation **Step 1: Identify the reactant structure** The starting material has: - A C=C double bond (alkene) with **one stereocenter** at the carbon bearing $CH_3$ and $H$ - A C=O group (carbonyl) that will form the oxime **Step 2: Count stereoisomers in the starting material** The alkene carbon has four different groups ($CH_3$, $H$, and two different substituents on the other carbon). This gives **2 stereoisomers** (E/Z isomers of the alkene). **Step 3: Analyze oxime formation** When $NH_2OH$ attacks the carbonyl carbon, it creates a **new stereocenter** at the C=O carbon (now bearing $OH$ and $NH$ groups). **Step 4: Count stereoisomers in the product** The oxime has: - **One stereocenter from the original alkene**: 2 configurations (E/Z) - **One stereocenter from oxime formation**: 2 configurations (E/Z around the C=N double bond) $$\text{Total stereoisomers} = 2 \times 2 \times 2 = 8$$ (The factor of 2 also comes from the possibility of both E and Z configurations around the C=N double bond independently for each alkene isomer) **Answer: (D) 8** ✓