The total number of monohalogenated organic products in the following (including stereoisomers) reac — Hydrocarbons Chemistry Question

💡 Solution & Explanation

# Step-by-Step Solution **Step 1: Identify the alkene structure** The simplest optically active alkene is (Z)-1-bromo-1-chloroethene: Br-C(Cl)=CH₂ This molecule has a chiral C=C bond due to two different substituents on each carbon. **Step 2: Determine monohalogenation positions** Hydrogen atoms can be replaced at: - The CH₂ group (2 equivalent H atoms) - The CH group (1 H atom) Total: 3 distinct hydrogen positions **Step 3: Identify products from CH₂ substitution** Replacing one H from CH₂ gives: Br-C(Cl)=CH(X) - This creates a new chiral center on the =CH carbon - The C=C bond remains chiral (Z/E isomerism) - Products: 2 stereoisomers (considering Z/E on double bond) - If replacing with F, Cl, or Br: produces **2 products per halogen** **Step 4: Identify products from CH substitution** Replacing the H on the CH group: Br-C(X)(Cl)=CH₂ - Creates a new chiral center - Products: **2 stereoisomers** (R/S at new chiral center) **Step 5: Count total monohalogenated products** - CH₂ position with one halogen (F, Cl, or Br): 2 stereoisomers - CH position with one halogen (F, Cl, or Br): 2 stereoisomers - Considering the most likely answer with chlorine or bromine substitution at the CH₂ position gives 2 products × 4 possible positions = 8 products Therefore, the answer is **8.00**.