See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze the product: The product is 1-bromo-2,3-dichloropropane. Starting from propene (CH2=CH-CH3), we need to introduce two chlorines across the double bond and one bromine at the allylic/terminal position. Step 2 - Identify the source of each halogen: The two chlorines are on adjacent carbons (C2 and C3), which is characteristic of addition of Cl2 across a double bond (anti-addition via Cl2/CCl4). The bromine is on C1 (the terminal carbon, formerly =CH2), which is the allylic position relative to the double bond before any addition. Step 3 - Determine the order of reactions: If we first do allylic bromination with NBS/hv (step 1), NBS selectively brominates the allylic position of propene. The allylic position in propene is C3 (the methyl group, CH3), giving allyl bromide: BrCH2-CH=CH2 (3-bromopropene). Then in step 2, Cl2/CCl4 adds across the remaining double bond of allyl bromide in an electrophilic addition, giving BrCH2-CHCl-CH2Cl, which is 1-bromo-2,3-dichloropropane — exactly the target product. Step 4 - Verify option D: Step 1 = NBS/hv (allylic bromination of propene at C3 giving allyl bromide), Step 2 = Cl2/CCl4 (electrophilic addition across double bond giving the 1,2-dichloride). This matches the product perfectly. Step 5 - Why other options fail: (a) Cl2/CCl4 first adds Cl2 across the double bond giving 1,2-dichloropropane, then Br2 would give a tetrahalide — not the correct product. (b) HBr first adds to the double bond (Markovnikov: CH3-CHBr-CH3, 2-bromopropane), then Cl2/CCl4 cannot add across a double bond that no longer exists and would give substitution — not the correct product. (c) Cl2/CCl4 adds across double bond giving 1,2-dichloropropane, then NBS/hv would do allylic bromination but there is no longer an allylic position — not the correct product. Therefore, the correct answer is D.