See image — Haloalkanes and Haloarenes Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material and its structure. C3H7I is propyl iodide. Since KOH/alc. (alcoholic KOH) is a dehydrohalogenation (E2 elimination) reagent, it converts an alkyl halide to an alkene. Starting from 1-iodopropane (n-propyl iodide, CH3CH2CH2I), elimination gives propene: CH3-CH=CH2 (compound A). Even from 2-iodopropane, elimination gives propene CH3-CH=CH2. Step 2: A (propene, CH2=CH-CH3) reacts with NBS (N-bromosuccinimide) under heat (Delta). NBS at elevated temperature performs allylic bromination via a radical mechanism. The allylic position in propene is the CH3 group (C3), giving allyl bromide: CH2=CH-CH2Br (compound B). Step 3: B (allyl bromide, CH2=CH-CH2Br) reacts with KCN in alcoholic solution. KCN is a nucleophile (CN-) that undergoes SN2 substitution on the primary allylic bromide. The CN- displaces Br- to give: CH2=CH-CH2CN (allyl cyanide, 3-butenenitrile). This is compound C. Step 4: Verify against options: (a) (CH3)2CHCN - this would require a secondary substrate; not consistent with allyl bromide + KCN. (b) CH2=CH-CH2CN - this matches the product of SN2 of CN- on allyl bromide. CORRECT. (c) Br-CH=CH-CN - this would require addition, not substitution; not the product here. (d) CH2=CH-CH(CN)(Br) - this has both Br and CN retained, which is not the product of simple SN2 substitution. Therefore, the correct answer is B.