See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Concept: Addition of HI to an alkene with an allylic iodide present, using excess HI in CCl4. Step 1: Identify the starting material. The compound is allyl iodide: H2C=CH-CH2-I (3-iodoprop-1-ene). It contains both a carbon-carbon double bond and an allylic C-I bond. Step 2: First equivalent of HI adds to the double bond. By Markovnikov's rule, H adds to the terminal carbon (CH2=) and I adds to the internal carbon (=CH-), giving: CH3-CHI-CH2-I (1,2-diiodopropane) as the initial addition product. Step 3: However, with excess HI present, consider what happens next. The allylic iodide (C-I bond at the allylic position) can undergo substitution or elimination. More importantly, note that the first addition of HI to H2C=CH-CH2-I following Markovnikov's rule gives CH3-CHI-CH2-I. Step 4: Re-examine: With excess HI in CCl4 (non-polar solvent, ionic conditions suppressed), the reaction proceeds via addition followed by elimination of HI or direct substitution. The allylic C-I in the starting material can be displaced, and the double bond reacts with HI. Step 5: Actually, the key insight is that allyl iodide (H2C=CH-CH2-I) with HI first adds across the double bond by Markovnikov's rule: H+ adds to CH2= giving CH3-CH(+)-CH2-I (secondary carbocation), then I- attacks to give CH3-CHI-CH2-I. This is option (a). Step 6: But with EXCESS HI, the geminal or vicinal diiodide can lose HI to regenerate an alkene, or more relevantly: the 1,2-diiodide (option a) reacts further. CH3-CHI-CH2-I with excess HI: the C-I bonds are present, and the product can undergo dehydroiodination to form CH3-CH=CHI or similar, then re-addition. Step 7: Reconsidering the mechanism with excess HI - the correct interpretation for this classic problem: H2C=CH-CH2-I + HI (first addition, Markovnikov) → CH3-CHI-CH2-I. Then the second equivalent of HI reacts with the allylic iodide via an SN2 or the diiodo compound loses I2 or HI, ultimately giving 2-iodopropane (CH3-CHICH3). The allylic iodide under HI undergoes: the vinyl/allylic system equilibrates, and the thermodynamic product with excess HI is 2-iodopropane (CH3-CH(I)-CH3), a secondary alkyl iodide, which is the most stable product. Step 8: The mechanism - allyl iodide + HI gives allylic cation [CH3-CH=CH2]+ (after loss of I from allylic position assisted by HI) which is resonance stabilized, then I- attacks at the more substituted carbon (secondary) to give CH3-CH(I)-CH3 as the major product. Why other options fail: - Option (a): Would be the kinetic Markovnikov addition product but not the major product with excess HI. - Option (c): 1-iodopropane is anti-Markovnikov, not favored. - Option (d): 1,3-diiodopropane has no reasonable mechanism here. Therefore, the correct answer is B.