See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1: Identify the starting material and first reaction. 2,4-hexadiyne is HC≡C-C≡C-CH2CH3... wait, 2,4-hexadiyne is CH3-C≡C-C≡C-CH3 (hex-2,4-diyne). Formula C6H6 confirms this (6 carbons, 2 triple bonds: 6H). Step 2: Reaction with Li in NH3(liq) — this is a dissolving metal (Birch-type) reduction of an alkyne, specifically a partial reduction (one equivalent of Li/NH3 reduces one triple bond via trans-addition of H2, giving a trans-alkene). With 2,4-hexadiyne (internal diyne), Li/NH3 reduces one of the triple bonds to give a trans (E)-enyne: CH3-CH=CH-C≡C-CH3 (i.e., (E)-hex-3-en-1-yne or more precisely (E)-hex-2-en-4-yne). The dissolving metal reduction of internal alkynes gives trans-alkenes. So the product is (E)-CH3-C≡C-CH=CH-CH3, i.e., (E)-hex-4-en-2-yne: CH3-C≡C-CH=CH-CH3. Actually, with a symmetric diyne CH3C≡CC≡CCH3, Li/NH3 reduces one triple bond to give (E)-CH3C≡C-CH=CHCH3. Step 3: Treat with 1 equivalent Cl2 in CCl4. Cl2 in CCl4 adds electrophilically across a double or triple bond. The molecule has both a triple bond (C2≡C3) and a double bond (C4=C5, trans). With 1 equivalent Cl2: - Addition to the double bond (trans/anti addition): anti addition of Cl2 to (E)-CH=CH gives the anti-addition product. For (E)-alkene, anti addition gives the (R,S) = meso or threo product — specifically the (3R,4S) + (3S,4R) pair for the dichloro product, or a specific diastereomer. - Addition to the triple bond: anti addition gives a (E)-1,2-dichloroalkene (trans-vinyl dichloride). Step 4: Identify constitutional isomers from products. The two possible constitutional isomers arise from Cl2 adding to either the C=C double bond or the C≡C triple bond of (E)-CH3C≡C-CH=CHCH3: - Addition to C=C (positions 4,5): gives CH3-C≡C-CHCl-CHCl-CH3 → 4,5-dichlorohex-2-yne... but this is a 1,2-dichloro compound on the former double bond portion. - Addition to C≡C (positions 2,3): gives CH3-CCl=CClC-CH=CHCH3 → (E or Z)-3,4-dichlorohex-3-en... Wait, let me re-examine the structures in the image more carefully in terms of carbon skeleton and Cl positions for a 6-carbon chain with one double bond and two Cl atoms. Structure (I): gem-dichloride at one carbon adjacent to a double bond — this would be a 1,1-addition or rearrangement, not typical. Structure (III): vicinal dichloride with Cl on C3 and C4 with double bond at C1-C2 (or C2-C3 region). Given the answer is (d) I and III, these correspond to the two constitutional isomers: one from Cl2 addition to the triple bond giving a vinylidene dichloride type (gem) product (I) and one from Cl2 addition to the double bond (III), OR both from addition at different sites giving different connectivity. The key concept: (E)-hex-2-en-4-yne reacts with 1 eq Cl2 at either the double bond or the triple bond, giving two different constitutional isomers. Structure I (gem-dichloro vinyl type from triple bond addition giving CCl2=) and Structure III (vicinal dichloro from double bond addition) are the two possible constitutional isomers, which matches answer D. Therefore, the correct answer is D.