See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

We need to determine which reactions produce 1,3-butadiene (CH2=CH-CH=CH2) as the major product. Reaction (a): 1,4-dibromobutane (Br-CH2-CH2-CH2-CH2-Br) treated with 2 moles of (CH3)3COK (potassium tert-butoxide) in (CH3)3COH. Potassium tert-butoxide is a strong, bulky base that favors E2 elimination. With 2 moles of base and a 1,4-dibromobutane substrate, two successive E2 eliminations occur — one at each end of the chain — removing HBr from each carbon bearing bromine. This gives CH2=CH-CH=CH2, i.e., 1,3-butadiene, as the major product. The bulky base ensures elimination rather than substitution. Reaction (b): 1,4-butanediol (HO-CH2-CH2-CH2-CH2-OH) treated with concentrated H2SO4. Concentrated H2SO4 at high temperature promotes dehydration (elimination of water). With a 1,4-diol and 2 equivalents of water removed (one from each end), double elimination yields 1,3-butadiene as the major product. This is an acid-catalyzed double dehydration. Reaction (c): Vinylacetylene (H2C=CH-C≡CH) treated with 1 mole of H2 using Ni2B (Lindlar-type catalyst or P-2 nickel). Ni2B (P-2 nickel) is a semi-hydrogenation catalyst that selectively reduces a triple bond to a cis-double bond (similar to Lindlar's catalyst). Addition of 1 mole of H2 to the triple bond of vinylacetylene (H2C=CH-C≡CH) gives H2C=CH-CH=CH2, which is 1,3-butadiene. The catalyst selectively stops at the alkene stage and does not over-reduce. The product is 1,3-butadiene. Since all three reactions (a), (b), and (c) produce 1,3-butadiene as the major product, the answer is (d) All of these. Therefore, the correct answer is D.