JEE Advanced level
A JEE Advanced-level example on Hydrocarbons
Question. Compound X is an acyclic hydrocarbon with molecular formula C₅H₈. X decolorizes bromine water and also gives a white precipitate with ammoniacal silver nitrate solution. When X is treated with excess HBr, compound Y is formed. The total number of hydrogen atoms in Y is:
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- Calculate degree of unsaturation for C₅H₈: (2×5 + 2 - 8)/2 = 2, indicating one triple bond
- White precipitate with ammoniacal AgNO₃ confirms terminal alkyne (forms silver acetylide). So X is pent-1-yne: CH₃-CH₂-CH₂-C≡CH
- Terminal alkynes add 2 molecules of HBr following Markovnikov rule. Both Br atoms attach to the same carbon (the one originally bearing the triple bond hydrogen)
- First HBr addition: CH₃CH₂CH₂C≡CH → CH₃CH₂CH₂CBr=CH₂. Second HBr addition: → CH₃CH₂CH₂CBr₂CH₃
- Product Y is 2,2-dibromopentane with molecular formula C₅H₁₀Br₂
Answer: 10
This example was solved twice independently and both solutions agree.