An unknown alkene 'A' is subjected to complete reductive ozonolysis ( followed by ) and yields exact — Hydrocarbons Chemistry Question
Question
An unknown alkene 'A' is subjected to complete reductive ozonolysis ($O_3$ followed by $Zn/H_2O$) and yields exactly two moles of an aldehyde having a molar mass of 44 g/mol as the sole organic product. What is the IUPAC name of alkene 'A'?
💡 Solution & Explanation
An aldehyde with a molar mass of 44 g/mol is acetaldehyde/ethanal ($CH_3CHO$: $12 + 1 + 12 + 1 + 16 = 44$). Since reductive ozonolysis produces exactly two moles of $CH_3CHO$ and no other products, the original alkene must be symmetrical and contain two $CH_3-CH=$ groups. Rejoining these fragments at the double bond yields $CH_3-CH=CH-CH_3$, which is But-2-ene.