Ozonolysis Of Alkenes

Reductive ozonolysis cleaves the carbon-carbon double bond and converts each doubly bonded carbon into a carbonyl group, with $Zn/H_{2}O$ ensuring aldehydes are obtained rather than acids. To work backwards, join the carbonyl carbons of the fragments with a double bond. Since only propanal, $C{H}_{3}C{H}_{2}CHO$, is produced, the alkene must give two identical propanal fragments, so the double bond connected two $C{H}_{3}C{H}_{2}CH=$ units. Reconnecting them gives $C{H}_{3}C{H}_{2}CH=CHC{H}_{2}C{H}_3$, which is hex-3-ene, a symmetrical alkene consistent with a single carbonyl product. Hex-1-ene would give methanal and pentanal, two different products, so it is wrong. Hex-2-ene would give ethanal and butanal, again two different aldehydes, so it is incorrect. 2-Methylpent-2-ene would give a ketone (propanone) plus an aldehyde, not propanal alone, so it does not fit. The logic relies on the fact that ozonolysis is a cleavage reaction: each carbon of the former double bond becomes the carbonyl carbon of a separate fragment, so identifying the fragments and rejoining their carbonyl carbons reconstructs the parent alkene. The use of a mild reductive workup with $Zn$ is crucial because it stops oxidation at the aldehyde stage and prevents over-oxidation to carboxylic acids. This retro-analysis is exactly the NCERT method for locating double bonds in an unknown structure. A sanity check: a single carbonyl product demands a symmetrical alkene, which only hex-3-ene satisfies.