Q.An alkene 'A' contains three C–C, eight C–H σ bonds and one C–C π bond. 'A' on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of 'A'.
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Start your 14-day free trial to unlock the full solution →Ozonolysis cleaves the double bond to give two aldehyde fragments; working backward from the product (ethanal, ) and the bond count reveals that A is but-2-ene.
The problem hands us structural clues—bond counts and an ozonolysis product—and asks us to reconstruct the alkene. Ozonolysis is the key: it cuts a double bond and converts each carbon of that double bond into a carbonyl group. When both fragments are aldehydes, the original double bond must have been between two carbons (each bearing one hydrogen).
Start by identifying the aldehyde product. We're told it has molar mass . The simplest aldehyde is methanal (, ), the next is ethanal (, ). So the aldehyde is ethanal, .
Now reconstruct the alkene:
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Ozonolysis gives two moles of the same aldehyde.
That means the alkene is symmetrical about the double bond. Each half, when oxidised, becomes .
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Reverse the ozonolysis.
Ethanal has the carbonyl carbon bonded to a methyl group: . In the original alkene, that carbonyl carbon was part of the double bond. Remove the oxygen and join the two aldehyde fragments at their carbonyl carbons:
This is but-2-ene.
- Verify the bond counts. …
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