Question
But-2-ene undergoes ozonolysis. Identify the products formed in: (a) Reductive workup (with Zn/H₂O or dimethyl sulfide) (b) Oxidative workup (with H₂O₂)
Solution — Step by Step
But-2-ene is CH₃–CH=CH–CH₃. The double bond sits between C2 and C3. Both carbons of the double bond carry one methyl group each — this symmetry is the key observation here.
Ozone (O₃) attacks the π bond and forms a molozonide, which rearranges into an ozonide. The C=C bond is completely cleaved — each carbon of the original double bond becomes a carbonyl carbon. Think of ozonolysis as a pair of scissors cutting the double bond and putting an oxygen on each end.
Reductive workup converts each carbonyl fragment without further oxidation:
- A disubstituted carbon (R–CH=) gives an aldehyde (R–CHO)
- A trisubstituted carbon (R₂C=) gives a ketone (R₂C=O)
For but-2-ene, both C2 and C3 are monosubstituted (each carries one –CH₃ and one –H). So both fragments become acetaldehyde (ethanal, CH₃CHO).
Products (reductive): 2 mol of CH₃CHO
Oxidative workup takes it one step further — aldehydes get oxidised to carboxylic acids. Ketones survive (they can’t be oxidised further under these conditions).
Since both fragments here are aldehydes (CH₃CHO), both get oxidised to acetic acid (ethanoic acid, CH₃COOH).
Products (oxidative): 2 mol of CH₃COOH
But-2-ene has 4 carbons. Each product (acetaldehyde or acetic acid) has 2 carbons. Two fragments × 2 carbons = 4 carbons. ✓ Carbon count is always a quick sanity check.
Why This Works
Ozonolysis is one of the cleanest ways to locate a double bond in an unknown compound — you cleave it, identify the fragments, and work backwards. This is why it’s a favourite in structure determination problems, both in JEE and NEET.
The choice of workup is what determines the final product. Reductive workup (Zn/H₂O, Me₂S, or PPh₃) stops at the aldehyde/ketone stage. Oxidative workup (H₂O₂, followed by acidic workup) pushes aldehydes all the way to carboxylic acids — ketones are unaffected because there’s no hydrogen on that carbonyl carbon to remove.
But-2-ene is symmetric, so both sides of the cleavage give identical fragments. This is actually a useful reverse-logic tool: if ozonolysis gives only one product, suspect a symmetric alkene.
Reverse engineering trick: If a question tells you ozonolysis gives only CH₃CHO with reductive workup, the alkene must be but-2-ene (or some symmetric CH₃CH=CHCH₃ variant). PYQs love asking this in the reverse direction — “identify the alkene from its ozonolysis products.”
Alternative Method — The Retrosynthetic Shortcut
Instead of drawing the mechanism, use the reconnection method:
- Write the products side by side: CH₃CHO and CH₃CHO
- Replace the C=O of each with a C– bond end
- Join the two C– ends with a double bond
This reverse approach is faster under exam pressure, especially when the question gives you the ozonolysis products and asks for the parent alkene (appeared in JEE Main 2022 Paper 1).
| Fragment type | Reductive workup | Oxidative workup |
|---|---|---|
| R–CH= (one H on C) | Aldehyde (R–CHO) | Carboxylic acid (R–COOH) |
| R₂C= (no H on C) | Ketone (R₂C=O) | Ketone (R₂C=O) — unchanged |
Common Mistake
Don’t confuse which workup gives which product. A very common error: students write carboxylic acid for reductive workup or stop at aldehyde for oxidative workup. Remember — reductive = gentle = stops at carbonyl. Oxidative = harsh = aldehydes get pushed to acids. If you mix these up, you lose the mark even if the rest of the reasoning is correct. In JEE Main 2022, this exact distinction was the deciding factor between options B and C.