Oxidation and ozonolysis

Alkenes don't just add things across their double bond — they can be torn apart by it. Pass ozone gas through an alkene, work up the fragile ozonide with zinc and water, and the C=C bond snaps cleanly in two, leaving a carbonyl group sitting on each of the carbons that used to be joined. Run the same alkene past potassium permanganate instead, and the outcome depends entirely on how hard you push: cold and dilute leaves the carbon skeleton intact and merely dresses the double bond in two new –OH groups; hot and concentrated rips the skeleton apart at the same spot ozone would have chosen. Both reactions matter for the same reason — the cleavage products are a fingerprint of exactly where the double bond sat, which is why chemists use them to reverse-engineer the structure of an unknown alkene from what's left after the reaction.

The core insight

Ozonolysis (O3, then Zn/H2O) and hot concentrated KMnO4 both cleave the alkene's C=C bond into two carbonyl carbons at the exact site of unsaturation — cold dilute KMnO4 stops short and only adds a diol — so the cleavage products reveal the original double bond's position and substitution.

The core idea

Ozonolysis: the clean cleavage. An alkene first adds ozone across its double bond to form a five-membered cyclic ozonide, a strained and explosive intermediate that is never isolated — it is decomposed immediately in a reductive work-up, typically zinc dust in water (or dimethyl sulfide), which prevents the fragments from over-oxidising to acids. The double bond disappears entirely, and each carbon that bore it becomes a carbonyl carbon:

R2C=CR2′→(i) O3 (ii) Zn/H2OR2C=O+O=CR2′R_2C=CR_2' \xrightarrow{(i)\ O_3\ (ii)\ Zn/H_2O} R_2C=O + O=CR_2'

where: R,R′R, R' stand for H or alkyl groups on the two alkene carbons. A carbon that carried at least one H becomes an aldehyde; a carbon carrying two alkyl groups (no H left to lose) becomes a ketone, since it cannot be pushed further to a carboxylic acid. This H-vs-no-H split is the entire logic of reading ozonolysis products backward to the parent alkene.

KMnO4: two very different reagents in one bottle. Concentration, temperature and pH turn permanganate from a gentle reagent into a destructive one.

  • Cold, dilute, alkaline KMnO4 (Baeyer's reagent) performs syn-dihydroxylation: it adds two –OH groups across the double bond without touching the carbon skeleton, giving a vicinal diol. The purple colour fades as the reaction proceeds — this decolourisation is the classical Baeyer's test for a C=C bond.
CH2=CH2→cold, diluteKMnO4/OH−CH2OH−CH2OHCH_2=CH_2 \xrightarrow[\text{cold, dilute}]{KMnO_4/OH^-} CH_2OH-CH_2OH
  • Hot, concentrated, acidified KMnO4 is a strong enough oxidant to cleave the double bond outright, exactly as ozonolysis does, but it pushes each fragment one step further: a carbon bearing at least one H is oxidised all the way to a carboxylic acid (not just an aldehyde), a carbon with two alkyl groups stops at a ketone, and a terminal =CH2=CH_2 carbon is oxidised past formic acid all the way to CO2CO_2 and water.
R−CH=CH2→Δhot, conc. KMnO4/H+R−COOH+CO2+H2OR-CH=CH_2 \xrightarrow[\Delta]{\text{hot, conc. } KMnO_4/H^+} R-COOH + CO_2 + H_2O

Polymerisation, in one line. Given heat, pressure and a catalyst, many alkene molecules can add to each other repeatedly through their double bonds, chaining into a long polymer — ethene into polyethene is the textbook case. ISC Class 11 asks only that you know this happens; no mechanism or kinetics is required at this level.

Figure
Two reagents attacking the same alkene, mild KMnO4 adding a diol on top while ozonolysis and hot KMnO4 cleave the double bond into two carbonyl fragments below
Same alkene, three fates depending on the reagent R2C=CR2' cold dilute KMnO4/OH- R2C(OH)-C(OH)R2' vicinal diol, skeleton intact O3 then Zn/H2O R2C=O + O=CR2' clean cleavage, two carbonyls hot conc. KMnO4/H+ R-COOH + ketone/CO2 cleavage plus further oxidation Cleavage always happens at the C=C bond, so the fragments locate it exactly. A carbon with an H on it goes to aldehyde (ozonolysis) or acid (hot KMnO4); a fully alkyl carbon stops at ketone either way.
Derivation AReconstructing an alkene from its ozonolysis fragments

The goal is to justify, from the cleavage mechanism itself, the rule every ISC student is expected to apply in reverse — given the aldehyde and/or ketone products of ozonolysis, rebuild the exact structure and double-bond position of the parent alkene by undoing the bond-breaking step one carbon at a time.

Step 1. Start from the forward reaction. Ozonolysis breaks only the C=C pi bond and, on work-up, replaces it with a C=O at each of the two carbons that were doubly bonded. Every other bond in the molecule — the sigma framework, all C–H and C–alkyl connections — is left completely untouched. This means each carbonyl product is a faithful record of exactly one half of the original alkene, substituents and all.

Step 2. Read off what each carbonyl carbon must have looked like before cleavage. If a product is an aldehyde, R−CHOR-CHO, its carbonyl carbon carried one H and one R group in the alkene, so that carbon was =CH−R=CH-R. If a product is a ketone, R−CO−R′R-CO-R', its carbonyl carbon carried two alkyl groups and no H, so that carbon was =CR−R′=CR-R'. Formaldehyde, HCHOHCHO, is the special case where both remaining bonds are H, so that carbon was a bare terminal =CH2=CH_2.

Step 3. Reassemble by reversing Step 1: strip the "=O" from each carbonyl carbon and join the two exposed carbons back together with a C=C double bond, keeping every substituent exactly where Step 2 placed it. As a check, apply this to formaldehyde (HCHOHCHO) plus acetone (CH3COCH3CH_3COCH_3): formaldehyde's carbon was a terminal =CH2=CH_2, and acetone's carbon carried two methyl groups, =C(CH3)2=C(CH_3)_2. Joining them gives (CH3)2C=CH2(CH_3)_2C=CH_2, 2-methylpropene.

Step 4. Confirm the reconstruction is self-consistent by re-cleaving it mentally: ozonolysis of (CH3)2C=CH2(CH_3)_2C=CH_2 would break the double bond exactly between the fully substituted carbon and the terminal carbon, regenerating acetone and formaldehyde — matching the products we started from. This forward-then-backward consistency check is the whole method; it works for any pair of carbonyl fragments, which is why ozonolysis is such a reliable tool for locating an unknown double bond.

Worked example

Problem: An alkene of molecular formula C5H10C_5H_{10} undergoes ozonolysis to give acetone (CH3COCH3CH_3COCH_3) and acetaldehyde (CH3CHOCH_3CHO) as the only two products. Deduce the structure of the alkene.

Step 1. Apply the reconstruction rule: remove "=O" from each carbonyl product and rejoin the two carbons with a double bond.

Step 2. Acetone, CH3−CO−CH3CH_3-CO-CH_3, is a ketone, so its carbonyl carbon carried two methyl groups and no H — this end of the double bond was fully substituted, =C(CH3)2=C(CH_3)_2.

Step 3. Acetaldehyde, CH3−CHOCH_3-CHO, is an aldehyde, so its carbonyl carbon carried one methyl group and one H — the other end of the double bond was =CH−CH3=CH-CH_3.

Step 4. Joining the two fragments at the former carbonyl carbons gives (CH3)2C=CH−CH3(CH_3)_2C=CH-CH_3, which is 2-methylbut-2-ene (C5H10C_5H_{10}) — matching the given molecular formula and confirming the structure.

Competitive stretch

Problem: But-1-ene (CH2=CH−CH2−CH3CH_2=CH-CH_2-CH_3) is treated separately with (a) cold, dilute alkaline KMnO4 and (b) hot, concentrated acidified KMnO4. Give the organic product(s) in each case.

Step 1 (mild). Cold dilute alkaline KMnO4 adds –OH to both double-bond carbons without breaking the skeleton: CH2=CH−CH2CH3→CH2OH−CHOH−CH2CH3CH_2=CH-CH_2CH_3 \to CH_2OH-CHOH-CH_2CH_3, butane-1,2-diol; the purple colour decolourises (Baeyer's test).

Step 2 (harsh). Hot concentrated acidified KMnO4 cleaves the C=C bond. The terminal =CH2=CH_2 carbon (two H atoms) is oxidised fully to CO2+H2OCO_2 + H_2O; the internal =CH−=CH- carbon (one H, plus the ethyl chain) is oxidised to a carboxylic acid, CH3CH2COOHCH_3CH_2COOH. Answer: butane-1,2-diol from the mild route; propanoic acid (plus CO2 and water) from the harsh route.

Common traps

  • Mixing up the mild product (a diol, cold dilute alkaline KMnO4) with the harsh cleavage products (acids/ketones/CO2, hot concentrated acidified KMnO4) — the conditions completely change the outcome, not just its extent.
  • Misreading which carbonyl a carbon becomes: two alkyl groups, no H → ketone; at least one H → aldehyde (ozonolysis) or acid (hot KMnO4). Get this backward and the reconstructed alkene is wrong.
  • Skipping the Zn/H2O work-up after ozonolysis — without it the ozonide can over-oxidise, and the clean aldehyde/ketone products never form.
  • Forgetting a terminal =CH2=CH_2 under hot concentrated KMnO4 goes all the way to CO2+H2OCO_2 + H_2O, not to formic acid.

Ask the tutor: if reconstructing an alkene from a pair of ozonolysis fragments still feels shaky, ask the AI tutor for a few more fragment pairs to practise on and check your reasoning.

How do I know whether a product will be an aldehyde or a ketone from ozonolysis?

Look at how many alkyl groups sat on that particular alkene carbon before cleavage. If it carried one H and one alkyl group (or two H atoms for a terminal carbon), it becomes an aldehyde, because that H remains available to oxidise further to a carbonyl H. If it carried two alkyl groups and no H, it becomes a ketone, since there is no H left for the carbonyl carbon to lose — a fully substituted carbon cannot go further than a ketone under these conditions.

Why does hot concentrated KMnO4 give an acid where ozonolysis stops at an aldehyde?

Both reagents cleave the same C=C bond at the same position, but they differ in oxidising power after that cleavage. The Zn/H2O work-up in ozonolysis is deliberately mild and stops right at the aldehyde or ketone stage. Hot, concentrated, acidified KMnO4 is a much stronger oxidant that does not stop there — it pushes any aldehyde-stage carbon one step further, all the way to a carboxylic acid, and pushes a terminal CH2 carbon all the way to CO2 and water.

Is the diol from cold KMnO4 the same kind of product as the diol you get from other addition reactions?

Conceptually yes — it is still an addition across the C=C bond with the carbon skeleton left intact, unlike ozonolysis or hot KMnO4, which both cleave the skeleton. What's specific here is the geometry and mechanism: cold dilute alkaline KMnO4 adds both new –OH groups to the same face of the double bond (syn-addition) via a cyclic manganate ester intermediate, giving a vicinal diol (two –OH groups on adjacent carbons) rather than a diol formed some other way.

Before this

Where this leads