Stir butan-1-ol with PCC (pyridinium chlorochromate) in dry dichloromethane and you get butanal, an aldehyde. Stir the same alcohol with Jones reagent, a chromium reagent in aqueous sulfuric acid, and you get butanoic acid. Both reagents contain chromium(VI), chromium in its +6 oxidation state. The difference that decides the product is the water.
This section explains that result. It recaps how far each kind of alcohol can be oxidized, shows how chromium removes a hydrogen, and shows why water carries a primary alcohol past the aldehyde. It then sorts the common reagents by where they stop.
What each class of alcohol can become
As Oxidation levels in organic chemistry showed, oxidizing an alcohol trades a C–H bond on the carbinol carbon (the carbon that carries the OH) for a second bond to oxygen, so each hydrogen on that carbon allows one step up. Butan-1-ol, a primary alcohol, can go up twice: first to butanal, then to butanoic acid. Butan-2-ol, a secondary alcohol, goes up once, to butanone. 2-Methylpropan-2-ol, a tertiary alcohol, has no hydrogen on the carbinol carbon, so it is not oxidized.
| Substrate | Anhydrous oxidant PCC, Swern, DMP | Aqueous Cr(VI) Jones, CrO₃/H₂SO₄ |
|---|---|---|
| Primary alcohol | Aldehyde | Carboxylic acid |
| Secondary alcohol | Ketone | Ketone |
| Tertiary alcohol | No reaction | No reaction |
Only the primary row depends on the reagent. A ketone has no hydrogen left on its carbonyl carbon, so it cannot be oxidized further under these conditions. Both kinds of reagent therefore give the same ketone from a secondary alcohol.
How chromium(VI) removes the hydrogen
In Jones reagent the chromium is supplied as CrO₃. In the aqueous acid it becomes chromic acid, H₂CrO₄. CrO₃ itself has no OH to lose as water, so the acid form is the one drawn. Chromic acid oxidizes a primary alcohol, R–CH₂–OH, in two steps. Oxidizing an aldehyde showed the same two steps acting on an aldehyde hydrate.
In the first step, a lone pair on the alcohol oxygen attacks the chromium, and the electrons of one Cr=O bond move onto its oxygen. A proton moves, and a water molecule leaves the chromium. The product is a chromate ester, R–CH₂–O–CrO₂OH, in which the carbon is joined through an oxygen to chromium.
In the second step, a water molecule removes one hydrogen from the carbinol carbon. The two electrons of that C–H bond become the new C=O bond. At the same moment the O–Cr bond breaks, and both of its electrons go to chromium. A C–H bond and a bond to a leaving group break in one step, as in an E2 elimination (Substitution & Elimination), so this step is called E2-like.
The second step is where a tertiary alcohol fails. Its oxygen can still attack chromium, and the chromate ester can still form. But the carbinol carbon has no hydrogen for water to remove, so the ester cannot collapse to a C=O, and no oxidation happens.
Chromium is reduced as the carbon is oxidized. The oxidation state of the carbinol carbon rises by two, and chromium’s falls by two, from +6 to +4. Further steps take the Cr(IV) on to Cr(III). Cr(VI) is orange and Cr(III) is green, so the mixture turns from orange to green as the reagent is used up.
PCC and PDC (pyridinium dichromate) are chromium(VI) reagents used without water. They reach a chromate ester by a different first step, and the E2-like second step is the same.
Why water decides where a chromium oxidation stops
Go back to butan-1-ol. With either PCC or Jones reagent, the two steps above turn it into butanal first. What happens next depends on whether there is water in the flask.
In water, some of the aldehyde adds water across its C=O to give its hydrate, R–CH(OH)₂, as Hydrates & cyanohydrins showed. The hydrate carbon holds an H and an OH, just as the carbinol carbon of an alcohol does. So chromic acid oxidizes the hydrate by the same two steps, and the product is the carboxylic acid.
In a dry solvent, no hydrate can form. The aldehyde carbon has an H but no OH, so there is no alcohol oxygen to form a chromate ester, and the oxidation stops at the aldehyde.
The hydrate is never isolated and never appears in the answer, but it does not need to be the main form of the aldehyde either. As chromium uses up the hydrate, more aldehyde adds water to replace it, until all of it has been oxidized. That is also why using less Jones reagent, or stopping the reaction early, does not reliably leave the aldehyde: in water, the aldehyde starts to be oxidized as soon as it forms.
The reagents worth knowing
- PCC, in dry dichloromethane. It is Cr(VI) with no water present, so a primary alcohol stops at the aldehyde and a secondary alcohol gives the ketone. PCC is mildly acidic. It is the standard answer when a question wants an aldehyde and offers a chromium reagent.
- PDC is PCC’s near twin. It is also anhydrous Cr(VI), and it also stops at the aldehyde.
- Jones reagent: CrO₃ in aqueous sulfuric acid, usually with acetone as the solvent. It is Cr(VI) with water, so a primary alcohol goes on to the carboxylic acid, and a secondary alcohol gives the ketone. It is strongly acidic. You will also see it written as CrO₃/H₂SO₄ or as Na₂Cr₂O₇/H₂SO₄, which behave the same way.
- Swern oxidation: oxalyl chloride and DMSO (dimethyl sulfoxide) at −78 °C, then triethylamine. It uses no metal at all and stops cleanly at the aldehyde. It is mild enough for sensitive molecules, which is why it is common in real synthesis.
- Dess–Martin periodinane (DMP): a hypervalent iodine reagent, meaning an iodine compound in which iodine carries more bonds than the usual one. It works at room temperature under nearly neutral conditions and also stops at the aldehyde. Like the Swern, it is metal-free, and it is easier to run because it needs no cooling to −78 °C.
- KMnO₄, usually hot and basic, with acid added afterwards. Manganese(VII) in water behaves like aqueous chromium: a primary alcohol goes on to the carboxylic acid, and a secondary alcohol gives the ketone. It is the reagent to recognize when a question wants the acid without naming chromium. Purple Mn(VII) turns into brown MnO₂ as it is used up.
- MnO₂ is the selective one. It oxidizes only allylic alcohols, whose OH sits on a carbon next to a C=C, and benzylic alcohols, whose OH sits on a carbon attached to a benzene ring. It leaves ordinary saturated alcohols alone. So MnO₂ can tell two alcohols in one molecule apart, and it does so by their position, not by their class.
Set MnO₂ aside and sort the rest by outcome. The list is short. PCC, PDC, the Swern and DMP give aldehydes; Jones reagent, other aqueous Cr(VI) and hot KMnO₄ give carboxylic acids.
Give the product of each. Classify the alcohol first, then count the hydrogens on its carbinol carbon, then check the reagent for water.
(a) 2-methylbutan-1-ol + PCC, CH₂Cl₂. Primary, with two carbinol hydrogens. PCC is anhydrous, so the oxidation stops one level up: 2-methylbutanal, CH₃CH₂CH(CH₃)CHO.
(b) 2-methylbutan-1-ol + CrO₃, H₂SO₄, H₂O. The same alcohol with an aqueous reagent. The aldehyde forms its hydrate and is oxidized a second time: 2-methylbutanoic acid, CH₃CH₂CH(CH₃)CO₂H.
(c) cyclohexanol + Jones. Secondary, with one carbinol hydrogen, so it goes up one level and stops, whatever the reagent: cyclohexanone. The water in the Jones reagent makes no difference for a secondary alcohol.
(d) 2-methylbutan-2-ol + Jones, in excess, overnight. Tertiary, so the carbinol carbon has no C–H for the second step to break. No reaction, and neither excess reagent nor extra time changes that.
The pattern: ask two questions, in this order. How many hydrogens are on the carbon that carries the OH? Is there water in the flask? The first sets how far the alcohol can go; the second says whether a primary alcohol gets all the way there.
Three traps
When no reagent on the list is selective
The molecule is 4-(hydroxymethyl)cyclohexan-1-ol. It has a CH₂OH group, which is a primary alcohol, and an OH on the ring, which is a secondary alcohol. The target is the ketone at the ring carbon, with the CH₂OH kept as it is.
Step 1: how far up? The secondary alcohol goes one level up, to the ketone. Every reagent on the list stops there.
Step 2: what else is in the molecule? The primary alcohol. PCC, PDC, the Swern and DMP turn it into an aldehyde, and Jones reagent and KMnO₄ turn it into a carboxylic acid. So every reagent on the list oxidizes the CH₂OH along with the ring OH, as the figure shows for PCC.
Step 3: is there a selective reagent? MnO₂ is the only one on the list that picks between alcohols, and it picks by position: allylic and benzylic only. Neither alcohol here is allylic or benzylic, so MnO₂ does nothing useful. A few reagents outside this course do favor one class. TEMPO (a stable radical used as a catalyst) with bleach, for example, oxidizes a primary alcohol much faster than a secondary one, but here the primary alcohol is the one you want to keep.
Step 4: reagent choice cannot solve it. The fix is to cover the primary alcohol first, oxidize, and then uncover it. A temporary cover like this is called a protecting group. This step is a preview; Synthesis & Retrosynthesis teaches protecting groups. The usual cover for an alcohol is a silyl ether, in which a silicon group replaces the H of the OH. TBSCl (tert-butyldimethylsilyl chloride) with imidazole (a mild base) puts it on, and it reacts faster with a primary OH than with a secondary one, so it covers the CH₂OH. TBAF, a fluoride source, takes it off at the end.
The transferable move is step 4, not the reagent names. Suppose two alcohols in one molecule would both react, but only one may react, and neither is allylic or benzylic. Then no oxidant on this list solves the problem. Seeing that the problem now needs a protecting group is the real skill.
What carries forward
Each of these oxidations has a reverse. Reducing carbonyls, the next section, uses hydride reagents (sources of H−) to take carbonyl compounds back down to alcohols. Knowing both directions is what makes synthesis planning possible. A primary alcohol and its aldehyde are one oxidation level apart, so either can be made from the other in one step, as long as the reagent does not overshoot.