Oxidizing an aldehyde — Organic Chemistry Notes

Stir butanal with an oxidizing agent and you get butanoic acid. Stir butanone with the same reagent and nothing happens. Leave butanal in a half-empty bottle for a few months and it turns into butanoic acid anyway, with air as the only oxidant.

This section explains that difference. It then works through the reagents that oxidize an aldehyde, the color tests built on them, and why aldehydes go bad on the shelf.

The difference is one hydrogen

Oxidizing a carbon means trading a C–H bond for a bond to oxygen. Butanal's carbonyl carbon holds one H. An oxidant replaces that H with an OH, and the aldehyde becomes a carboxylic acid.

Look at what else each carbonyl carbon holds. Only butanal has an H there to trade for an O.

Butanone's carbonyl carbon holds two carbons and no H. The only way to put another oxygen on it is to break a C–C bond, and ordinary oxidants do not do that. So a ketone stops where an aldehyde keeps going.

Alcohols follow the same count. A primary alcohol has two C–H bonds on the carbon that carries the OH and can be oxidized twice: first to the aldehyde, then to the acid. A secondary alcohol has one and is oxidized once, to a ketone. A tertiary alcohol has none and is not oxidized.

Chromium(VI) oxidizes the hydrate

Stir butanal with Jones reagent and you get butanoic acid. Jones reagent is chromium(VI), chromium in its +6 oxidation state, supplied as CrO₃ or Na₂Cr₂O₇, in aqueous sulfuric acid, usually with acetone as the solvent. The orange Cr(VI) turns green as it is used up.

A chromium(VI) oxidant needs a carbon that holds both an H and an OH, as Hydrates & cyanohydrins showed. The aldehyde carbon has the H but no OH. In water, though, some of the aldehyde adds water to give its hydrate, R–CH(OH)₂. The hydrate carbon holds an H and an OH, so the hydrate is what chromium attacks.

The attack takes two steps. First, an OH oxygen of the hydrate bonds to the chromium of chromic acid, H₂CrO₄, which is what CrO₃ becomes in water. A proton moves, and a water molecule leaves the chromium. The product is a chromate ester: the carbon is now joined through an oxygen to chromium, C–O–Cr.

Second, a water molecule removes the hydrogen on the 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, which drops from +6 to +4. The Cr(IV) goes on, in further steps, to green Cr(III). A C–H bond and a bond to a leaving group break in one step, as in an E2 elimination (Substitution & Elimination).

Read the panels in order. The hydrogen that water removes in the third panel is the one the aldehyde had all along.

The hydrate needs water. That is why PCC, a Cr(VI) reagent used in dry dichloromethane, oxidizes a primary alcohol to the aldehyde and stops there: with no water, no hydrate forms. Jones reagent, in water, takes a primary alcohol through the aldehyde and this same hydrate all the way to the acid; Oxidation & Reduction, the next chapter, compares the two reagents in full.

Chlorite adds to the C=O itself

Treat (E)-hex-4-enal with sodium chlorite, NaClO₂, in a mildly acidic buffer (NaH₂PO₄, in tert-butanol and water). You get (E)-hex-4-enoic acid, and the C=C is untouched. This is the Pinnick oxidation.

The buffer is there for a reason. In mildly acidic water, some chlorite ion, ClO₂⁻, picks up a proton to give chlorous acid, HClO₂, and chlorous acid is the actual oxidant. One of its oxygens attacks the carbonyl carbon, as in the nucleophilic additions earlier in this chapter. A proton then moves, so the old carbonyl oxygen ends up as OH. The carbon now holds R, H, OH and an O–Cl–O group.

Five atoms of that adduct then form a ring: H, C, O, Cl and the end O. The end oxygen takes the hydrogen from carbon. The pair that held that H now makes a double bond from carbon to the oxygen that links it to chlorine. The O–Cl bond breaks, and its electrons stay on chlorine. Out come the carboxylic acid and hypochlorous acid, HOCl. Chlorine drops from +3 to +1.

HOCl would cause trouble if it stayed. It destroys chlorite, and it adds across C=C bonds. So the flask also holds a large excess of 2-methylbut-2-ene, a cheap alkene that reacts with the HOCl first. A reagent added to use up a by-product like this is called a scavenger. The scavenger is why the C=C in hex-4-enal survives.

The second panel is the oxidation. Follow the highlighted H: it leaves carbon and ends up on the far oxygen of the chlorine group.

No hydrate is involved. So the rule from the last section, an H and an OH on one carbon, belongs to chromium(VI); it is not a rule for every oxidant.

Permanganate: strong and unselective

Warm butanal with potassium permanganate, KMnO₄, in aqueous base, then add acid. You get butanoic acid. In the basic solution the product forms as its carboxylate ion, RCOO⁻; the acid workup (the acid wash after the reaction) puts the H back on. The purple permanganate turns into a brown solid, manganese dioxide, MnO₂.

The H on the carbonyl carbon becomes an OH. The color change shows the permanganate being used up.

KMnO₄ is cheap, and it oxidizes the aldehyde completely. It also attacks much else. Hot, concentrated KMnO₄ cuts a C=C in two, as Alkenes & Alkynes showed. It also oxidizes a benzylic carbon, one attached directly to a benzene ring; Aromatic Follow-Through, later in the course, covers that.

Choosing the reagent

Go back to (E)-hex-4-enal. You want the acid and you want to keep the C=C. Pinnick conditions, or silver oxide, give exactly that. Hot KMnO₄ oxidizes the aldehyde too, but it also cuts the C=C, and two different acids come out.

One substrate, two outcomes. The aldehyde carbon ends as a COOH in both; only the C=C decides between them.

The table sets out the four standard reagents. All four oxidize the aldehyde. What separates them is the last column.

Reagent	Conditions	What else it attacks
Jones reagent	CrO₃ in aqueous H₂SO₄ and acetone	Primary and secondary alcohols; groups that acid destroys, such as acetals
KMnO₄	Warm, basic; then an acid workup	C=C bonds, benzylic carbons, alcohols
NaClO₂ (Pinnick)	NaH₂PO₄ buffer, tert-butanol and water, 2-methylbut-2-ene	Very little; C=C bonds and alcohols survive
Ag₂O	Aqueous base	Very little; C=C bonds, alcohols and acid-sensitive groups such as acetals survive. Silver is expensive.

So the question is never whether the aldehyde will oxidize. It is what else in the molecule you can afford to lose.

Tollens’ reagent and the silver mirror

Warm a few drops of butanal with Tollens’ reagent in a clean test tube, and the inside of the glass turns into a mirror. Tollens’ reagent is the diamminesilver(I) ion, Ag(NH₃)₂⁺, in aqueous ammonia, which is basic.

RCHO + 2 Ag(NH₃)₂⁺ + 3 HO⁻ → RCOO⁻ + 2 Ag + 4 NH₃ + 2 H₂O

The aldehyde is oxidized. Because the solution is basic, it ends as the carboxylate ion. Its two electrons go to two silver ions, one each, and each Ag⁺ becomes silver metal. The metal plates out on clean glass as a silver mirror.

The left panel is the same for every test in this section; only the metal on the right changes.

Tollens’ reagent is a weak oxidant. It leaves alcohols and C=C bonds alone. That makes it a good test: a mirror points to an aldehyde, or to something that can turn into one (the sugars below). It is not used to make acids, though. The solution must be made fresh, and if it is left standing it can deposit explosive silver compounds, so it is never stored or scaled up. Silver oxide in base, a solid that can be weighed out, is the version chemists use to make acids (table above).

Fehling’s and Benedict’s solutions

Both are deep blue solutions of copper(II), Cu²⁺, in base. On its own, Cu²⁺ would drop out of a basic solution as Cu(OH)₂. A chelating ligand, one that grips the metal through several atoms at once, keeps it dissolved. An aldehyde reduces the copper(II) to copper(I), which comes out as a brick-red solid, Cu₂O.

RCHO + 2 Cu²⁺ + 5 HO⁻ → RCOO⁻ + Cu₂O + 3 H₂O

The copper half of the reaction. The aldehyde half is the same as in the Tollens’ figure.

The two copper tests share a blind spot. An aliphatic aldehyde, one whose CHO is attached to an alkyl carbon, like butanal, gives the red solid. An aromatic aldehyde, one whose CHO sits on a benzene ring, like benzaldehyde, gives a negative or unreliable result. Tollens’ detects both.

Reagent	Cu(II) complexed by	Notes
Fehling’s	tartrate	Reliable for aliphatic aldehydes; unreliable with aromatic ones. Mixed fresh from two solutions.
Benedict’s	citrate	Keeps as a single solution; the version long used clinically. Same blind spot for aromatic aldehydes.

Why glucose tests positive and sucrose does not

Glucose gives a silver mirror and a brick-red Benedict’s test. Yet over 99% of glucose in water is a ring with no aldehyde group. The ring is a cyclic hemiacetal, a carbon that carries both an OH and an OR, as Acetals & hemiacetals showed.

The hemiacetal carbon is C1, the old aldehyde carbon. It is called the anomeric carbon. It is bonded to two oxygens: the ring oxygen and an OH. The ring can open at C1 and give the aldehyde back. In base, HO⁻ first takes the H from the C1 OH. That O⁻ then forms a C=O. The bond from C1 to the ring oxygen breaks, and its electrons go to the ring oxygen, which takes H⁺ from water.

Hydrogens and stereochemistry are left off. The two highlighted oxygens on the right were joined to C1 in the ring.

Well under 0.1% of the glucose is open at any moment. That is enough. The oxidant uses up the open form, and more ring opens to replace it, until all of the glucose has been oxidized. Sugars that do this are called reducing sugars, named for what they do to the metal ion.

Fructose tests positive too, which looks wrong at first. Fructose is a ketose, a sugar whose carbonyl is a ketone, at C2. Its open chain has no aldehyde. But the tests run in base, and base moves a hydrogen between C1 and C2. The step in between is an enediol: a C=C with an OH on each of its two carbons. From the enediol, a second hydrogen move gives an aldose, a sugar with an aldehyde at C1 (glucose, or its C2 partner mannose). The aldose then reduces the metal. Enolate Chemistry, later in the course, treats these hydrogen moves in full.

Read left to right; base runs each step both ways. R is C3 to C6. The highlights mark where a hydrogen leaves (first panel) and where one arrives (second and third).

So what decides a positive test is a free anomeric carbon: one that still carries its OH, so that its ring can open. Biomolecules, later in the course, uses this as the definition of a reducing sugar.

Two sugars join when the anomeric OH of one reacts with an OH of the other to give an acetal. That C–O–C link is a glycosidic bond; Acetals & hemiacetals called the product a glycoside. In maltose, C1 of one glucose is bonded through an oxygen to C4 of a second glucose. The second glucose keeps its anomeric OH, so its ring can open, and maltose tests positive. Lactose, galactose joined to C4 of glucose, is built the same way and also tests positive.

In sucrose, the glycosidic bond joins C1 of glucose to C2 of fructose, the anomeric carbons of both. Neither ring has an anomeric OH left, so neither can open, and sucrose tests negative.

The dots mark the anomeric carbons. Other OH and CH₂OH groups, hydrogens and stereochemistry are left off.

Worked example — reading a pair of test results

An unknown gives a silver mirror with Tollens’ and no brick-red solid with Fehling’s. What can you conclude, and what would you do next?

Step 1 — what a positive Tollens’ rules in. An aldehyde, or something that turns into one in base: a sugar with a free anomeric carbon, fructose included. A plain ketone gives no mirror.

Step 2 — what a negative Fehling’s would normally mean. On its own, no aldehyde. That contradicts step 1, so one of the two tests must have a blind spot. Fehling’s has one: it is unreliable with aromatic aldehydes.

Step 3 — reconcile. The pair is not a contradiction; it is a diagnosis. An aromatic aldehyde is positive to Tollens’ and negative to Fehling’s. Benzaldehyde is the standard example.

Step 4 — confirm. Do not settle it with Benedict’s, which shares the blind spot. Oxidize the compound with Ag₂O and identify the acid, benzoic acid if the reading is right, by its melting point. The trap here is treating a negative result from a reagent with a known blind spot as proof of absence.

The consequence on the shelf

An old bottle of benzaldehyde often has white crystals around the cap. They are benzoic acid, made by the air. Such contamination is a common cause of unexplained failed reactions. So aldehydes are stored under nitrogen or argon, and are often made just before use.

Air oxidizes the aldehyde by a radical chain, a repeating cycle of one-electron steps like the halogenation chains in How Reactions Happen. The chain starts at the aldehyde C–H. That bond is weak: about 89 kcal/mol, against 96 to 101 kcal/mol for the C–H bonds of an alkane chain. Removing its H leaves an acyl radical, R–C(•)=O, which its own C=O helps to stabilize.

Initiation. Light, a trace of metal, or a stray radical (X•) takes the aldehyde H, leaving the acyl radical.
Oxygen adds. O₂ is a diradical: it has two unpaired electrons. One of them pairs with the acyl radical's electron, giving an acylperoxy radical, RC(=O)OO•.
The chain carries on. The acylperoxy radical takes the aldehyde H from a second molecule. That gives a peroxy acid, RC(=O)OOH, and a fresh acyl radical, which goes back to the step before.
Finish. The peroxy acid oxidizes one more aldehyde, with no radicals involved. Two molecules of carboxylic acid come out.

Each fishhook moves one electron. The third panel makes a new acyl radical, which goes back to the second panel: that loop is the chain. Lone pairs are left off.

A ketone has no C–H on its carbonyl carbon, so this chain has nowhere to start. A ketone keeps on the shelf.

A ketone can be oxidized, but only by breaking a C–C bond, and both routes this course covers do exactly that. The Baeyer–Villiger oxidation (Carboxylic Acids & Derivatives, a later chapter) puts an oxygen into one of those bonds to give an ester. Hot, concentrated KMnO₄ cuts straight through. Neither happens by accident in a bottle.

What carries forward

Aldehyde to carboxylic acid is a one-step move in any synthesis, with the reagent chosen from the table above. Read the other way, the same ease is a warning: a route that makes an aldehyde and leaves it standing has a problem, because air will finish the job.