Chloroform and iodoform

Open a bottle of chloroform in a school lab and it smells sweet, almost pleasant — nothing like the poison it can quietly become. Left in a clear bottle, in sunlight, with air trapped above the liquid, $CHCl_3$ slowly turns into phosgene, $COCl_2$ , the same gas once used as a chemical weapon. That single fact — an everyday solvent decaying into a war gas — is why chloroform bottles are amber, filled to the brim, and dosed with a splash of ethanol. Chloroform and its close cousin iodoform ( $CHI_3$ ) both belong to a family called trihalomethanes, and both are made by the same underlying trick: take a compound with a $CH_3CO$ - or $CH_3CH(OH)$ - fragment and hit it with excess halogen and base. This article works through how each is prepared, why chloroform needs careful storage, and the structures of two other halogenated compounds the syllabus pairs with them — the refrigerant gas freon and the insecticide DDT.

The core insight

Chloroform and iodoform are both made by exhaustive halogenation of a methyl ketone or ethanol followed by hydrolysis with base — the haloform reaction — and chloroform's defining hazard is its slow aerial oxidation to toxic phosgene, guarded against by amber, brim-filled, ethanol-dosed storage bottles.

The core idea

Chloroform, preparation. Chloroform is made industrially or in the lab from ethanol or acetone by controlled chlorination followed by hydrolysis — the haloform reaction run with chlorine. A simple classroom route uses bleaching powder ( $CaOCl_2$ ), which first oxidises ethanol to acetaldehyde, then chlorinates it exhaustively, then hydrolyses the trichloro-compound to release chloroform:

CH_3CH_2OH \xrightarrow{CaOCl_2} CH_3CHO \xrightarrow{3\,Cl_2} CCl_3CHO \xrightarrow{Ca(OH)_2} CHCl_3

where: the first arrow is oxidation of the alcohol to an aldehyde, the second is exhaustive chlorination of the three hydrogens on the methyl carbon to give trichloroacetaldehyde (chloral), and the third is base-catalysed cleavage that splits off chloroform and leaves calcium formate, $(HCOO)_2Ca$ , behind. The same sequence starting from acetone gives the identical product, since acetone also carries the required $CH_3CO$ - group.

Chloroform, key property. Chloroform is a colourless, heavy, sweet-smelling liquid, but on standing in light and air it slowly oxidises to phosgene, a highly toxic and corrosive gas:

2\,CHCl_3 + O_2 \xrightarrow{\text{light}} 2\,COCl_2 + 2\,HCl

Bottles are therefore stored dark (amber glass, to cut the light that drives the reaction), completely filled (to starve the reaction of $O_2$ ), and dosed with about one percent ethanol, which mops up any phosgene that does form: $COCl_2 + 2\,C_2H_5OH \rightarrow (C_2H_5O)_2CO + 2\,HCl$ , converting it to harmless diethyl carbonate.

Iodoform, preparation. Iodoform is made by the iodoform reaction: treating ethanol, a methyl ketone, or any compound that oxidises to give a $CH_3CO$ - or $CH_3CH(OH)$ - fragment, with iodine and sodium hydroxide (which generate $NaOI$ in situ). With acetone:

CH_3COCH_3 + 3\,I_2 + 4\,NaOH \rightarrow CHI_3\!\downarrow + CH_3COONa + 3\,NaI + 3\,H_2O

Iodoform separates as a pale-yellow crystalline solid with a sharp, characteristic antiseptic smell, once used itself as a wound antiseptic.

Freons and DDT — structure, for context. Freons are chlorofluorocarbons: alkanes in which hydrogens are replaced by both $Cl$ and $F$ . Freon-12, $CCl_2F_2$ (dichlorodifluoromethane), is the classic example — a tetrahedral carbon bearing two chlorines and two fluorines — once widely used as a refrigerant because it is inert, non-toxic and non-flammable. DDT, $(4\text{-}ClC_6H_4)_2CH$ - $CCl_3$ , has two benzene rings each chlorinated at the para position, joined through a carbon that also carries a $CCl_3$ group; it was a widely used organochlorine insecticide before its environmental persistence became clear.

Figure

Reaction scheme comparing chloroform and iodoform formation from a common methyl carbonyl fragment through excess halogen and base

Why it works

The haloform reaction works because a methyl group attached directly to a carbonyl (or a carbon that will be oxidised to one) has three acidic hydrogens sitting right next to an electron-withdrawing $C=O$ . Base or hypohalite deprotonates one at a time, and each carbanion is instantly halogenated; this repeats three times because each new halogen makes the remaining hydrogens on that same carbon even more acidic, so the reaction does not stop after one or two substitutions — it runs all the way to the trihalomethyl ketone, $CX_3CO$ -. That trihalomethyl group is a superb leaving group once attacked by hydroxide at the carbonyl carbon, so the carbon-carbon bond next to it cleaves cleanly, releasing $CHX_3$ (chloroform or iodoform) and leaving a carboxylate salt behind. This mechanism explains why the reaction is so selective. Only a $CH_3CO$ - ketone, or an alcohol that first oxidises to give one ( $CH_3CH(OH)$ -, which $NaOI$ or $CaOCl_2$ readily oxidises), has the three hydrogens on the right carbon to start the sequence. A primary alcohol like propan-1-ol oxidises instead to an aldehyde with no methyl group on the carbonyl carbon ( $CH_3CH_2CHO$ ), so there is nothing to exhaustively halogenate and no haloform is released — which is exactly why the iodoform test distinguishes ethanol and methyl ketones from other alcohols and aldehydes.

Worked example

Show, with equations, how iodoform forms from propan-2-ol, and explain why propan-1-ol gives a negative iodoform test.

Step 1. $NaOI$ , generated from $I_2$ and $NaOH$ , first oxidises the secondary alcohol propan-2-ol to acetone: $CH_3CH(OH)CH_3 + NaOI \rightarrow CH_3COCH_3 + NaI + H_2O$ .

Step 2. Acetone carries the required $CH_3CO$ - group, so it undergoes the full haloform sequence with excess $NaOI$ : $CH_3COCH_3 + 3\,I_2 + 4\,NaOH \rightarrow CHI_3\!\downarrow + CH_3COONa + 3\,NaI + 3\,H_2O$ , giving a yellow precipitate of iodoform — a positive test.

Step 3. Propan-1-ol, by contrast, oxidises under the same conditions to propanal, $CH_3CH_2CHO$ . The methyl group here is one carbon away from the carbonyl, not attached to it, so there are no acidic hydrogens on the carbonyl carbon itself for the halogenation-cleavage sequence to act on.

Step 4. No trihalomethyl fragment can therefore form, and no iodoform precipitates. The test is positive for propan-2-ol (via acetone) and negative for propan-1-ol, confirming that the iodoform test requires a methyl group directly on the carbinol or carbonyl carbon, not merely somewhere in the molecule.

Common traps

Treating the iodoform test as a test for "any alcohol" — it only fires for ethanol and secondary alcohols/ketones with a methyl group directly on the carbinol or carbonyl carbon.
Naming chlorine gas or HCl as chloroform's dangerous decomposition product — the correct answer is phosgene, $COCl_2$ , formed by slow aerial oxidation in light.
Forgetting any one of the three storage precautions — dark bottle, brim-full, ethanol added — each blocks a different part of the phosgene-forming pathway (light, oxygen, or residual phosgene itself).
Mixing up freons (chlorofluorocarbons, refrigerants) with DDT (an organochlorine insecticide) — both are chlorinated organics from the same syllabus clause but have unrelated structures and uses.

Ask the tutor: want a walkthrough of exactly why the haloform mechanism accelerates with each substitution, or a side-by-side sketch of the Freon-12 and DDT structures?

Why does the haloform reaction always stop exactly at three halogen substitutions, not two or four

Each hydrogen removed from the methyl carbon is replaced by a halogen that withdraws more electron density, making the remaining hydrogens on that carbon even more acidic and easier to remove. So the reaction accelerates as it proceeds — once the third halogen goes in there are no hydrogens left on that carbon to remove, and the trihalomethyl ketone formed is now primed for the separate base-hydrolysis step that snips off the haloform. There is no natural stopping point at one or two substitutions because those partially halogenated intermediates react faster, not slower, than the starting material.

Is iodoform still used as a medicine like it once was

Iodoform was historically used as a topical antiseptic and wound dressing because of its mild germicidal action, credited partly to the slow release of iodine. It has largely been replaced by more effective, less irritating modern antiseptics, but it remains a useful example in the syllabus because its preparation is a clean, visually distinctive test — the yellow precipitate with its sharp smell is unmistakable and is still used qualitatively to identify ethanol, methyl ketones and related secondary alcohols in the lab.

Do freons and DDT actually undergo the same haloform-type reactions as chloroform and iodoform

No — they are grouped with chloroform and iodoform only because the syllabus clause covers halogenated organics together, not because they share a reaction mechanism. Freons are simple substituted alkanes (like Freon-12, $CCl_2F_2$ ) valued for their chemical inertness as refrigerants, and DDT is an aromatic organochlorine insecticide with two chlorinated benzene rings. Neither has the $CH_3CO$ - or $CH_3CH(OH)$ - fragment the haloform reaction needs, so for exam purposes you only need their structures and general uses, not any preparation route linking them to chloroform or iodoform.

Chloroform and iodoform

The core idea

Why it works

Worked example

Common traps

Before this

Where this leads