Experiment: preparation of iodoform

How is iodoform prepared?

Where this experiment comes from. NCERT Chemistry Part II explains the haloform reaction on page 240. Its Class 12 Laboratory Manual, Unit 8, pages 88–89, describes the iodoform test for alcohols; this is distinct from isolating a preparative yield. The historical method below is from Ira Remsen’s public-domain An Introduction to the Study of the Compounds of Carbon; or, Organic Chemistry (1903), page 28. Check the official syllabus for your examination year for the required practical coverage.

In plain words: iodine and alkali make hypoiodite, which oxidises ethanol to ethanal (acetone needs no oxidation) and then swaps the three hydrogens of the CH₃ group next to C=O for iodine; the CI₃ group then breaks off as iodoform, which is sparingly soluble in water.

Haloform reaction of the CH_₃CO– group

• Methyl ketones and ethanal, the aldehyde with a CH₃CO– group, are oxidised by sodium hypohalite to the sodium salt of an acid with one carbon atom fewer, and the methyl group becomes haloform; with sodium hypoiodite the product is iodoform (NCERT page 240). Balanced, NCERT’s scheme for X = I reads (reasoned):

• A CH₃CH(OH)– group is first oxidised to CH₃CO–, so ethanol and alcohols such as propan-2-ol give iodoform too (NCERT page 240).
• For acetone, the overall equation is (reasoned):

For ethanol, the preliminary oxidation consumes one additional mole of iodine:

Iodoform from acetone or ethanol with iodine and alkali

• Remsen’s method (1903), from ethanol. “Dissolve 20g crystallized sodium carbonate in 100g water. Pour 10g alcohol into the solution, and, after heating to 60° to 80°, gradually add 10g iodine. The iodoform separates from the solution.” (Remsen, page 28.) Here the sodium carbonate supplies the alkali, and the temperatures are in degrees Celsius. This is a historical description, not a validated modern teaching-lab protocol: “crystallized” carbonate must not be silently replaced gram-for-gram by anhydrous carbonate. Actual work requires the instructor’s approved scale, reagent specifications and procedure.
• General method, from acetone (no fixed quantities): acetone reacts in the same way when warmed with iodine and an alkali (NCERT page 240; Wikipedia, “Iodoform”). Add the iodine gradually to the warm alkaline mixture, collect the yellow solid that separates by filtration, and wash it with cold water; iodoform can be recrystallised from alcohol, in which it dissolves (reasoned from Remsen’s solubilities, page 28).

Yellow crystals of iodoform

• Remsen describes iodoform as a solid, soluble in alcohol and ether but insoluble in water, crystallising in delicate, six-sided, yellow plates, melting point 119° (page 28). Beyond the NCERT text, Wikipedia gives pale yellow crystals with a characteristic, sweetish odour and a melting point of 119 °C (Wikipedia, “Iodoform”).
• A melting range near the reference value, 119 °C (Remsen, page 28; Wikipedia, “Iodoform”), supports identity and purity, but does not prove either by itself; record the range and use other observations (reasoned).
• NCERT notes that iodoform was used earlier as an antiseptic, its action being due to the free iodine it liberates, and that it has been replaced by other iodine formulations because of its objectionable smell (NCERT Unit 6, page 187).

Safety

• Use the safety data for the actual reagents and concentrations: iodine vapour is harmful and irritating, and alkali can damage skin and eyes. Signal words depend on the specific product; they are not a substitute for its handling instructions.
• The NCERT Laboratory Manual marks sodium hydroxide as corrosive, and ethanol and acetone as flammable (Unit 8, page 89; Unit 10, page 107).
• Do this preparation only in a well-ventilated laboratory under a teacher’s supervision, with eye protection, and avoid breathing iodine vapour; warm the mixture in a water bath, never over a flame (reasoned from those hazard labels).

Solved examples

Common mistakes to avoid

Quick revision summary

• Iodoform, CHI₃, forms when a CH₃CO– compound (acetone, ethanal) or a CH₃CH(OH)– alcohol (ethanol, propan-2-ol) is warmed with iodine and an alkali: the haloform reaction (NCERT page 240).
• Remsen’s 1903 ethanol/carbonate method is a historical illustration; actual preparation needs an approved modern laboratory procedure.
• Yellow plates, reference melting point about 119 °C; sparingly soluble in water, soluble in alcohol and ether.
• NCERT gives the reaction chemistry and a small-scale identification test; test observations and preparative yield are different objectives.

Questions students ask

What is the difference between iodoform and iodine?

Iodoform is a compound, CHI₃, made from iodine, while iodine, I₂, is the element; beyond the NCERT text, iodoform forms yellow crystals melting at 119 °C, and iodine violet-black crystals melting at about 114 °C that give off a violet vapour (PubChem). Nearly all of iodoform’s mass is iodine, as the polyhalogen compounds page works out, and it is made from iodine by the haloform reaction, which turns the CH₃ of a CH₃CO– group into CHI₃ (NCERT page 240). Heated strongly, iodoform decomposes and gives off iodine again (PubChem).

What is the difference between iodoform and chloroform?

Iodoform, CHI₃, and chloroform, CHCl₃, are both haloforms, but iodoform is a yellow crystalline solid melting at 119 °C and chloroform a colourless liquid boiling at about 61 °C, 334 K, a point beyond the NCERT text (PubChem; Wikipedia, “Chloroform”). Both come from the haloform reaction: chloroform was first made around 1831 from bleaching powder and ethanol, a route industry has since dropped (Wikipedia, “Chloroform”). The iodine version serves as the test because iodoform precipitates as a yellow solid that clouds the mixture (Wikipedia, “Haloform reaction”), where chloroform would be an easily missed colourless liquid (reasoned).

Can iodoform be made with potassium iodide instead of iodine?

Iodoform can be made from potassium iodide if an oxidising agent turns the iodide into hypoiodite, as sodium hypochlorite does in the first method of the NCERT Laboratory Manual’s iodoform test (Unit 8, pages 88–89). The ideal stoichiometry incorporates every iodine atom into the product (this does not guarantee 100% experimental yield): with acetone, three KI and three NaOCl give one CHI₃, as shown by CH₃COCH₃ + 3KI + 3NaOCl → CHI₃ + CH₃COONa + 3KCl + 2NaOH (balanced). With iodine and alkali only half do, because the other three of acetone’s six iodine atoms leave as NaI, and with ethanol only three of every eight (reasoned from the balanced equations).

How can iodoform be prepared from ethyne?

Iodoform can be prepared from ethyne in two steps: water adds to ethyne in the presence of H₂SO₄ and HgSO₄ to give ethanal (NCERT page 232), and ethanal, which has a CH₃CO– group, gives iodoform when warmed with iodine and an alkali (NCERT page 240; reasoned). Any other terminal alkyne, RC≡CH with R an alkyl or aryl group, works the same way, because by Markovnikov addition it hydrates to a methyl ketone, RCOCH₃ (reasoned).

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Why does propan-2-ol give iodoform although it is not a ketone?

   Show answerHide answer

   The hypoiodite first oxidises its CH₃CH(OH)– group to CH₃CO–, giving acetone, which then undergoes the haloform reaction (NCERT page 240).

2. Q2. How would you convert isopropyl alcohol to iodoform? (NCERT Unit 6, Exercise 6.19 (xiv), page 191.)

   Show answerHide answer

   Warm it with iodine and sodium hydroxide: CH₃CH(OH)CH₃ + 4I₂ + 6NaOH → CHI₃ + CH₃COONa + 5NaI + 5H₂O (reasoned from NCERT page 240).