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Say you have a carboxylic acid and you need the amine with one carbon fewer. The usual textbook route turns the acid into an amide and then runs the Hofmann bromamide reaction. The Schmidt reaction skips the amide. The acid, hydrazoic acid (HN3) and a strong acid go into one flask, and the amine comes out.
What does the Schmidt reaction do to a carboxylic acid?
The overall change is RCOOH + HN3 → RNH2 + CO2 + N2.
The Schmidt reaction is the reaction of hydrazoic acid with a carbonyl compound in the presence of a strong mineral acid. Hans Wolff's review of it in Organic Reactions describes it as "a convenient method for the preparation of amines from acids" (Wolff, Organic Reactions). The same review notes that other carbonyl compounds react differently: ketones give amides, and aldehydes give nitriles and formyl derivatives of amines.
The reaction is named after Karl Friedrich Schmidt. Wolff's review dates his first report to 1923, and his short paper "Über den Imin-Rest" appeared in Berichte der deutschen chemischen Gesellschaft in 1924 (Schmidt, 1924).
Count the atoms. Acetic acid has two carbons and methylamine has one. The carbon you lose is the carboxyl carbon, which leaves as CO2. The new nitrogen comes from HN3, and the other two nitrogen atoms leave as N2 gas.
What is the mechanism of the Schmidt reaction?
The textbook mechanism for carboxylic acids, as set out in Wikipedia's article on the Schmidt reaction, has four stages:
The acid is protonated and loses water, giving an acylium ion, R–C≡O+.
Hydrazoic acid adds to the acylium ion, giving a protonated acyl azide.
The R group moves from carbon to nitrogen as N2 leaves, giving a protonated isocyanate, R–N=C=O.
Water adds to the isocyanate. The resulting carbamic acid loses CO2, leaving the amine.
The same source notes that this acylium-ion route becomes more important in concentrated acid (above 90% sulfuric acid). Otherwise the acyl azide forms from the protonated acid, as in Fischer esterification. Because the mixture is strongly acidic, the amine is held as its ammonium salt until the mixture is made basic at the end.
How is the Schmidt reaction different from Hofmann?
The Hofmann rearrangement turns a primary amide into a primary amine with one less carbon, using bromine and sodium hydroxide, and it also goes through an isocyanate (Hofmann rearrangement, Wikipedia). Our explainer on Hofmann bromamide degradation walks through that route. The Curtius rearrangement from an acyl azide starts one step later, from an acyl azide you have already made.
Here is how Schmidt and Hofmann compare:
Starting compound: Schmidt starts from the carboxylic acid. Hofmann starts from the primary amide.
Reagents: Schmidt uses HN3 with H2SO4. Hofmann uses Br2 with NaOH.
Medium: Schmidt runs in strong acid. Hofmann runs in base.
Steps from the acid: Schmidt takes one. Hofmann needs the amide first, so it takes at least two (acid to amide, then amide to amine).
By-products: Schmidt gives off N2 and CO2. In Hofmann, the bromine ends up as bromide salt and the carbon as carbonate.
Why is hydrazoic acid so dangerous?
Hydrazoic acid boils at 37 °C, is volatile and highly toxic (comparable to hydrogen cyanide), and is dangerously explosive when undiluted (Hydrazoic acid, Wikipedia). The same source notes that it is usually made by acidifying an azide salt such as sodium azide. PubChem's hazard summary describes it as "extremely explosive" (PubChem: hydrazoic acid).
For you, the Schmidt reaction is a reaction to know on paper. Never try it outside a properly equipped research lab with trained supervision.
How do you use the Schmidt reaction in a conversion question?
Worked example: convert benzoic acid to aniline in one step.
Count carbons. Benzoic acid, C6H5COOH, has 7. Aniline, C6H5NH2, has 6. You are losing one carbon, and NH2 replaces COOH, so this is a step-down reaction.
Check the starting compound. It is an acid, not an amide, so Schmidt works in one step. Hofmann would need the amide first.
Write the equation: C6H5COOH + HN3 → C6H5NH2 + CO2 + N2.
Balance-check it. Each side has 7 C, 7 H, 3 N and 2 O.
Name the reagents correctly: HN3 with concentrated H2SO4, then basify to free the amine.
For the multi-step route from acetic acid, see our post on step-down conversion from acetic acid to methylamine.
If you mix up which step-down reaction starts from which compound, practise with the Fermi Problem Library. It has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. To see where these rearrangements sit beside the other ways of making amines, the Fermi Concept Library gives each concept its own lesson and its place in the map.
Next time a question asks for "one carbon less", count the carbons first, then look at what you are starting from.
