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Formation of a new benzene ring by joining carbon fragments, often followed by elimination and tautomerization that establish aromaticity.

hydroxide attacks an ester carbonyl to form a tetrahedral intermediate. Collapse expels ethoxide, and acid–base transfer gives the carboxylate and ethanol. Repeating this nucleophilic acyl substitution at the other ester produces the malonate dianion; acid workup gives $\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{L}{)}_2-\mathrm{C}{\mathrm{O}}_2\mathrm{H}$, with $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ as defined above.

On heating, one carboxylic acid group is lost through a six-membered cyclic proton-transfer/bond-cleavage arrangement. A carbonyl oxygen of the retained group accepts the proton from the departing carboxyl OH as its C–C bond breaks and carbon dioxide forms; electron redistribution gives the enol-like form of the retained acid. proton transfer/tautomerization then producesThe net decarboxylation is ${\mathrm{C}}_9{\mathrm{H}}_{12}{\mathrm{O}}_4\to {\mathrm{C}}_8{\mathrm{H}}_{12}{\mathrm{O}}_2+\mathrm{C}{\mathrm{O}}_2$. The central carbon gains a hydrogen, both allyl substituents survive, and only one carboxyl group is removed. It would be incorrect to draw hydrolysis directly giving the final acid without accounting for this carbon loss. The favourable cyclic pathway is characteristic of malonic acid decarboxylation.

The first reagent supplies a vinyl nucleophile, $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-$, rather than an allyl group. Two successive conjugate additions install vinyl at C2 and a 2-formylethyl chain at C1. Number the original nitro-bearing alkene carbon C1. Vinylmagnesium bromide attacks the beta carbon C2 of the nitroalkene, shifting the original C1=C2 electrons into a magnesium nitronate. The new C2–vinyl bond remains intact during the second stage; its terminal double bond is not the electrophilic site in that stage.

The carbon end of the nitronate at C1 then attacks the terminal beta carbon of acrolein. Move the acrolein C=C electrons towards its carbonyl and the C=O electrons onto oxygen. This forms the C1–CH2CH2CHO connectivity through an aldehyde enolate; protonation by methanol and tautomerization restore the aldehyde. The nitronate returns to a neutral nitro group as it makes the second C–C bond. Thus the chain is attached through the carbon formerly beta to the acrolein carbonyl, rather than by a direct addition to its aldehyde carbon.

After the first addition, C2 bears the vinyl group while the nitronate carbon C1 remains approximately planar. Approach of the second electrophile to the less crowded face, opposite the adjacent vinyl substituent, gives the new formylethyl chain trans to vinyl. Consequently nitro and vinyl are cis, as depicted: at C1 the chain and nitro group occupy opposite faces. This is a substrate-controlled diastereoselective tandem Michael addition. The reagents and starting material contain no chiral source, so the depicted product represents a racemic pair rather than an enantioselective synthesis. Steric facial selection explains the relative configuration without requiring a uniquely established magnesium aggregate or transition-state conformation.

Both permitted alternatives follow the same condensation–activation–migration sequence. sodium acetate liberates a useful concentration of hydroxylamine from its hydrochloride while providing a buffered medium. nitrogen of NH2OH attacks the ketone carbonyl; proton transfer gives a carbinolamine. protonation of its carbon-derived OH and loss of water, with nitrogen lone-pair donation, then gives the oxime after deprotonation. The two C=N geometries can be separated.

For the B-to-D route, oxime O-tosylation begins when oxime oxygen attacks sulfur of tosyl chloride and chloride leaves; triethylamine accepts the released proton. The resulting O-tosyl oxime has an excellent leaving group. The $K-R_a$ bond, anti to N–OTs, migrates to nitrogen while N–O cleaves, giving a ring-expanded nitrilium ion. Its local connectivity is $R_s-K\equiv N^{+}-R_a$, with the existing scaffold still joining the two paths. water attacks the nitrilium carbon $K$. proton transfers and imidic-acid/amide tautomerization produce D, with nitrogen inserted on the aromatic side of the old carbonyl.

For the C-to-E route, the same O-tosylation activates the other oxime geometry. Now the anti bond is $K-R_s$; its migration gives $R_a-K\equiv N^{+}-R_s$, and aqueous trapping produces E, with nitrogen inserted on the spiro-junction side. Each Beckmann rearrangement expands the ketone's six-membered ring to a seven-membered lactam. The original carbonyl carbon remains the amide carbonyl carbon; nitrogen is the inserted atom. In this mechanism the ketone oxygen is lost during oxime condensation, the oxime oxygen leaves with the sulfonate group, and water supplies the new amide oxygen.

Suspend starch in water, warm it to make a paste, and heat with a dilute acid, such as dilute sulfuric acid, to cause hydrolysis of the glycosidic bonds. The chemical reaction passes through shorter carbohydrates; sufficient hydrolysis ultimately gives glucose. The usual repeat-unit accounting isThis idealized polymer formula omits end-group details. The acid catalyses cleavage and is not consumed in the displayed equation. Merely using amylase from malt should not be described as necessarily producing pure glucose; it can leave maltose and shorter polysaccharides.

When sulfuric acid is used, neutralization with calcium carbonate removes the acidity and much of the sulfate as calcium sulfate. Filter out the insoluble residue, then concentrate the chemical solution gently and crystallize the glucose; further recrystallization is required to distinguish a reasonably pure chemical substance from a crude syrup. Overheating and unnecessarily prolonged strong acid treatment degrade the sugar. Hydrolysis of sucrose is a less selective preparative answer because it gives fructose alongside glucose.

For the historical copper test, use freshly mixed Fehling's reagent: an aqueous copper(II) sulfate portion and an alkaline tartrate portion. The tartrate keeps copper(II) in coordination complexes instead of allowing immediate bulk precipitation of copper(II) hydroxide. Add a suitably prepared sample to a small test portion and warm in a hot water bath. A yellow-orange to brick-red precipitate of copper(I) oxide indicates reduction of the copper(II) reagent. Run a known glucose sample and a water blank alongside it. In a simple sample, a visible difference between these controls helps distinguish a true positive from a deteriorated reagent or contamination.^[ref]

The key redox accounting on the copper side isThe required electrons come from oxidation of the sugar. This half-equation deliberately omits the changing tartrate ligands and does not assign a unique organic product. The often printed claim that hot alkaline glucose simply becomes gluconate in this test is too simple: competing isomerization, oxidation and fragmentation make the organic chemistry more complicated.^[ref]

Fructose, lactose, maltose and various other reducing substances can also give a positive Fehling's test. In particular, fructose can form aldoses by tautomerization in alkaline solution; an initial ketone functional group does not make it reliably negative.^[ref] Fermentation by suitable yeast, with carbon dioxide formation, and optical rotation can supply further historical evidence, but neither is individually specific to glucose. These are anticipated diagnostic observations, not measurements supplied by the paper. In present laboratory practice an appropriately validated enzymatic glucose assay can establish glucose more selectively; the old reducing-sugar test is not by itself a diagnosis of diabetes.^[ref]