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An aldohexose is a six-carbon aldose: its open-chain reference form has an aldehyde group at C1. Glucose is an example.^[ref]

The unmodified open-chain skeleton has four stereocenters, C2–C5, giving $2^4=16$ stereoisomers, or eight enantiomer pairs. Fixing the C5 hydroxy group on the right leaves $2^3=8$ D forms; D-glucose's right/left/right/right pattern selects one. An aldohexose label therefore specifies less than a particular sugar name, and a molecular formula specifies less still. The existing glucose article checks its chemical atom inventory; monosaccharide ring-chain interconversion follows those same numbered carbon positions into a ring.

An aldopentose is a five-carbon aldose: its open-chain reference form has an aldehyde group at C1. Ribose is an example; deoxyribose is its 2-deoxy derivative.^[ref]

For the unmodified open-chain aldopentose skeleton, count one $\mathrm{CHO}$, three $\mathrm{CHOH}$ and one $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ group: ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5$. The three interior carbon positions, C2–C4, are stereocenters. Each allows two configurations, giving $2^3=8$ stereoisomers, grouped into four enantiomer pairs. The unlike chain ends prevent a meso compound identification. D-ribose selects the right/right/right hydroxy group pattern in the Fischer projection; the word aldopentose alone does not select that pattern. Compare the existing ribose and deoxyribose diagrams for the separate question of which oxygen is present.

An amino acid combines carboxylic acid and amine functionality in the same molecule; the name alone does not specify their separation. An alpha-amino acid has its amino nitrogen attached to the carbon immediately adjacent to the carboxyl carbon, the amino acid alpha carbon. For example, ${\mathrm{H}}_2\mathrm{N}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COOH}$ is glycine, whereas ${\mathrm{H}}_2\mathrm{N}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{COOH}$ has an extra intervening carbon and is not an alpha-amino acid. These chemical formulas describe connectivity, not the predominant protonation state in water.

Most common proteinogenic amino acids share the skeleton ${\mathrm{H}}_2\mathrm{N}-\mathrm{CH}(R)-\mathrm{COOH}$, with $R$ the amino acid side chain. In glycine, $R$ is hydrogen; in alanine, $R=\mathrm{C}{\mathrm{H}}_3$. Proline is a useful exception to the displayed primary-amine skeleton: its amino nitrogen is part of a ring and has a second carbon substituent. The shared backbone permits assembly into polypeptides, while different amino acid side chains give the resulting proteins diverse chemistry. ^[ref]

In a chemical solution, protonation changes the electric charges without changing this backbone. Free glycine near neutral pH is predominantly ${\mathrm{H}}_3{\mathrm{N}}^{+}-\mathrm{C}{\mathrm{H}}_2-\mathrm{CO}{\mathrm{O}}^{-}$, a zwitterion. Its positive and negative groups are joined within one molecule; zero net electric charge does not mean absence of local electric charges, nor does it mean the uncharged drawing ${\mathrm{H}}_2\mathrm{N}-\mathrm{C}{\mathrm{H}}_2-\mathrm{COOH}$ predominates. Glycine protonation explains how lowering or raising pH changes the balance. ^[ref]

Building a protein joins the carboxyl carbon of one amino acid to the amino nitrogen of another through a peptide bond, an amide linkage. The formal net condensation loses the elements of water; cellular translation uses activated substrates at a ribosome, rather than simply mixing free amino acids in water. Incorporated units are amino acid residues. The interior amide groups no longer have the same acid/base behavior as the two free terminal groups of glycine, so its free-molecule glycine titration cannot be copied independently at every chain position. ^[ref]

Chirality distinguishes most common proteinogenic amino acids from glycine. The other nineteen members of the conventional twenty have a stereocenter at their amino acid alpha carbon; ordinary translation incorporates their L forms. Glycine's two identical hydrogen substituents make it achiral. Thus being an amino acid, or even an alpha-amino acid, does not by itself imply chirality. ^[ref]

The anomeric carbon is the former carbonyl group carbon that becomes a stereocenter on ring closure: C1 of glucose. It is an existing chemical atom, not an added one.

An allene has a $\mathrm{C}=\mathrm{C}=\mathrm{C}$ unit whose terminal substituents lie in perpendicular planes. In $\mathrm{abC}=\mathrm{C}=\mathrm{Ccd}$, with $a\ne b$ and $c\ne d$, the two terminal pairs can define enantiomers without any tetrahedral carbon stereocenter. The groups need be different within each end pair, not all four mutually different. This is axial chirality.^[ref]

Chirality means that a molecule cannot be superposed on its mirror image. A carbon attached to four different groups is a common source of chirality; the definition concerns the whole molecule, rather than requiring this particular structural feature.

The test is spatial: apply a translation and a rotation to the entire molecule until corresponding chemical atoms coincide, keeping their identities and chemical bonds. A mirror operation is a reflection; physically turning a model is a rotation. They are different operations. If no such superposition exists, the two forms are enantiomers; if one succeeds, the molecule is achiral. A different-looking flat drawing alone establishes neither conclusion.

For a simple molecule with exactly one tetrahedral stereocenter and four constitutionally different substituents, the mirror forms cannot superpose. Compare entire attached groups, including paths around a ring, rather than just the first chemical atom. Several tetrahedral stereocenters require a whole-molecule check: a meso compound can be achiral. Conversely, axial chirality needs no tetrahedral stereocenter.^[ref]

Why care about this distinction? Two enantiomers can fit differently into an already chiral molecule, just as two opposite hands fit differently into one glove. Thus a protein's spatial arrangement can distinguish a mirror pair even when an achiral measurement cannot. Polarimetry provides a different kind of test: it measures a sample's net optical rotation, which depends on the populations of its enantiomers.

In the usual DNA context, deoxyribose means 2-deoxy-D-ribose, a five-carbon monosaccharide with molecular formula ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_4$. Relative to D-ribose, the hydroxy group at C2 is replaced by hydrogen. The locant 2 matters: it identifies which position loses its oxygen. See Favre and Powell, IUPAC Blue Book, P-102.5.3: deoxy sugars and IUPAC: nucleic acids and their component sugars.

The paired Fischer projections in ribose make the change concrete. C2 has hydrogen on both sides, while C3 and C4 retain their hydroxy groups on the right. C2 is therefore not a stereocenter: exchanging its two identical hydrogen substituents cannot produce a distinct configuration. The unchanged C3 and C4 still carry stereochemical information.

Replacing $\mathrm{OH}$ by $\mathrm{H}$ removes one oxygen but leaves the total hydrogen count unchanged: the hydrogen belonging to the removed hydroxy group is replaced by a hydrogen bonded directly to carbon. Thus ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_5$ becomes ${\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_4$. This comparison describes structures, rather than a complete chemical reaction or its mechanism.

In DNA, deoxyribose is incorporated as a ring in each nucleotide, with phosphate connections building the backbone. Its C2′ bears hydrogen where RNA's ribose bears a hydroxy group. This small local difference lets the sugar structure identify the two kinds of nucleic acid; the nucleobases supply their sequence information. See Alberts et al.: nucleotides as the subunits of DNA and RNA.

Epimerization interconverts diastereomers differing at one stereocenter. It can change a product ratio after the initial bond-forming chemical reaction.

In a Fischer projection, each cross represents a carbon position: horizontal chemical bonds point toward the viewer and vertical chemical bonds point away. For the open-chain aldehyde sugars illustrated here, place $\mathrm{CHO}$ at the top and ${\mathrm{CH}}_2\mathrm{OH}$ at the bottom. This is a convention for stereochemistry, not a measured molecular shape. See Moss, IUPAC Recommendations 1996: Fischer projection.

In these Fischer projections, D specifies that the hydroxy group at the lowest stereocenter lies on the right: C4 for ribose and deoxyribose, C5 for glucose. D does not specify every other stereocenter, and it does not specify the direction of rotation of polarized light. See McMurry: the meaning of D and L.

Glucose is a monosaccharide with molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. Its familiar biological form, D-glucose, is an aldohexose. In the open-chain Fischer projection, $\mathrm{CHO}$ is at the top and ${\mathrm{CH}}_2\mathrm{OH}$ at the bottom; the hydroxy groups at C2, C3, C4, and C5 point right, left, right, right. This full pattern distinguishes D-glucose from other six-carbon aldehyde sugars. See McMurry, Organic Chemistry, 25.4: configurations of the aldoses.

The molecular formula can be checked directly: one $\mathrm{CHO}$ group, four $\mathrm{CHOH}$ groups, and one ${\mathrm{CH}}_2\mathrm{OH}$ group give six carbon, twelve hydrogen, and six oxygen chemical atoms. All four interior carbon positions are stereocenters in this open-chain structure. Neither their orientations nor the arrangement of chemical bonds can be recovered from the molecular formula alone.

In water, glucose exists mainly in six-membered rings in chemical equilibrium with a small amount of open-chain glucose. Ring closure connects the hydroxy group at C5 to the aldehyde carbon at C1; the ring contains five carbon chemical atoms and one oxygen, with C6 outside it. No chemical atoms are lost, so the molecular formula remains ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. See McMurry, Organic Chemistry, 25.5: cyclic structures of monosaccharides.

Glucose is useful both as a substrate in cellular metabolism and as a building block of polysaccharides. Cellulose chains assembled from glucose units help support plant cell walls. The same small sugar thus connects chemical fuel to a structural material; its role depends on how it is transformed or linked. See Cooper, The Cell: A Molecular Approach, the molecular composition of cells.

A meso compound is an achiral molecule despite having stereocenters. Tartaric acid, $\mathrm{HOOC}-\mathrm{CH}(\mathrm{OH})-\mathrm{CH}(\mathrm{OH})-\mathrm{COOH}$, has two tetrahedral carbon stereocenters. Its $(2R,3R)$ and $(2S,3S)$ forms are enantiomers, but $(2R,3S)$ and $(2S,3R)$ name the same meso form after equivalent ends are identified. A symmetric spatial arrangement has a mirror plane exchanging its two halves. Thus there are three stereoisomers, not four.^[ref]

Opposite R and S configurations alone do not guarantee a meso compound: the rest of the molecule must permit mirror superposition. Likewise, a meso compound is one achiral chemical compound, whereas a racemate is a mixture of two chiral molecule populations. Both can give zero optical rotation for different reasons.

A monosaccharide is a single-unit carbohydrate that cannot be split into smaller sugars by hydrolysis. Ribose, deoxyribose, and glucose are examples. The unit is a whole molecule containing many chemical atoms; a disaccharide instead contains two linked monosaccharide units.^[ref]

Classification combines carbon count with open-chain connectivity. An aldose has an aldehyde group; a ketose has a ketone group. Thus glucose is an aldohexose, while fructose is a six-carbon ketose. These names describe the open-chain reference forms even when cyclic forms predominate in water.^[ref]

Check the distinction by counting the groups in the drawings. Open-chain glucose contains one $\mathrm{CHO}$, four $\mathrm{CHOH}$ and one $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ group. Open-chain fructose contains two $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$, three $\mathrm{CHOH}$ and one $\mathrm{C}=\mathrm{O}$ group. The respective hydrogen totals are $1+4(2)+3=12$ and $2(3)+3(2)=12$; each structure also has six carbon and six oxygen chemical atoms. Their shared molecular formula ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$ therefore cannot identify the carbonyl group position or the stereochemistry.

The beauty is that a familiar sugar name can embrace an interconverting family of forms. Monosaccharide ring-chain interconversion connects those drawings to mutarotation, a change visible with polarimetry. Spatial arrangement also matters to biology: an enzyme's active site encounters a particular arrangement of hydroxy groups, rather than merely an inventory of chemical atoms. The existing enzyme and chirality articles explain that recognition principle.

In water, D-glucose mainly occupies cyclic forms. The C5 hydroxy group adds to the C1 aldehyde, forming a cyclic hemiacetal; reversible opening restores the aldehyde. Reclosure can give either anomer. Their changing proportions cause mutarotation, the approach of optical rotation to an equilibrium value. This pathway involves changing chemical bonds, rather than simply rotating a hydroxy group about a single chemical bond.^[ref]

In this conventional Haworth projection, C5's $\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$ group points up. The alpha anomer's C1 hydroxy group points down, on the opposite face; the beta anomer's points up, on the same face. C2/C3/C4 retain down/up/down hydroxy groups. These are above/below relationships to a reference ring plane, not a claim that the ring is physically flat.^[ref]

The anomeric carbon is C1, the existing carbonyl group carbon that becomes a new stereocenter; no extra carbon chemical atom appears. The two anomers differ at that centre while retaining the other configurations.^[ref]

Trace the ring as C1–C2–C3–C4–C5–O5–C1, where O5 denotes the oxygen originally in the C5 hydroxy group. It contains five carbon chemical atoms and one oxygen; C6 remains attached outside the ring. The C1 carbonyl group oxygen becomes the C1 hydroxy group oxygen. This accounts for every oxygen: O5 in the ring and five hydroxy groups at C1, C2, C3, C4 and C6. Ring closure preserves ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$. A six-membered ring and a six-carbon sugar are thus different counts.

The first base equivalent removes O–H; the second removes the alpha C–H. The first equivalent of lithium diisopropylamide converts the carboxylic acid to its lithium carboxylate. A further equivalent removes the remaining hydrogen on the carbon between the two allyl groups, giving a carboxylic acid dianion:The second negative charge has enolate-like delocalization into the carboxyl group; lithium coordination and aggregation are not represented by this simple charge-bookkeeping structure. The carbon nucleophile attacks methyl iodide with C–I bond cleavage, forming the new C–Me bond. acid workup reprotonates the carboxylate, giving $F=\mathrm{H}{\mathrm{O}}_2\mathrm{C}-\mathrm{C}(\mathrm{Me})(\mathrm{L}{)}_2$. The two equivalents are therefore chemically necessary for the stated carbon alkylation rather than simply an arbitrary excess. A single equivalent is consumed predominantly by the much more acidic carboxylic proton. The product has formula ${\mathrm{C}}_9{\mathrm{H}}_{14}{\mathrm{O}}_2$ and still has two identical allyl groups, so this methylation does not create a stereocenter.

The malonate carbon receives two allyl groups. With $L=\mathrm{C}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, attached through its first carbon, the product issodium hydride removes a methylene proton from diethyl malonate; molecular hydrogen formation accompanies production of an enolate stabilized by resonance in chemistry. Its carbon attacks the bromomethylene carbon of allyl bromide in an SN2 reaction, expelling bromide. The monoalkylated malonate still has one hydrogen between the carbonyls. Excess base removes that hydrogen, and a second equivalent of allyl bromide produces a second carbon–carbon bond. The final malonate carbon has no remaining hydrogen and cannot undergo a third analogous deprotonation there.

The two alkenes are preserved because the substitution occurs at the primary carbon bearing bromine. Counting the introduced groups gives ${\mathrm{C}}_{13}{\mathrm{H}}_{20}{\mathrm{O}}_4$ for the diallylated diethyl ester. Its two identical allyl groups also mean that the central carbon is not a stereocenter. This stage is the alkylation portion of the malonic ester synthesis.

A and C are chiral; B is achiral. In A, a quaternary ammonium stereocenter occurs because nitrogen has four different substituents: phenyl, methyl, ethyl, and the sulfinylethyl chain. It has no nitrogen lone pair available for ordinary pyramidal inversion of an amine. The sulfoxide sulfur is also stereogenic: its pyramidal arrangement distinguishes oxygen, methyl, the nitrogen-containing chain, and the lone-pair direction. There are two independent stereocenters and no internal symmetry, giving four configurations in two pairs of enantiomers. This is chirality of the cation; a bromide counterion does not remove it.

B can adopt a trans-1,4-dimethylcyclohexane chair with both methyl groups equatorial. An inversion centre at the chair centre exchanges its opposite carbon atoms and methyl groups. Its mirror image is therefore superposable: trans substituents in a flat ring drawing do not by themselves prove chirality. ring inversion gives the less favourable diaxial conformer without changing the trans configuration.

In C, tetrahedral phosphorus has four different substituent directions: oxo, methoxy, ethoxy, and phenoxy. C has a phosphorus stereocenter. Delocalization in the P–O bonds does not make these four ligand environments identical. A tetrahedral drawing and its mirror cannot be superposed without exchanging two ligands.

Write the reacting terminus as $R-\mathrm{C}{\mathrm{H}}_2-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$, where $R$ includes the pre-existing stereocentres and the intact $\mathrm{OC}{\mathrm{H}}_2\mathrm{C}(=\mathrm{O})\mathrm{NM}{\mathrm{e}}_2$ substituent. ozone undergoes 1,3-dipolar cycloaddition across the terminal alkene to give a molozonide. In a resonance description of ozone as ${\mathrm{O}}^{-}-{\mathrm{O}}^{+}=\mathrm{O}$, the alkene pi pair bonds to one terminal oxygen while the other terminal oxygen forms the second C–O bond and the ozone pi bond shifts onto its central oxygen. Cleavage of this unstable primary ozonide produces a carbonyl compound and a carbonyl oxide, which recombine to form the secondary ozonide. Reductive workup breaks the peroxide bonds and gives the carbonyl fragments $R-\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ and formaldehyde.

Excess sodium borohydride also reduces the newly formed aldehyde: hydride attacks its carbonyl carbon, shifting the pi electrons onto oxygen, and the final acidic workup protonates the alkoxide. Thus the retained chain becomes $R-\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$. The terminal alkene carbon becomes formaldehyde and can be reduced to methanol; it is no longer part of the main product. This accounts for the one-carbon shortening of the retained skeleton.

The original amide carbonyl is present in both H and I and is not reduced under the ordinary borohydride workup. The existing stereocenters and benzyl ether are retained. Stopping the mechanism at an aldehyde, or deleting the amide C=O when reading the smaller printed scheme, would give the wrong product. The new terminal alcohol carbon is CH2 and therefore adds no stereocenter.

The initial nucleophile attacks silicon, not the nitrile or an alkene carbon. Each trimethylsilyl group has an electrophilic Si center. A methyl group from methyllithium attacks Si, passing through a pentacoordinate silicon arrangement; Si–O cleavage releases $\mathrm{SiM}{\mathrm{e}}_4$ and the lithium salt of the oxygen-containing fragment. Cleaving the silyl enol ether generates a lithium enolate; cleaving the aryl silyl ether generates a phenoxide. Neither operation needs a new deprotonation at a carbon stereocenter.Here $R-O^{-}$ represents either the phenoxide or the oxygen-centered enolate contributor. Excess reagent permits both cleavages. The pre-existing methyl/nitrile stereocenter retains its configuration: it has no hydrogen to remove, and neither bond attached to that center is broken.

After HMPA is added, each anionic oxygen attacks the carbonyl carbon of allyl chloroformate. The C=O pi electrons move to the carbonyl oxygen, giving a tetrahedral intermediate; collapse expels chloride. The enolate becomes an allyl enol carbonate, and the phenoxide becomes an aryl allyl carbonate, each containing $\mathrm{O}-\mathrm{C}(\mathrm{O})-\mathrm{OC}{\mathrm{H}}_2\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$. oxygen acylation accounts for preservation of the enol double bond and for the two newly installed carbonate groups. carbon attack would give a different carbon skeleton. The two oxygens are trapped because they were selectively unmasked by desilylation, rather than because every ether oxygen in the molecule is equally reactive.

Lithium diisopropylamide removes an alpha proton from the ethyl side of the ketone. The C–H electrons form the C=C part of a lithium enolate, while the C=O pi electrons move to oxygen. The phenyl side has no competing alpha C–H site. Diisopropylamine is the conjugate-acid byproduct.The carbon end of this enolate attacks the primary carbon bearing iodine in $\mathrm{IC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{OMe}$. The C–I bond breaks to iodide in an SN2 reaction, and carbonyl reformation givesThus a three-carbon methoxypropyl chain is attached at the ketone alpha carbon. The remote methyl ether survives. The hindered amide base is chosen to generate the enolate rather than to add to the carbonyl. No absolute stereochemical preference is specified for the new stereocenter under these achiral conditions.

Acid protonates one methoxy group of 2,2-Dimethoxypropane. methanol departure gives the oxocarbenium species $\mathrm{M}{\mathrm{e}}_2\mathrm{C}(\mathrm{OMe}{)}^{+}$, which is captured by one diol oxygen. After proton transfer, protonation and loss of the remaining methoxy group generate a second oxocarbenium center. Intramolecular attack by the other diol oxygen closes the five-membered ring; deprotonation regenerates acid. The net transacetalization exchanges two methoxy groups for the two diol oxygens, releasing two methanol molecules.

The two starred carbon centers retain their original configurations, and the two carbon side chains on the acetonide are trans. No C–O bond at either original stereocenter is broken; reaction takes place at the reagent-derived CMe${}_2$ acetal carbon. The apparent change in wedge/dash presentation comes from redrawing the backbone into the five-membered ring, not inversion at a starred center. In the planar pentagon used in the diagram, draw the alkynyl-bearing side chain behind and the $\mathrm{C}{\mathrm{H}}_2\mathrm{OPMB}$ side chain in front. The unstarred methyl-bearing center also retains its configuration. The acetal carbon itself is not stereogenic because its two methyl substituents are identical.