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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.

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.

Diastereomers are different stereoisomers that are not enantiomers. The meso compound form of tartaric acid is a diastereomer of either member of its enantiomer pair.

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.

Tartaric acid has two hydroxy groups and two carboxylic acid groups. Its meso compound form illustrates why two stereocenters need not make a chiral molecule.