Skip to main content

Stereoisomers of tartaric acid; meso form

Short answer

Tartaric acid has three stereoisomers: (2R,3R), (2S,3S), and meso (2R,3S). The meso form exists because the two chiral carbons have the same four groups, and the (2R,3S) configuration creates an internal plane of symmetry, making it optically inactive by internal compensation.

Write all the possible stereoisomers of tartaric acid. Does tartaric acid has meso form? Explain with reason.

Tartaric acid HOOC–CH(OH)–CH(OH)–COOH has two chiral carbons with the same four groups, so it has three stereoisomers (not four).

   COOH         COOH         COOH
H—C—OH       HO—C—H       H—C—OH
HO—C—H       H—C—OH       H—C—OH
   COOH         COOH         COOH
 (2R,3R)      (2S,3S)      meso (2R,3S)
 L-(+)        D-(−)        inactive
  • (2R,3R) and (2S,3S) are enantiomers (optically active, equal and opposite rotation).
  • The meso form has a plane of symmetry (internal mirror) between C2 and C3, so it is optically inactive by internal compensation. It is a diastereomer of both active forms.
  • An equimolar mixture of the (+) and (−) forms is the racemic (±) acid.

Does tartaric acid have a meso form? Yes. Both chiral carbons carry the same four groups (H, OH, COOH, CH(OH)COOH). In the (2R,3S) form the configuration at C2 is the mirror of C3, so a plane of symmetry runs through the middle of the C2–C3 bond. The rotation of the upper half is cancelled by the lower half (internal compensation), so this form is optically inactive and not resolvable.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Engineering Chemistry chapter 9 questions →