Tartaric acid, HOOC-CH(OH)-CH(OH)-COOH, has two chiral carbons, leading to specific stereoisomers. The number of possible enantiomeric pairs is 1 due to the presence of a meso form. This compound is key in organic chemistry for understanding optical activity.

Compound Structure

HOOC-CH(OH)-CH(OH)-COOH features a four-carbon chain with carboxylic acids at both ends and hydroxyl groups on the middle carbons. Each -CH(OH)- carbon attaches to four different groups: H, OH, COOH, and CH(OH)COOH, confirming two chiral centers. For n chiral centers without symmetry, 2^n = 4 stereoisomers form, but meso compounds reduce optically active pairs.​

Total Stereoisomers

Two chiral centers predict 4 stereoisomers: (R,R), (S,S), (R,S), and (S,R). The (R,S) and (S,R) configurations are identical due to a plane of symmetry bisecting the C2-C3 bond, creating the meso-tartaric acid form. Meso-tartaric acid is achiral and optically inactive, leaving (R,R)-tartaric acid and (S,S)-tartaric acid as the enantiomeric pair.​

Enantiomeric Pairs Explained

An enantiomeric pair consists of two mirror-image enantiomers that are non-superimposable and rotate plane-polarized light oppositely. Here, only one such pair exists: the (R,R) and (S,S) forms. The meso form does not form an enantiomeric pair with anything, as it is its own mirror image and superimposable.​

Why Not More Pairs?

Options implying 2 or 3 pairs overlook the meso compound’s internal symmetry, which halves optically active stereoisomers. Without the identical end groups (COOH), all 4 would be distinct, yielding 2 pairs—but symmetry enforces 3 total stereoisomers (1 meso + 1 pair). Thus, the answer is exactly 1 enantiomeric pair.​