Animated Solution for Chemistry - Organic Chemistry: The absolute configuration of

When dealing with stereochemistry, the Fischer projection is a powerful tool, but it comes with its own set of strict rules. In this problem, we are given a molecule of tartaric acid and asked to determine the absolute configuration (R or S) of its two chiral centers, $C_2$ and $C_3$.

To do this, we must rely on the Cahn-Ingold-Prelog (CIP) priority rules. The fundamental principle is simple: atoms directly attached to the chiral center are ranked based on their atomic number. The higher the atomic number, the higher the priority.

Assigning Priorities

Let's break down the top chiral center, $C_2$. It is bonded to four groups: $-\text{OH}$, $-\text{COOH}$, $-\text{CH(OH)COOH}$ (the rest of the chain), and $-\text{H}$.

1. Priority 1: The oxygen atom in $-\text{OH}$ has the highest atomic number (8), so it takes the crown. 2. Priority 2: We have a tie between the carbon of $-\text{COOH}$ and the carbon of the lower chain. We must look at what these carbons are attached to. The $-\text{COOH}$ carbon is bonded to three oxygens (due to the double bond counting twice). The lower chain carbon is bonded to one oxygen, one carbon, and one hydrogen. Since $(O,O,O)$ beats $(O,C,H)$, the $-\text{COOH}$ group gets priority 2. 3. Priority 3: The lower chain $-\text{CH(OH)COOH}$ naturally takes priority 3. 4. Priority 4: Hydrogen, with an atomic number of 1, is always dead last.

The Horizontal Rule

Now, we trace the path from Priority 1 $\to$ 2 $\to$ 3. For $C_2$, this path goes from right to top to bottom, which is counter-clockwise. Normally, counter-clockwise means an S configuration.

However, there is a golden rule in Fischer projections: If the lowest priority group (Hydrogen) is on a horizontal bond, you must reverse the final result! Since our hydrogen is pointing towards us (horizontal), the $S$ configuration flips to R.

Applying the exact same logic to the bottom chiral center, $C_3$: - Priority 1: $-\text{OH}$ (left) - Priority 2: $-\text{COOH}$ (bottom) - Priority 3: Top chain - Priority 4: $-\text{H}$ (right)

Tracing 1 $\to$ 2 $\to$ 3 gives us a counter-clockwise path ($S$). But once again, hydrogen is on the horizontal axis. We flip the $S$ to an R.

Therefore, the absolute configuration of the molecule is (2R, 3R). This makes it an optically active enantiomer of tartaric acid, unlike its meso counterpart which would be $(2R,3S)$.