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

Imagine you are looking at a molecule from a very specific angle, where the horizontal bonds are reaching out to hug you, and the vertical bonds are pointing away into the distance. This is the essence of a Fischer projection!

In this problem, we are tasked with finding the absolute configuration—the exact 3D spatial arrangement—of a molecule with two chiral centers, $C_2$ and $C_3$.

To unlock this, we need our trusty Cahn-Ingold-Prelog (CIP) priority rules. Let's dive in!

The CIP rules are the universal language of stereochemistry. The core principle is simple: Atomic number is king.

When looking at the atoms directly attached to a chiral center, the one with the highest atomic number gets the highest priority (Priority 1).

If there is a tie—for example, if two attached atoms are both Carbons—we don't panic. We simply look at what those Carbons are attached to next, moving down the chain until the tie is broken.

Finally, we have the golden rule of Fischer projections: If the lowest priority group (usually Hydrogen) is sitting on a horizontal bond, the perspective is flipped.

This means whatever configuration we trace (Apparent $R$ or $S$), we must reverse it to get the actual configuration.

Let's focus our attention on the top chiral center, $C_2$.

The four groups attached to it are $-\text{OH}$, the entire $C_3$ group below it, $-\text{COOH}$ above it, and $-\text{H}$.

Oxygen has an atomic number of 8, which is the highest here, so $-\text{OH}$ proudly takes Priority 1.

Now for the tie-breaker between the $C_3$ carbon and the $-\text{COOH}$ carbon. The $C_3$ carbon is directly attached to a heavy Chlorine atom (atomic number 17). The $-\text{COOH}$ carbon is attached to Oxygen (atomic number 8).

Since 17 is greater than 8, the $C_3$ group wins Priority 2, leaving $-\text{COOH}$ with Priority 3. Hydrogen, as always, is Priority 4.

Now, we trace the path from Priority 1 to 2 to 3. The path goes from right, to down, to up. This is a clockwise direction, which gives us an apparent $R$ configuration.

But wait! Our lowest priority Hydrogen is on a horizontal bond. Applying our golden rule, we reverse the result. The apparent $R$ becomes an actual $S$.

Therefore, the absolute configuration at $C_2$ is $2S$.

Moving down to the second chiral center, $C_3$.

The groups attached are $-\text{Cl}$, the $C_2$ group above it, $-{\text{CH}}_3$ below it, and $-\text{H}$.

Chlorine is the heavyweight champion here with an atomic number of 17, so it easily takes Priority 1.

Next, we compare the $C_2$ group and the $-{\text{CH}}_3$ group. The $C_2$ carbon is attached to an Oxygen atom, while the $-{\text{CH}}_3$ carbon is only attached to tiny Hydrogens.

Oxygen beats Hydrogen, so the $C_2$ group gets Priority 2, and $-{\text{CH}}_3$ gets Priority 3. Hydrogen is Priority 4.

Tracing the path from 1 to 2 to 3, we move from right, to up, to down. This is a counter-clockwise direction, giving us an apparent $S$ configuration.

Once again, our Hydrogen is on a horizontal bond. We must reverse the result. The apparent $S$ becomes an actual $R$.

Therefore, the absolute configuration at $C_3$ is $3R$.

We have successfully decoded both chiral centers!

Combining our results, the absolute configuration of the entire molecule is $(2S,3R)$.

This perfectly matches option (a). Stereochemistry might look intimidating at first, but once you master the CIP rules and the horizontal bond trick, it becomes a highly rewarding puzzle to solve!