Animated Solution for Chemistry - Organic Chemistry: Given The compound(s), which on reaction with HNO${}_3$ will give the product having degree of rotation, $[\alpha {]}_D=-52.7^{\circ }$ is (are)

When dealing with carbohydrates, understanding their chemical reactivity is just as important as knowing their structure. In this problem, we are presented with D-glucose and asked to analyze its reaction with concentrated nitric acid (${\text{HNO}}_3$).

Nitric acid is a powerful oxidizing agent. Unlike milder reagents like bromine water, which only oxidize the aldehyde group to a carboxylic acid (yielding gluconic acid), ${\text{HNO}}_3$ is strong enough to oxidize both the terminal aldehyde ($\text{-CHO}$) and the primary alcohol (${\text{-CH}}_2\text{OH}$) groups. The result is a dicarboxylic acid known as an aldaric acid. In the case of D-glucose, this product is specifically called D-glucaric acid, or saccharic acid.

The problem states that the resulting D-glucaric acid, labeled as compound P, has a specific rotation of $[\alpha {]}_D=+52.7^{\circ }$. Specific rotation is a physical property that measures how much a chiral molecule rotates plane-polarized light.

The core of the question asks us to identify which of the given options will react with ${\text{HNO}}_3$ to produce a compound with a specific rotation of $[\alpha {]}_D=-52.7^{\circ }$.

What does a specific rotation of exactly $-52.7^{\circ }$ imply? It means the target product must be the exact enantiomer of compound P. Enantiomers are non-superimposable mirror images of each other. They share identical physical properties (like melting point and boiling point) but rotate plane-polarized light by exactly the same magnitude in opposite directions.

To find our target molecule, we must first draw the enantiomer of P. In a Fischer projection, drawing an enantiomer is straightforward: you simply invert every single chiral center.

For compound P (D-glucaric acid), the hydroxyl ($\text{-OH}$) groups from top to bottom (C2 to C5) are positioned Right, Left, Right, Right. Therefore, in its enantiomer (L-glucaric acid), the $\text{-OH}$ groups must be positioned Left, Right, Left, Left.

Now, we systematically oxidize each option with ${\text{HNO}}_3$ and compare the resulting dicarboxylic acid to our target enantiomer.

Option (C): When we oxidize the terminal groups of option (C), the resulting structure has its $\text{-OH}$ groups positioned Left, Right, Left, Left. This is a perfect, direct match for the enantiomer of P. Thus, (C) is undoubtedly a correct answer.

Option (D): When we oxidize option (D), the resulting structure has its $\text{-OH}$ groups positioned Right, Right, Left, Right. At first glance, this does not match our target enantiomer. However, we must be incredibly careful with Fischer projections.

Fischer projections are 2D representations of 3D molecules. Because of the specific conventions used to draw them (horizontal lines are wedges coming out at you, vertical lines are dashes going away from you), there are strict rules about how you can manipulate them on paper.

One of the most critical rules is that rotating a Fischer projection by $180^{\circ }$ in the plane of the paper yields the exact same molecule.

Let's apply a $180^{\circ }$ rotation to the oxidized product of option (D). The top $\text{-COOH}$ becomes the bottom $\text{-COOH}$, and vice versa. More importantly, the chiral centers swap positions and apparent orientations. Carbon-2 moves to the Carbon-5 position, and its right-facing $\text{-OH}$ now faces left.

After performing this $180^{\circ }$ rotation on the entire molecule, the resulting projection has its $\text{-OH}$ groups positioned Left, Right, Left, Left. Astoundingly, this perfectly superimposes on the enantiomer of P!

Therefore, the product from option (D) is identical to the product from option (C). Both compounds yield the enantiomer of P upon oxidation, making both (C) and (D) the correct answers.