The structure of D-(+)-glucose is given. The structure of L(-)-glucose is:
(A) Image A
(B) Image B
(

Q: Why can't L-glucose just be D-glucose with only the bottom OH group flipped to the left?

Flipping only one chiral center creates a diastereomer (specifically an epimer), not an enantiomer. If you only flip C-5 of D-glucose, you get L-idose, which is a completely different sugar with different chemical and physical properties. The 'L' in L-glucose means it is the exact mirror image of D-glucose.

Q: Does the 'L' in L-glucose mean it is levorotatory (rotates light to the left)?

Not necessarily! The 'D' and 'L' designations refer to the relative spatial configuration compared to glyceraldehyde, not the optical rotation. While L-glucose happens to be levorotatory (indicated by the (-) sign), other L-sugars can be dextrorotatory (+). Optical rotation must be determined experimentally.

Q: What happens to the top $-CHO$ and bottom $-CH_2OH$ groups when we take the mirror image?

In a Fischer projection, the vertical bonds represent groups pointing away from you. Because the $-CHO$ and $-CH_2OH$ groups are achiral (they don't have four different groups attached), their internal symmetry means they look identical in the mirror. We just leave them as they are on the vertical axis.

Q: Are L-sugars found in nature, or are they just theoretical?

While D-sugars (like D-glucose and D-fructose) overwhelmingly dominate biological systems, L-sugars do exist in nature! For example, L-arabinose and L-fucose are important components of plant cell walls and bacterial glycoproteins.

Q: How can I quickly check if two Fischer projections are enantiomers during an exam?

Simply go down the carbon chain and check every horizontal pair. If **every single** left-right pair is perfectly inverted between the two structures, they are enantiomers. If even one pair is the same while others are inverted, they are diastereomers.

Question

The structure of D-(+)-glucose is

The structure of L(–)-glucose is

Sigma Percentile · Accepted Answer

### The Beauty of Stereochemistry

Stereochemistry is the study of how molecules are arranged in three-dimensional space. When we look at complex biomolecules like carbohydrates, their spatial arrangement isn't just a minor detail—it dictates their entire biological function. The human body, for instance, is perfectly tuned to metabolize D-glucose, while its mirror image, L-glucose, passes through our system largely unrecognized by our enzymes.

To represent these 3D molecules on a flat piece of paper, chemists use **Fischer projections**. In a Fischer projection, the vertical lines represent bonds going away from you (into the page), and the horizontal lines represent bonds coming towards you (out of the page).

### Understanding the D and L Nomenclature

When we classify a sugar as 'D' or 'L', we are looking at a very specific part of the molecule. We examine the **highest-numbered chiral center**—which is the chiral carbon furthest away from the carbonyl group (the aldehyde or ketone at the top).

In the case of glucose, an aldohexose, this is carbon-5 (C-5). If the hydroxyl ($-OH$) group on this specific carbon points to the **right**, the sugar belongs to the **D-series**. If it points to the **left**, it belongs to the **L-series**. The question provides us with D-(+)-glucose, and as expected, the $-OH$ group on C-5 is clearly on the right.

### The Enantiomeric Relationship

Here is the critical conceptual leap: The D and L forms of a specific sugar name (like glucose) are **enantiomers**.

Enantiomers are pairs of molecules that are **non-superimposable mirror images** of each other. Think of your left and right hands—they are mirror images, but no matter how you turn them, you cannot perfectly align them on top of one another.

Because L-glucose is the enantiomer of D-glucose, we cannot simply flip the $-OH$ group on C-5 and call it a day. Doing so would only change one chiral center, creating a *diastereomer* (specifically, L-idose), not an enantiomer. To create the true mirror image, we must invert **every single chiral center** in the molecule.

### Constructing the Mirror Image

Let's systematically build L-glucose by placing a virtual mirror next to D-glucose:

1. **The Achiral Backbone:** The aldehyde group ($-CHO$) at the top and the primary alcohol ($-CH_2OH$) at the bottom lie on the vertical axis. Because they do not possess chirality themselves, their internal symmetry means they remain unchanged in our mirror projection.

2. **Inverting C-2:** In D-glucose, the $-OH$ is on the right. In the mirror, it reflects to the **left**.

3. **Inverting C-3:** In D-glucose, the $-OH$ is on the left. In the mirror, it reflects to the **right**.

4. **Inverting C-4:** In D-glucose, the $-OH$ is on the right. In the mirror, it reflects to the **left**.

5. **Inverting C-5:** In D-glucose, the $-OH$ is on the right. In the mirror, it reflects to the **left**.

### The Final Match

By reading our newly constructed L-glucose from top to bottom (C-2 to C-5), the sequence of hydroxyl groups is **Left, Right, Left, Left**.

When we carefully scan the given options, we see that **Option (A)** perfectly matches this exact sequence. It is a beautiful demonstration of how visualizing a simple mirror plane can effortlessly solve what might initially look like a complex memorization problem.

Animated Solution for Chemistry - Organic Chemistry: The structure of D-(+)-glucose is The structure of L(–)-glucose is

Select Answer:

Visualized Solution

$D -Glucose Structure$

$Relationship between D and L Isomers$

$Drawing the Mirror Image$

$Inverting Chiral Centers$

$Finalizing L-Glucose$

The Sigma Insight: Biomolecules

The Beauty of Stereochemistry

Understanding the D and L Nomenclature

The Enantiomeric Relationship

Constructing the Mirror Image

The Final Match

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