Carbohydrates - Janice smith Flashcards

Question

Answer 1

To convert a 3D wedge-dash structure to a Fischer projection, follow these steps: Identify the chiral center – The carbon attached to four different groups. Orient the molecule correctly: The most oxidized group (CHO, COOH, etc.) should be at the top. The longest carbon chain should be vertical. Rearrange the bonds: The horizontal bonds (side groups) must come forward (wedges). The vertical bonds must go behind (dashes). Flatten the structure into a cross-like representation: Place the CHO at the top. Place the CH₂OH at the bottom. Arrange the OH and H groups horizontally. Check for proper orientation – If necessary, rotate by 180° (not 90°!) to maintain correct stereochemistry.

Answer 2

In monosaccharide chemistry, Fischer projections are drawn with the most oxidized carbon (CHO or COOH) at the top to maintain a consistent structural convention. This allows for easy comparison between different sugars. It also helps to assign D- or L- configuration correctly.

Answer 3

Two Fischer projections are enantiomers if: They are mirror images of each other. All chiral centers have opposite configurations (e.g., R → S or S → R). The OH and H groups on each chiral center are swapped left ↔ right.

Answer 4

A 90° rotation swaps the stereochemistry, effectively converting the molecule into its enantiomer. This is incorrect because it inverts the configuration at the chiral center. Only 180° rotations preserve stereochemistry. Rule of thumb: 90° rotation → Wrong! (changes enantiomer) 180° rotation → Correct! (keeps the same molecule)

Answer 5

The D/L notation for sugars is determined by the chiral center farthest from the aldehyde/ketone group. If the OH group is on the right → D-sugar. If the OH group is on the left → L-sugar.

Answer 6

To correctly convert a wedge-dash structure into a Fischer projection, follow these steps: Identify the chiral center – The carbon bonded to four different groups. Position the molecule correctly: The most oxidized group (CHO, COOH, etc.) is placed at the top. The longest carbon chain is placed vertically. Convert the bonds: The horizontal bonds should come forward (wedges). The vertical bonds should go behind (dashes). Draw the cross structure: The top and bottom remain fixed (CHO at the top, CH₂OH at the bottom). The side groups (OH, H, Br, Cl, etc.) are arranged based on their 3D position. ✔ Tip: If you find it difficult, redraw the molecule as a tetrahedral structure first, then tilt it into a Fischer projection.

Answer 7

Fischer projections are widely used in organic and biological chemistry because they: ✅ Simplify complex molecules – They provide a 2D representation of chiral centers. ✅ Help compare stereochemistry – They make it easy to distinguish between enantiomers and diastereomers. ✅ Are essential for carbohydrate chemistry – Sugars (glucose, fructose, ribose) are often represented using Fischer projections. ✅ Aid in D/L classification – They determine whether a molecule belongs to the D- or L-family of biomolecules. ✅ Prevent misinterpretation – Unlike wedge-dash notation, Fischer projections always follow a fixed orientation, reducing confusion

Answer 8

Fischer projections are designed to show stereochemistry in a clear 2D way. The chiral center is positioned in the middle and treated like a cross. The horizontal bonds are drawn as wedges (coming forward) because they are closer to the viewer in tetrahedral notation. The vertical bonds are dashed (going behind) because they are farther from the viewer. ✔ This convention ensures that the molecule is always viewed from the same perspective, preventing misinterpretation.

Answer 9

A 90° rotation inverts the stereochemistry, converting the molecule into its enantiomer. This is because side groups swap positions, changing the (R) or (S) configuration of the chiral center. Correct rotations: 180° rotation → No change in stereochemistry. 90° rotation → Changes enantiomer (incorrect). ✔ Always rotate Fischer projections by 180° if needed, but never by 90°!

Answer 10

The number of stereoisomers for a molecule with n n chiral centers is calculated as: 2 n 2 n Since an aldohexose has 4 stereogenic centers, the number of stereoisomers is: 2 4 = 16 2 4 =16 These 16 stereoisomers form 8 pairs of enantiomers. Each pair consists of a D-isomer and its mirror image, the L-isomer.

Answer 11

Fischer projections simplify the representation of sugars with multiple chiral centers. The molecule is drawn with: A vertical backbone (carbon chain). Horizontal bonds projecting forward (wedges). Vertical bonds projecting behind (dashes). ✔ Key Rule: In Fischer projections of sugars, all horizontal bonds come forward as wedges.

Answer 12

The D/L system is determined by the stereochemistry of the highest-numbered chiral center (farthest from the carbonyl group). D-glucose: OH on the right → D-configuration. L-glucose: OH on the left → L-configuration. ✔ Most naturally occurring sugars are in the D-form.

Answer 13

3D Wedge-Dash Model Shows the actual spatial arrangement. Highlights tetrahedral geometry. Fischer Projection Converts the structure into a 2D cross format. Easier to compare stereochemistry in multiple sugars.

Answer 14

The D/L system is an older naming convention for sugars, separate from the R/S system. This system was originally based on glyceraldehyde, the simplest chiral sugar. (R)-glyceraldehyde → D-isomer (S)-glyceraldehyde → L-isomer

Answer 15

The configuration of the stereogenic center farthest from the carbonyl (C=O) group determines whether a sugar is D or L. If the OH group on this chiral carbon is on the right in a Fischer projection, the sugar is D. If the OH group on this chiral carbon is on the left, the sugar is L.

Answer 16

D-sugar (e.g., D-glyceraldehyde) OH group on the right at the last chiral center. L-sugar (e.g., L-glyceraldehyde) OH group on the left at the last chiral center.

Answer 17

Fischer projections align the carbon backbone vertically. The OH position at the last chiral center (C5 in glucose) tells us if it’s D or L. Right → D Left → L

Answer 18

Diastereomers that differ at only one chiral center. Example: D-Glucose vs. D-Mannose (C2 epimers). D-Glucose vs. D-Galactose (C4 epimers).

Answer 19

1. Sweet Taste All monosaccharides are sweet, but their sweetness varies. Relative sweetness comparison: Fructose → Sweetest. Glucose → Less sweet than fructose. Galactose → Least sweet among common monosaccharides. ✔ Fructose is used in sweeteners due to its high sweetness. 2. High Melting Points Monosaccharides are polar compounds. Due to strong hydrogen bonding, they have high melting points compared to nonpolar organic compounds. ✔ Stronger intermolecular forces = higher melting points. 3. Water Solubility Highly soluble in water due to multiple -OH (hydroxyl) groups. Hydrogen bonding between -OH groups and water molecules increases solubility. ✔ This is why glucose dissolves easily in blood. 4. Insolubility in Organic Solvents Monosaccharides are too polar to dissolve in nonpolar solvents like diethyl ether. Most organic compounds dissolve in organic solvents, but monosaccharides do not because of their hydrophilic nature. ✔ Monosaccharides dissolve in polar solvents, not nonpolar ones. Key Takeaways ✔ Monosaccharides are sweet, with fructose being the sweetest. ✔ They have high melting points due to strong hydrogen bonding. ✔ They are water-soluble but insoluble in nonpolar solvents like diethyl ether.

Answer 20

Pyranose Ring (6-membered): Formed when the hydroxyl (-OH) at C5 reacts with the carbonyl (C=O) at C1. More stable for aldohexoses like glucose. Example: D-Glucose forms D-Glucopyranose. Furanose Ring (5-membered): Formed when the hydroxyl (-OH) at C4 reacts with the carbonyl (C=O) at C1. More common in ketohexoses like fructose. Example: D-Fructose forms D-Fructofuranose. ✔ Glucose prefers a pyranose form, while fructose often forms a furanose form.

Answer 21

When the hydroxyl group reacts with the carbonyl carbon, a new chiral center forms at C1. This creates two possible configurations: α-Anomer: The OH at the anomeric carbon (C1) is drawn down. β-Anomer: The OH at the anomeric carbon (C1) is drawn up. ✔ α and β anomers are stereoisomers that differ at the anomeric carbon

Answer 22

Which OH group reacts? The OH at C5 is the right distance to react with the C1 carbonyl to form a six-membered ring (pyranose). This positions the oxygen from C5 in the upper right of the ring. ✔ This is why glucose forms a pyranose ring rather than a furanose ring. Key Takeaways ✔ Monosaccharides cyclize into pyranose (6-membered) or furanose (5-membered) rings. ✔ Glucose forms a pyranose ring via C5-OH attacking C1. ✔ Cyclization creates a new chiral center, leading to α- and β-anomers.

Answer 23

D-Glucose exists in an acyclic (open-chain) form and a cyclic (hemiacetal) form. Cyclization occurs when the hydroxyl (-OH) at C5 nucleophilically attacks the aldehyde carbon (C1). This reaction creates a new chiral center at C1, called the anomeric carbon. The result is two possible anomers: α-D-Glucose (α-D-Glucopyranose) → OH at C1 is down. β-D-Glucose (β-D-Glucopyranose) → OH at C1 is up. ✔ Cyclization forms a six-membered pyranose ring (glucopyranose).

Answer 24

The transformation from a linear Fischer projection to a cyclic Haworth projection follows these steps: Rotate the Fischer projection so that the hydroxyl (-OH) groups are aligned properly. Rearrange the structure to show the ring formation. Convert the Fischer projection into a Haworth projection: Groups on the right of the Fischer projection → point down in the Haworth projection. Groups on the left of the Fischer projection → point up in the Haworth projection. ✔ D-Glucose follows this convention to form α- and β-D-glucopyranose.

Answer 25

What are anomers? Anomers are diastereomers that differ only at the anomeric carbon (C1). α-D-Glucose and β-D-Glucose are anomers. Since they differ at only one stereogenic center, they are not enantiomers but diastereomers. ✔ α-D-Glucose and β-D-Glucose are diastereomers.

Answer 26

The six-membered cyclic form of glucose (glucopyranose) is drawn in Haworth projections. These flat ring representations help visualize the α and β anomers. Key convention: α-anomer → The OH on C1 is down. β-anomer → The OH on C1 is up. ✔ Haworth projections provide a clear view of ring structures in carbohydrates.

Answer 27

Glycosides are acetals formed when a monosaccharide reacts with an alcohol in the presence of acid (e.g., HCl). Only the anomeric hydroxyl (-OH) group of the hemiacetal reacts, while other hydroxyl groups remain unchanged. The reaction converts a hemiacetal (1 -OH + 1 -OR group) into an acetal (2 -OR groups). ✔ Key distinction: Glycosides have no free hemiacetal group, meaning they do not undergo mutarotation like normal monosaccharides.

Answer 28

Example: Treating α-D-Glucose with methanol (CH₃OH) and HCl. The result is two glycosides: α-Glycoside → The new -OCH₃ group is down. β-Glycoside → The new -OCH₃ group is up. ✔ This follows the same α/β labeling rules as anomers: α-glycoside = OR group is trans to CH₂OH. β-glycoside = OR group is cis to CH₂OH.

Answer 29

Protonation (H⁺) of the hemiacetal OH → Creates a better leaving group. Loss of water (H₂O) → Forms a planar carbocation at the anomeric carbon. Nucleophilic attack by CH₃OH from either face: Attack from below → Forms the α-glycoside. Attack from above → Forms the β-glycoside. Final deprotonation → Forms a stable acetal glycoside. ✔ Both α- and β-glycosides form in equal amounts since the carbocation is planar and can be attacked from both sides.

Answer 30

✔ Glycosides form when an alcohol reacts with a monosaccharide’s hemiacetal carbon. ✔ Acid catalysis (HCl) is needed for glycoside formation. ✔ The reaction produces α- and β-glycosides in equal amounts due to carbocation formation. ✔ Glycosides do not undergo mutarotation since they lack a free hemiacetal. Would you like a detailed step-by-step mechanism drawing for glycoside formation? 🚀

Answer 31

This image illustrates the glycoside formation mechanism from β-D-glucose using HCl and methanol (CH₃OH). Step 1-2: Protonation and Formation of a Planar Carbocation Protonation of the hemiacetal OH group (C1) using HCl: The OH group at C1 gets protonated (H⁺ attaches to oxygen). This makes the OH group a better leaving group. Loss of water (H₂O) forms a resonance-stabilized carbocation: The C1-O bond breaks, releasing water (H₂O), creating a planar carbocation at C1. This intermediate is now highly reactive. ✔ Key concept: The planar structure at C1 allows for nucleophilic attack from either side. Step 3-4: Nucleophilic Attack and Deprotonation Methanol (CH₃OH) attacks the carbocation from both sides: Attack from above → forms β-glycoside. Attack from below → forms α-glycoside. Deprotonation: The oxygen in CH₃OH loses an H⁺, forming a stable glycoside (acetal). ✔ Since the carbocation is planar, α- and β-glycosides form in equal amounts. Final Products: α-Glycoside (OCH₃ is down at C1). β-Glycoside (OCH₃ is up at C1). HCl is regenerated, completing the reaction. ✔ The final glycoside product is stable and does not undergo mutarotation. Key Takeaways ✔ Glycosides form through hemiacetal activation and nucleophilic substitution. ✔ The reaction goes through a planar carbocation, allowing for α/β product formation. ✔ Once formed, glycosides do not revert to the open-chain form (no mutarotation).

Answer 32

A glycoside is an acetal formed when a monosaccharide reacts with an alcohol, replacing the hemiacetal -OH with an -OR group."

Answer 33

"A glycoside has two -OR groups at the anomeric carbon, whereas a hemiacetal has one -OH and one -OR group."

Answer 34

"HCl protonates the hemiacetal -OH, making it a better leaving group and facilitating the formation of a planar carbocation."

Answer 35

,"Because the reaction goes through a planar carbocation intermediate, the nucleophile (CH₃OH) can attack from either side, forming both α- and β-glycosides."

Answer 36

,An acid-catalyzed nucleophilic substitution reaction.

Answer 37

,"It is replaced by an -OCH₃ group, forming a stable acetal."

Answer 38

"Because the hemiacetal group is converted into an acetal, which does not open up to the linear form."

Answer 39

A planar carbocation at the anomeric carbon. How does CH₃OH contribute to glycoside formation?,"It acts as a nucleophile, attacking the planar carbocation to form a new C-O bond."

Answer 40

,"Deprotonation of the newly formed acetal, regenerating HCl and stabilizing the glycoside.

Answer 41

,"The mechanism begins with the protonation of the hemiacetal OH group by HCl, making it a better leaving group. This results in the loss of water, forming a planar carbocation at the anomeric carbon. Methanol (CH₃OH) then acts as a nucleophile, attacking the carbocation from either side, producing both α- and β-glycosides. Finally, deprotonation stabilizes the acetal product."

Answer 42

,"The anomeric carbon in a hemiacetal is attached to both an oxygen and a hydroxyl group, making it reactive. Upon protonation, the OH group leaves as water, creating a resonance-stabilized carbocation that is highly electrophilic and ready for nucleophilic attack."

Answer 43

,"Unlike hemiacetals, which can open and close in equilibrium, glycosides are acetals, meaning they have two -OR groups at the anomeric carbon. Acetals are stable in neutral and basic conditions, preventing mutarotation and conversion back to the open-chain form."

Answer 44

,"Since the reaction proceeds via a planar carbocation, nucleophilic attack can occur from either the top or bottom face. This results in a mixture of both α- and β-glycosides in equal amounts."

Answer 45

,"In α-glycosides, the -OCH₃ group is trans to the CH₂OH group, while in β-glycosides, it is cis. Both are formed from the same planar carbocation intermediate, but the direction of nucleophilic attack determines which anomer is produced.

Answer 46

,"HCl is a strong acid that effectively protonates the hemiacetal OH group, increasing its leaving group ability. This facilitates the formation of a stable carbocation intermediate, allowing nucleophilic attack to proceed efficiently."

Answer 47

"Nucleophilic attack by an alcohol (e.g., CH₃OH) on the planar carbocation results in the formation of a stable acetal. This attack can happen from either side, leading to the formation of both α- and β-glycosides."

Answer 48

,"A hemiacetal has one -OH and one -OR group, allowing for equilibrium with the open-chain form. An acetal (glycoside) has two -OR groups, making it stable and unable to revert to the open-chain form."

Answer 49

","If ethanol (C₂H₅OH) were used instead of methanol (CH₃OH), the reaction would still proceed in the same way, but the final glycoside would have an -OC₂H₅ group instead of -OCH₃. The mechanism remains unchanged, only the identity of the alkoxy group differs."

Answer 50

","Glycoside formation is essential in biological systems for linking monosaccharides into disaccharides and polysaccharides (e.g., sucrose, cellulose, and starch). Glycosidic bonds stabilize carbohydrate structures and regulate biological recognition processes."