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Organic Chemistry 2 > Carbohydrates (Chapter 24) > Flashcards

Carbohydrates (Chapter 24) Flashcards

(58 cards)

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1
Q

Empirical Formula of Carbohydrates

A

Cm(H2O)n

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2
Q

Monosaccharide

Simple Sugar

A

An aldehyde/ketone containing at least three carbon atoms and two hydroxyl groups.

A monosaccharide can be either an aldose OR a ketose.

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3
Q

Linkage Connecting Two Monosaccharides

A

Ether Linkage

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4
Q

Aldose vs. Ketonse

A
  • Aldose: A monosaccharide containing an aldehyde group.
  • Ketose: A monosaccharide containing a ketone group.
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5
Q

Nomenclature: Monosaccharide Chain Length

A
  • Triose: 3 Carbons
  • Tetraose: 4 Carbons
  • Pentose: 5 Carbons
  • Hexose: 6 Carbons
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6
Q

Numbering: Aldose vs. Ketose

A
  • Aldose: Numbering begins at the aldehydic carbon.
  • Ketose: Numbering begins at the terminal carbon closest to the ketone.
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7
Q

Monosaccharides: L-Enantiomers

A

The highest-numbered hydroxyl group substituent is located on the left side of the projection.

The highest-numbered carbon stereocenter has the S-conformation.

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8
Q

Monosaccharides: D-Enantiomers

A

The highest-numbered hydroxyl group substituent is located on the right side of the projection.

The highest-numbered carbon stereocenter has the R-conformation.

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9
Q

Five-Membered Cyclic Hemiacetal

A

Furanose

The five-membered furanose ring is more stable than the six-membered pyranose ring.

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10
Q

Six-Membered Cyclic Hemiacetal

A

Pyranose

The six-membered pyranose ring is more stable than the five-membered furanose ring.

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11
Q

Stereoselectivity of Pyranose

A

The pyranose stereoisomer that places the C1 hydroxyl group at the equitorial position (i.e. the R-isomer) will be the dominant product.

The placement of the C1 hydroxyl group at the equitorial position minimizes steric effects.

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12
Q

Stereochemsitry of Cyclic Hemiacetal Formation

A

Hemiacetal cyclization forms a pair of diasteromers (i.e. anomers) due to stereoisomerization at the aldehydic (C1) carbon.

The new stereocenter of hemiacetal cyclization is formed at the former aldehydic carbon.

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13
Q

Anomer

A

A stereoisomer of a cyclic hemiacetal

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14
Q

Mutarotation

A

The interconversion between the α-anomer and β-anomer of a cyclic hemiacetal.

The mutarotation interconversions occur while the aldoses are in the acyclic form. (The α-anomer and β-anomer of the cyclic aldose are in equilibrium in solution).

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15
Q

α-Anomer vs. β-Anomer

Cyclic Hemiacetals

A
  • α-Anomer: The C1 hydroxyl group and highest-numbered hydroxyl group are placed on the same side of the cyclic projection.
  • β-Anomer: The C1 hydroxyl group and highest-numbered hydroxyl group are placed on opposite sides of the cyclic projection.

The stereochemistry/configuration at the anomeric carbon is the ONLY difference between the α-anomer and β-anomer.

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16
Q

Anomeric Carbon

Cyclic Hemiacetals

A

C1 Carbon

The anomeric carbon is the hemiacetal carbon of a cyclic monosaccharide.

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17
Q

Stereoisomer Priority Trend

A

—OH > —COOH > —CH2OH

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18
Q

Acyclic Monosaccharide → Cyclic Hemiacetal

A

Hemiacetal Cyclization

The hemiacetal cyclization reaction can occur under basic conditions OR acidic conditions.

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19
Q

Cyclic Hemiacetal → Acyclic Monosaccharide

A

Cyclic Hemiacetal Ring-Opening

The hemiacetal cyclization reaction can is reversible.

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20
Q

Reagents: Hemiacetal Cyclization

A
  • Acid-Catalyzed: H+, H2O
  • Base-Catalyzed: OH–, H2O

The hemiacetal cyclization reaction CAN occur under neutral conditions (but the cyclization rate is much slower without a catalyst).

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21
Q

Enantiomer-Switching

D-Saccharide → L-Saccharide

A

All chiral centers within the aldose/ketose are inverted to the opposite stereoisomer configuration.

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22
Q

Fischer Projections: Monosaccharides

A
  • Carbon atoms are numbered top-to-bottom.
  • Horizontal bonds are depicted as wedges.
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23
Q

Determining Anomer Stereochemistry

A
  • α-Anomer: The anomeric/hemiacetal carbon and the highest-numbered carbon have OPPOSITE stereoisomer configuration.
  • β-Anomer: The anomeric/hemiacetal carbon and the highest-numbered carbon have the SAME stereoisomer configuration.
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24
Q

Mutarotation: Interconversion Explanation

Study These Flashcards
A

Interconversion results from the varying plane-of-attack of the alcohol/oxide group to form the cyclic hemiacetal. (Attack of the alcohol/oxide group from above the molecule plane OR below the molecule plane creates differing anomeric forms of the cyclic monosaccharide.)

Both cyclic anomeric forms of the monosaccharide are in equilibrium with the acyclic aldose conformation.

25
*Plane Analysis:* Determining Anomer Stereochemistry ## Footnote This *plane analysis* technique functions ONLY when the endocyclic Oxygen is bonded to the highest-numbered carbon.
* **α-Anomer:** The *anomeric hydroxyl group* and the *terminal CH2OH group* are on OPPOSITE planes of the cyclic configuration. * **β-Anomer:** The *anomeric hydroxyl group* and the *terminal CH2OH group* are on the SAME plane of the cyclic configuration.
26
Strength of Aldose Oxidants
Bromine-Water < Nitric Acid < Periodic Acid ## Footnote * **Bromine-Water:** Oxidation of Terminal Aldehyde ONLY. * **Nitric Acid:** Oxidation of Terminal Aldehyde AND Terminal Alcohol. * **Periodic Acid:** Oxidation of ALL Carbon-Carbon Bonds
27
Aldose → Aldonic Acid
Bromine-Water Oxidation | Weak Oxidation ## Footnote The Bromine-Water Oxidation reaction converts the *terminal aldehyde group* to a *carboxylic acid group*.
28
Aldonic Acid
A sugar-acid monosaccharide containing one terminal carboxylic acid group.
29
Aldose → Aldaric Acid
Nitric Acid Oxidation | Medium Oxidation ## Footnote The Nitric Acid Oxidation reaction converts the *terminal aldehyde group* AND *terminal alcohol group* to *carboxylic acid groups*.
30
Aldose → Carboxylic Acid + Aldehyde
Periodic Acid Oxidation | Strong Oxidation ## Footnote The Periodic Acid Oxidation reaction oxidizes ALL carbon-carbon (C—C) bonds to C—OH or C=O bonds
31
Ketose → Caboxylic Acid + Aldehyde + CO2
Periodic Acid Oxidation | Strong Oxidation ## Footnote The Periodic Acid Oxidation reaction oxidizes ALL carbon-carbon (C—C) bonds to C—OH or C=O bonds
32
Aldaric Acid
A sugar-acid monosaccharide containing two terminal carboxylic acid groups.
33
*Reagents:* Bromine-Water Oxidation
Br2, H2O
34
*Reagents:* Nitric Acid Oxidation
HNO3, H2O
35
*Reagents:* Periodic Acid Oxidation
HIO4 (Excess)
36
*Product Identities:* Periodic Acid Oxidation
* **Ketones** → CO2 * **Aldehydes** → Formic Acid * **2° Alcohols** → Formic Acid * **1° Alcohols** → Aldehydes
37
Formic Acid vs. Formaldehyde
* **Formic Acid:** 1-Carbon Carboxylic Acid * **Formaldehyde:** 1-Carbon Aldehyde
38
Aldose → Alditol
NaBH4 Reduction ## Footnote The NaBH4 Reduction reaction reduces the *terminal aldehyde group* to an *alcohol group*.
39
Ketose → Alditol
NaBH4 Reduction ## Footnote The NaBH4 Reduction reaction reduces the *ketone group* to a *2° alcohol group*.
40
*Reagents:* NaBH4 Reduction
NaBH4, CH3OH (Solvent)
41
Mild/Weak Oxidants
* **Bromine-Water:** Br2, H2O * **Silver:** Ag(I) * **Copper:** Cu(II)
42
Why are 2° alcohol groups NOT oxidized during Nitric Acid Oxidation?
The 2° alcohol groups are too *sterically hindered* to interact with the nitric acid oxidant.
43
*Reaction:* Sugar Chain Extension
Cyanohydrin Formation-Reduction ## Footnote * The Cyanohydrin Formation-Reduction reaction creates *two* new sugars (due to formation of a stereocenter) that are one carbon *longer* than the starting aldose.
44
*Reagents:* Cyanohydrin Formation-Reduction ## Footnote **Starting Material:** Aldose
1. HCN 2. H2, Pd 3. H2O
45
*Mechanism:* Cyanohydrin Formation-Reduction
1. HCN Attack at Aldehydic Carbon 2. Hydrogenation across C≡N Bond 3. Hydration of Imine Group
46
Pyranose → Multi-Pyranoside | Furanose → Multi-Furanoside
Base-Catalyzed William Ether Synthesis ## Footnote The base-catalyzed William Ether Synthesis reaction converts ALL *hydroxyl groups* (on the pyranose) to *methoxy groups*.
47
Pyranose → Mono-Pyranoside | Furanose → Mono-Furanoside
Acid-Catalyzed Anomeric Hydroxyl Methylation ## Footnote The acid-catalyzed Anomeric Hydroxyl Methylation reaction converts ONLY the *anomeric hydroxyl group* to a *methoxy group* (i.e. it converts the *hemiacetal* group to an *acetal* group).
48
*Reagents:* Base-Catalyzed William Ether Synthesis | Saccharide Acetal Formation
NaOH, (CH3)2SO4
49
*Reagents:* Acid-Catalyzed Anomeric Hydroxyl Methylation | Saccharide Acetal Formation
CH3OH, H3O+
50
Glycoside
Acetal-Containing Monosaccharide
51
Stability of Glycoside Acetals
* The acetal group is **stable** under *basic conditions* AND *neutral conditions*. * The acetal group is **cleaved** under *acidic conditions*. ## Footnote * Mutarotation of glycoside acetals does NOT occur under *basic conditions* or *neutral conditions*. * Mutarotation of glycoside acetals DOES occur under *acidic conditions* (after the acetal is converted to a hemiacetal).
52
*Reagents:* Cleavage of Glycoside Acetal
H3O+, Δ (Heat)
53
Stability of Cyclic Acetals
Cyclic acetals are **stable** under *basic conditions* AND *neutral conditions*.
54
1,2-Diol → Cyclic Acetal
Ketone, H3O+
55
Methods of Sugar Hydoxyl Protection
* Cyclic Acetal Synthesis via 1,2-Diol * Base-Catalyzed William Ether Synthesis * Acid-Catalyzed Anomeric Hydroxyl Methylation
56
Disaccharide → Monosaccharide
Ether Linkage Cleavage ## Footnote The α-anomer AND β-anomer of the monopyranose are formed. (The anomer with the anomeric hydroxyl group placed at the *equitorial* position will predominate.)
57
*Reagents:* Ether Linkage Cleavage
H3O+
58
Why are *hemiacetal —OH groups* more reactive than *simple —OH groups* under acidic conditions? | Pyranose Cleavage
Reaction of the *hemiacetal —OH group* with acid (i.e. protonation) results in a *more stable* oxocarbenium intermediate form. ## Footnote Reaction of the *simple —OH group* with acid results in a *less stable* carbocation intermediate form.

Organic Chemistry 2 (12 decks)

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