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Carbohydrates (Chapter 24)

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Total cards: 24•Content version: v4•Public deck•Updated: 8/1/2026

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#1
Front (Question)

Empirical Formula of Carbohydrates

Back (Answer)

Cm(H2O)n

#2
Front (Question)

Monosaccharide Simple Sugar

Back (Answer)

An aldehyde/ketone containing at least three carbon atoms and two hydroxyl groups. A monosaccharide can be either an aldose OR a ketose.

#3
Front (Question)

Linkage Connecting Two Monosaccharides

Back (Answer)

Ether Linkage

#4
Front (Question)

Aldose vs. Ketonse

Back (Answer)

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

#5
Front (Question)

Nomenclature: Monosaccharide Chain Length

Back (Answer)

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

#6
Front (Question)

Numbering: Aldose vs. Ketose

Back (Answer)

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

#7
Front (Question)

Monosaccharides: L-Enantiomers

Back (Answer)

The highest-numbered hydroxyl group substituent is located on the left side of the projection. The highest-numbered carbon stereocenter has the S-conformation.

#8
Front (Question)

Monosaccharides: D-Enantiomers

Back (Answer)

The highest-numbered hydroxyl group substituent is located on the right side of the projection. The highest-numbered carbon stereocenter has the R-conformation.

#9
Front (Question)

Five-Membered Cyclic Hemiacetal

Back (Answer)

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

#10
Front (Question)

Six-Membered Cyclic Hemiacetal

Back (Answer)

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

#11
Front (Question)

Stereoselectivity of Pyranose

Back (Answer)

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.

#12
Front (Question)

Stereochemsitry of Cyclic Hemiacetal Formation

Back (Answer)

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.

#13
Front (Question)

Anomer

Back (Answer)

A stereoisomer of a cyclic hemiacetal

#14
Front (Question)

Mutarotation

Back (Answer)

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).

#15
Front (Question)

α-Anomer vs. β-Anomer Cyclic Hemiacetals

Back (Answer)

• 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.

#16
Front (Question)

Anomeric Carbon Cyclic Hemiacetals

Back (Answer)

C1 Carbon The anomeric carbon is the hemiacetal carbon of a cyclic monosaccharide.

#17
Front (Question)

Stereoisomer Priority Trend

Back (Answer)

—OH > —COOH > —CH2OH

#18
Front (Question)

Acyclic Monosaccharide → Cyclic Hemiacetal

Back (Answer)

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

#19
Front (Question)

Cyclic Hemiacetal → Acyclic Monosaccharide

Back (Answer)

Cyclic Hemiacetal Ring-Opening The hemiacetal cyclization reaction can is reversible.

#20
Front (Question)

Reagents: Hemiacetal Cyclization

Back (Answer)

• 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).

#21
Front (Question)

Enantiomer-Switching D-Saccharide → L-Saccharide

Back (Answer)

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

#22
Front (Question)

Fischer Projections: Monosaccharides

Back (Answer)

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

#23
Front (Question)

Determining Anomer Stereochemistry

Back (Answer)

• 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.

#24
Front (Question)

Mutarotation: Interconversion Explanation

Back (Answer)

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.