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Chapter 25: Biomolecules: Carbohydrates

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 25: Biomolecules: Carbohydrates

25 problems · 36 figures

25-1 §25.1 source ↗

Classify each of the following monosaccharides:
(a)
(b)
(c)
(d)

Answer (4 parts)

(a)
Aldotetrose
(b)
Ketopentose
(c)
Ketohexose
(d)
Aldopentose

25-2 §25.2 source ↗

Convert each of the following Fischer projections into a tetrahedral representation, and assign R or S stereochemistry:
(a)
(b)
(c)

Answer (3 parts)

(a)
S
(b)
R
(c)
S

25-3 §25.2 source ↗

Which of the following Fischer projections of glyceraldehyde represent the same enantiomer?
(a)
(b)
(c)
(d)

Answer

A, B, and C are the same.

25-4 §25.2 source ↗

Redraw the following molecule as a Fischer projection, and assign R or S configuration to the chirality center (green = Cl):

Answer

25-5 §25.2 source ↗

Redraw the following aldotetrose as a Fischer projection, and assign R or S configuration to each chirality center:

Answer

25-6 §25.3 source ↗

Assign R or S configuration to each chirality center in the following monosaccharides, and tell whether each is a D sugar or an L sugar:
(a)
(b)
(c)

Answer (3 parts)

(a)
L-Erythrose; 2S,3S
(b)
D-Xylose; 2R,3S,4R
(c)
D-Xylulose; 3S,4R

25-7 §25.3 source ↗

(+)-Arabinose, an aldopentose that is widely distributed in plants, is systematically named (2R,3S,4S)-2,3,4,5-tetrahydroxypentanal. Draw a Fischer projection of (+)-arabinose, and identify it as a D sugar or an L sugar.

Answer

25-8 §25.4 source ↗

Only the D sugars are shown in Figure 25.4. Draw Fischer projections for the following L sugars:
(a)
L-Xylose
(b)
L-Galactose
(c)
L-Allose
Figure 25.4 in the book ↗
Configurations of D aldoses. The structures are arranged from left to right so that the –OH groups on C2 alternate right/left (R/L) across each series. Similarly, the –OH groups at C3 alternate two right/two left (2R/2L), the –OH groups at C4 alternate 4R/4L, and the –OH groups at C5 are to the right in all eight (8R). Each D aldose has a corresponding L enantiomer that is not shown.

Answer (3 parts)

(a)
(b)
(c)

25-9 §25.4 source ↗

How many aldoheptoses are there? How many are D sugars, and how many are L sugars?

Answer

16 D and 16 L aldoheptoses

25-10 §25.4 source ↗

The following model is that of an aldopentose. Draw a Fischer projection of the sugar, name it, and identify it as a D sugar or an L sugar.

Answer

25-11 §25.5 source ↗

Ribose exists largely in a furanose form, produced by addition of the C4 –OH group to the C1 aldehyde. Draw D-ribose in its furanose form.

Answer

25-12 §25.5 source ↗

Figure 25.6 shows only the β-pyranose and β-furanose anomers of D-fructose. Draw the α-pyranose and α-furanose anomers.
Figure 25.6 in the book ↗
Pyranose and furanose forms of fructose in aqueous solution. The two pyranose anomers result from addition of the C6 –OH group to the C2 carbonyl; the two furanose anomers result from addition of the C5 –OH group to the C2 carbonyl.

Answer

25-13 §25.5 source ↗

Draw β-D-galactopyranose and β-D-mannopyranose in their more stable chair conformations. Label each ring substituent as either axial or equatorial. Which would you expect to be more stable, galactose or mannose?

Answer

25-14 §25.5 source ↗

Draw β-L-galactopyranose in its more stable chair conformation, and label the substituents as either axial or equatorial.

Answer

25-15 §25.5 source ↗

Identify the following monosaccharide, write its full name, and draw its open-chain form as a Fischer projection.

Answer

α-D-Allopyranose

25-16 §25.6 source ↗

Draw the products you would obtain by reaction of β-D-ribofuranose with: (a) CH3I, Ag2O (b) (CH3CO)2O, pyridine

Answer

25-17 §25.6 source ↗

Reduction of D-glucose leads to an optically active alditol (D-glucitol), whereas reduction of D-galactose leads to an optically inactive alditol. Explain.

Answer

D-Galactitol has a plane of symmetry and is a meso compound, whereas D-glucitol is chiral.

25-18 §25.6 source ↗

Reduction of L-gulose with NaBH4 leads to the same alditol (D-glucitol) as reduction of D-glucose. Explain.

Answer

The −CHO end of L-gulose corresponds to the −CH2OH end of D-glucose after reduction.

25-19 §25.6 source ↗

D-Glucose yields an optically active aldaric acid on treatment with HNO3, but D-allose yields an optically inactive aldaric acid. Explain.

Answer

D-Allaric acid has a symmetry plane and is a meso compound, but D-glucaric acid is chiral.

25-20 §25.6 source ↗

Which of the other six D aldohexoses yield optically active aldaric acids on oxidation, and which yield optically inactive (meso) aldaric acids? (See Problem 25-19.)

Answer

D-Allose and D-galactose yield meso aldaric acids; the other six D-hexoses yield optically active aldaric acids.

25-21 §25.6 source ↗

What product(s) would you expect from Kiliani–Fischer reaction of D-ribose?

Answer

D-Allose + D-altrose

25-22 §25.6 source ↗

What aldopentose would give a mixture of L-gulose and L-idose on Kiliani–Fischer chain extension?

Answer

L-Xylose

25-23 §25.6 source ↗

Two of the four D aldopentoses yield D-threose on Wohl degradation. What are their structures?

Answer

D-Xylose and D-lyxose

25-24 §25.7 source ↗

Show how neuraminic acid can arise by an aldol reaction of N-acetylmannosamine with pyruvate (CH3COCO2–).

Answer

25-25 §25.8 source ↗

Show the product you would obtain from the reaction of cellobiose with the following reagents:
(a)
NaBH4
(b)
Br2, H2O
(c)
CH3COCl, pyridine

Answer (3 parts)

(a)
The hemiacetal ring is reduced.
(b)
The hemiacetal ring is oxidized.
(c)
All hydroxyl groups are acetylated.

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Problems and answers are adapted from Organic Chemistry: A Tenth Edition by John McMurry, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/organic-chemistry/pages/1-why-this-chapter. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.