/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Draw Fischer projections for (a)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 2

Draw Fischer projections for (a) L-mannose, (b) L-fucose (6-deoxy-L- galactose), (c) D-xylitol, and (d) D-iduronate.

Short Answer

Expert verified
The Fischer projections of L-mannose, L-fucose, D-xylitol, and D-iduronate are drawn by recognizing the structure and configuration of each carbon atom in the sugar or sugar derivative. The arrangement of -OH and -H around each carbon atom is key to depict the correct Fischer projection.

Step by step solution

01

Drawing L-mannose

The Fischer projection for L-mannose can be drawn as below: The first carbon atom from the top has an -OH group on the left. The second carbon atom has -OH on the right. The third and fourth carbon atoms' -OH groups are on the left. The fifth carbon atom's -OH is on the right. The lowest (sixth) carbon atom is denoted as CH2OH.
02

Drawing L-fucose

For L-fucose, take L-galactose as a reference. Start with L-galactose's Fischer projection and convert the -OH group at the fourth carbon atom from the top to -H (which simply means deleting the -OH group to make it a deoxy sugar), and the remaining structure will be L-fucose. The other atoms remain in the same configuration as L-galactose with the only difference being at the fourth carbon atom where the -OH is replaced by -H.
03

Drawing D-xylitol

D-xylitol is a sugar alcohol derived from xylose. The Fischer projection of D-xylitol can be drawn as follows: From the top, the first carbon atom has -OH on right, while the second and third carbon atoms have -OH groups on the left, and the fourth carbon atom has -OH on the right. The lowest (fifth) carbon atom is denoted as CH2OH.
04

Drawing D-iduronate

For D-iduronate, the first four carbons have their -OH groups on the right side, whereas the fifth carbon has -OH on the left. Lastly, an =O (indicating a carbonyl group) replaces -OH at the bottom carbon (in CH2OH) giving COOH.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Draw the structure of each of the following molecules and label each chiral carbon with an asterisk. (a) \(\alpha\)-D-Glucose 1-phosphate. (b) 2 -Deoxy- \(\beta\)-D-ribose 5 -phosphate. (c) D-Glyceraldehyde 3-phosphate. (d) L-Glucuronate.

A carbohydrate-amino acid polymer that is a potent inhibitor of influenza virus has been synthesized. The virus is thought to be inactivated when multiple sialyl groups bind to viral surface proteins. Draw the chemical structure of the carbohydrate portion of this polymer (below, where \(X\) represents the rest of the polymer). NeuNAc \(\alpha-(2 \rightarrow 3)\) Gal \(\beta-(1 \rightarrow 4)\) Glu \(\beta-(1 \rightarrow)-X\)

Identify each of the following: (a) Two aldoses whose configuration at carbons 3,4 , and 5 matches that of D-fructose. (b) The enantiomer of D-galactose. (c) An epimer of D-galactose that is also an epimer of D-mannose. (d) A ketose that has no chiral centers. (e) A ketose that has only one chiral center. (f) Monosaccharide residues of cellulose, amylose, and glycogen. (g) Monosaccharide residues of chitin.

In aqueous solution, virtually all D-glucose molecules \((>99 \%)\) are in the pyranose form. Other aldoses have a greater proportion of molecules in the open-chain form. D-Glucose may have evolved to be the predominant metabolic fuel because it is less likely than its isomers to react with and damage cellular proteins. Explain why D-glucose reacts less than other aldoses with the amino groups of proteins.

Describe the general structural features of glycosaminoglycans.
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.