Ch. 20 - The Organic Chemistry of Carbohydrates

Bruice8th EditionOrganic ChemistryISBN: 9780135213711Not the one you use?Change textbook

Bruice 8th Edition

Ch. 20 - The Organic Chemistry of Carbohydrates

Problem 44

Chapter 21, Problem 44

Treatment with sodium borohydride converts aldose A to an optically inactive alditol. Wohl degradation of A forms B, whose alditol is optically inactive. Wohl degradation of B forms D-glyceraldehyde. Identify A and B.

Verified step by step guidance

Step 1: Understand the problem. Sodium borohydride (NaBH₄) reduces aldoses to alditols by converting the aldehyde group (-CHO) into a primary alcohol (-CH₂OH). The fact that the alditol of A is optically inactive suggests that A is a meso compound, meaning it has internal symmetry and no net optical activity.

Step 2: Perform the first Wohl degradation. Wohl degradation involves converting an aldose into a smaller aldose by removing the terminal carbon atom as hydrogen cyanide (HCN). The fact that the alditol of B is also optically inactive suggests that B is also a meso compound.

Step 3: Perform the second Wohl degradation. The problem states that the second Wohl degradation of B produces d-glyceraldehyde. This is a key clue because d-glyceraldehyde is the simplest aldose with a single chiral center, and it is optically active.

Step 4: Work backward to identify B. Since B undergoes Wohl degradation to form d-glyceraldehyde, B must be a four-carbon aldose (aldotetrose) with internal symmetry. The only meso aldotetrose is erythrose, which has two chiral centers and internal symmetry.

Step 5: Work backward to identify A. Since A undergoes Wohl degradation to form B (erythrose), A must be a five-carbon aldose (aldopentose) with internal symmetry. The only meso aldopentose is ribose. Therefore, A is ribose, and B is erythrose.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

11m

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Aldoses and Alditols

Aldoses are a type of monosaccharide that contain an aldehyde group, while alditols are sugar alcohols formed by the reduction of aldoses. The conversion of an aldose to an alditol typically involves the addition of hydrogen to the carbonyl group, resulting in a compound that lacks optical activity if it has a symmetrical structure.

Wohl Degradation

Wohl degradation is a chemical reaction that involves the oxidative cleavage of aldoses to produce smaller aldehydes or ketones. This process is significant in carbohydrate chemistry as it allows for the identification and transformation of sugars, leading to the formation of simpler sugars like d-glyceraldehyde from more complex aldoses.

Optical Activity

Optical activity refers to the ability of a compound to rotate plane-polarized light, which is a characteristic of chiral molecules. A compound is optically inactive if it is either achiral or if it contains a plane of symmetry, resulting in equal contributions to light rotation from its enantiomers. Understanding optical activity is crucial for identifying the stereochemistry of sugars and their derivatives.

Recommended video:

Guided course

08:17

Mutorotation and Optical Activity

Related Practice

Textbook Question

Draw the mechanism for the interconversion of α-D-glucose and β-D-glucose in dilute HCl.

views

Textbook Question

Propose a mechanism for the formation of d-allose from d-glucose in a basic solution.

Name the following:

The disaccharide lactulose consists of a D-galactopyranose subunit and a D-fructofuranose subunit joined by a β-1,4′-glycosidic linkage. After treatment of lactulose with 1. excess CH₃I/Ag₂O, 2. HCl/H₂O, the d-galactopyranose subunit was found to have one nonmethylated OH group, whereas the D-fructofuranose subunit had two. Draw the structure of ⍺-lactulose.

Textbook Question

A hexose was obtained after (+)-glyceraldehyde underwent three successive Kiliani–Fischer syntheses. Identify the hexose from the following experimental information: oxidation with nitric acid forms an optically active aldaric acid; a Wohl degradation followed by oxidation with nitric acid forms an optically inactive aldaric acid; and a second Wohl degradation forms erythrose.

Textbook Question

A student isolated a monosaccharide and determined that it had a molecular weight of 150. Much to his surprise, he found that it was not optically active. What is the structure of the monosaccharide?

views