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Ch.5 - Stereochemistry
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh.5 - StereochemistryProblem 35a
Chapter 5, Problem 35a

For each structure,
1. draw all the stereoisomers.
2. label each structure as chiral or achiral.
3. give the relationships between the stereoisomers (enantiomers, diastereomers).
(a)

Verified step by step guidance
1
Step 1: Analyze the given structure. The molecule is a Fischer projection with an aldehyde group (CHO) at the top, two hydroxyl groups (OH) on the horizontal bonds, and a primary alcohol group (CH2OH) at the bottom. The central carbons are stereocenters.
Step 2: Determine the stereoisomers. Each stereocenter can have two configurations (R or S). For two stereocenters, the total number of stereoisomers is 2^n, where n is the number of stereocenters. Here, n = 2, so there are 4 stereoisomers.
Step 3: Draw all stereoisomers. To do this, assign different combinations of R and S configurations to the two stereocenters. For example, (R,R), (R,S), (S,R), and (S,S). Represent each configuration using Fischer projections.
Step 4: Label each structure as chiral or achiral. A molecule is chiral if it has a non-superimposable mirror image. Check for symmetry in each stereoisomer to determine chirality.
Step 5: Determine the relationships between the stereoisomers. Compare each pair of stereoisomers. If they are non-superimposable mirror images, they are enantiomers. If they are not mirror images but differ in configuration, they are diastereomers.

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Key Concepts

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

Stereoisomerism

Stereoisomerism refers to the phenomenon where compounds have the same molecular formula and connectivity of atoms but differ in the spatial arrangement of their atoms. This can lead to different physical and chemical properties. The two main types of stereoisomers are enantiomers, which are non-superimposable mirror images, and diastereomers, which are not mirror images of each other.
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Chirality

Chirality is a property of a molecule that makes it non-superimposable on its mirror image, much like left and right hands. A chiral molecule typically has at least one carbon atom bonded to four different substituents, creating two distinct forms (enantiomers). In contrast, achiral molecules can be superimposed on their mirror images and do not exhibit this property.
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Relationship between Stereoisomers

The relationship between stereoisomers is crucial for understanding their interactions and properties. Enantiomers have identical physical properties except for their interaction with polarized light and reactions in chiral environments. Diastereomers, on the other hand, differ in physical properties and reactivity, making their relationships more complex. Identifying these relationships helps in predicting the behavior of compounds in various chemical contexts.
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Related Practice
Textbook Question

3,4-Dimethylpent-1-ene has the formula CH2=CH—CH(CH3)—CH(CH3)2. When pure (R)-3,4-dimethylpent-1-ene is treated with hydrogen over a platinum catalyst, the product is (S)-2,3-dimethylpentane.

a. Draw the equation for this reaction. Show the stereochemistry of the reactant and the product.

b. Has the chiral center retained its configuration during this hydrogenation, or has it been inverted?

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Textbook Question

The specific rotation of (S)-2-iodobutane is +15.90°.

a. Draw the structure of (S)-2-iodobutane.

b. Predict the specific rotation of (R)-2-iodobutane.

c. Determine the percentage composition of a mixture of (R)- and (S)-2-iodobutane with a specific rotation of –7.95°.

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Textbook Question

(+)-Tartaric acid has a specific rotation Of +12.0°. Calculate the specific rotation of a mixture of 68% (+)-tartaric acid and 32% (–)-tartaric acid.

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Textbook Question

Calculate the specific rotations of the following samples taken at 25 °C using the sodium D line.

b. 0.050 g of sample is dissolved in 2.0 mL of ethanol, and this solution is placed in a 2.0-cm polarimeter tube. The observed rotation is clockwise 0.043°.

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Textbook Question

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

d. What is the relationship between the two isomeric products?

e. Will these two products be produced in identical amounts? That is, will the product mixture be exactly 50:50?

f. Will these two stereoisomers have identical physical properties such as boiling point, melting point, solubility, etc.? Could they be separated (theoretically, at least) by distillation or recrystallization?

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Textbook Question

Free-radical bromination of the following compound introduces bromine primarily at the benzylic position next to the aromatic ring. If the reaction stops at the monobromination stage, two stereoisomers result.

a. Propose a mechanism to show why free-radical halogenation occurs almost exclusively at the benzylic position.

b. Draw the two stereoisomers that result from monobromination at the benzylic position.

c. Assign R and S configurations to the asymmetric carbon atoms in the products.

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