Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

5. Chirality

Fischer Projection

Organic Chemistry

5. Chirality

Fischer Projection

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6:32 minutes

Problem 3

Textbook Question

(a) There is only one ketotriose, called dihydroxyacetone. Draw its structure.
(b) There is only one aldotriose, called glyceraldehyde. Draw the two enantiomers of glyceraldehyde.

Verified step by step guidance

Step 1: Understand the terms 'ketotriose' and 'aldotriose'. A 'triose' is a monosaccharide with three carbon atoms. A 'ketotriose' contains a ketone functional group, while an 'aldotriose' contains an aldehyde functional group.

Step 2: For part (a), dihydroxyacetone is the only ketotriose. Its structure consists of three carbon atoms: the middle carbon is part of a ketone group (C=O), and the other two carbons each have a hydroxyl group (-OH) attached. Draw the structure as follows: CH2OH-C(=O)-CH2OH.

Step 3: For part (b), glyceraldehyde is the only aldotriose. Its structure consists of three carbon atoms: the first carbon is part of an aldehyde group (-CHO), the second carbon has a hydroxyl group (-OH) attached, and the third carbon is part of a primary alcohol group (-CH2OH).

Step 4: Glyceraldehyde has two enantiomers (non-superimposable mirror images) because the second carbon is a chiral center. To draw the enantiomers, place the -OH group on the second carbon either to the right (D-glyceraldehyde) or to the left (L-glyceraldehyde) in a Fischer projection.

Step 5: Verify the structures by ensuring that the ketotriose (dihydroxyacetone) has no chiral centers and that the aldotriose (glyceraldehyde) has one chiral center, leading to two enantiomers. Label the enantiomers clearly as D- and L-glyceraldehyde.

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Key Concepts

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

Ketotriose

A ketotriose is a type of carbohydrate that contains three carbon atoms and a ketone functional group. The only ketotriose is dihydroxyacetone, which has the molecular formula C3H6O3. Its structure features a central carbon atom bonded to a ketone group (C=O) and two hydroxyl groups (–OH), making it a simple sugar that plays a role in metabolism.

Aldotriose

An aldotriose is a carbohydrate with three carbon atoms and an aldehyde functional group. Glyceraldehyde is the only aldotriose, characterized by its molecular formula C3H6O3. It exists in two enantiomeric forms, which are mirror images of each other, due to the presence of a chiral carbon atom, making it an important compound in biochemistry and metabolism.

Enantiomers

Enantiomers are a pair of molecules that are non-superimposable mirror images of each other, often differing in their optical activity. In the case of glyceraldehyde, the two enantiomers are D-glyceraldehyde and L-glyceraldehyde, which have identical physical properties except for their interaction with polarized light. Understanding enantiomers is crucial in organic chemistry, especially in the context of biological systems where chirality can affect the function of molecules.

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Related Practice

Textbook Question

In this attempt to convert the line angle drawing of d-erythrose (shown) to the Fischer projection (shown), by viewing it from a certain direction (shown), a mistake was made. What was the mistake?

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Textbook Question

Draw Fischer projections of the following molecules.

(c)

Textbook Question

Convert the following perspective formulas to Fischer projections.

(c)

(d)

Textbook Question

Convert the following Fischer projections to perspective formulas

(c)

(d)

Textbook Question

For each set of examples, make a model of the first structure, and indicate the relationship of each of the other structures to the first structure. Examples of relationships: same compound, enantiomer, structural isomer.

(c)

Textbook Question

Which of the following compounds are chiral? Draw each compound in its most symmetric conformation, star (*) any asymmetric carbon atoms, and draw any mirror planes. Label any meso compounds. You may use Fischer projections if you prefer.

(e) (R,R)-2,3-dibromobutane

(f)

Textbook Question

Draw a Fischer projection for each compound. Remember that the cross represents an asymmetric carbon atom, and the carbon chain should be along the vertical, with the IUPAC numbering from top to bottom.

(e)

Textbook Question

a. Are d-erythrose and l-erythrose enantiomers or diastereomers?

b. Are l-erythrose and l-threose enantiomers or diastereomers?

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