Draw the structure for a compound with molecular formula C 2 H 2 I 2 F 2 c. that is optically active.

Hello, everyone. Today, we have the following problem. A compound has a molecular formula of the following. Draw the structure of the compound if it is optically active. So chiral molecules are optically active and are able to rotate plane polarized light. And thus, we must draw the compound such that it contains at least one asymmetric center and has no plane of symmetry. So for possible structures, we have the following, we will use fisher projections for better visualization. So we can draw the two carbons as they are, we can also draw the two hydrogens bound to those carbons. And as for the bromine and the chlorine. Well, for one structure, we can have our bromine on the left and then we could have a chlorine on the right. And then we can switch those two groups for the second carbon. Now, for a different compound, we can have the saint the hydrogens in the same positions. But this time, we can have the chlorine on the right for the first carbon and for the second carbon and then the bromine on the left for both carbons. Now between these two molecules, we must determine which one has a plane of symmetry and go from there. Well, if we look at our second compound, this has a plane of symmetry, the top half is equal to the bottom half. And thus, we can say that our first compound that we drew that is asymmetric is optically active in our compound overall. I hope it helped. And until next time.

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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

Optical Activity

Organic Chemistry

5. Chirality

Optical Activity

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3:07 minutes

Problem 73c

Textbook Question

Draw the structure for a compound with molecular formula C₂H₂I₂F₂
c. that is optically active.

Verified step by step guidance

Step 1: Start by drawing a basic structure for the compound. Since there are two carbon atoms in the molecular formula, connect these two carbon atoms.

Step 2: Now, add the halogens (Iodine and Fluorine) to the carbon atoms. Each carbon atom should be connected to one Iodine and one Fluorine atom. This is because each carbon atom can form four bonds and two of these bonds are already taken by the other carbon atom and a hydrogen atom.

Step 3: To make the compound optically active, it must have a chiral center. A chiral center is a carbon atom that is connected to four different groups. In this case, each carbon atom is connected to a hydrogen, the other carbon, an iodine, and a fluorine. Therefore, each carbon atom is a chiral center.

Step 4: Draw the 3D structure of the compound. The hydrogen and the other carbon atom should be in the plane of the paper, the iodine should be coming out of the plane (use a wedge to represent this), and the fluorine should be going into the plane (use a dashed line to represent this).

Step 5: Finally, remember that the actual molecule is a combination of these two 3D structures, which are mirror images of each other. This is what makes the compound optically active.

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Key Concepts

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

Optical Activity

Optical activity refers to the ability of a chiral compound to rotate the plane of polarized light. A compound is chiral if it has no plane of symmetry and cannot be superimposed on its mirror image. This property is crucial for understanding how certain molecules interact with light and is a key factor in determining the behavior of organic compounds in biological systems.

Chirality and Stereocenters

Chirality is a geometric property of some molecules that makes them non-superimposable on their mirror images. A stereocenter, typically a carbon atom bonded to four different substituents, is essential for creating chirality. In the context of the given molecular formula, identifying potential stereocenters is vital for determining if the compound can exist in optically active forms.

Molecular Structure and Isomerism

The molecular structure of a compound defines its connectivity and spatial arrangement of atoms, which directly influences its chemical properties. Isomerism, particularly stereoisomerism, occurs when compounds have the same molecular formula but differ in the arrangement of atoms. For C2H2I2F2, understanding how different arrangements can lead to chiral centers is essential for drawing the correct structure that is optically active.

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Naproxen is a commercially available anti-inflammatory sold under the name Aleve.

(b) Given that naproxen rotates plane-polarized light in the clockwise direction, should it be referred to as (d) or (l)?

(d) What direction (d or l; + or −) would you expect the enantiomer of naproxen to rotate plane-polarized light?

Textbook Question

When the following substituted biphenyl was synthesized, it was found to have a specific rotation [α] of -23° at 25°C . When the specific rotation was measured at 100°C the compound had a specific rotation of 0° . Upon cooling back to 25°C the specific rotation was measured again, resulting in [α] = 0°. Explain these results.

Textbook Question

(R)-Limonene is a cyclic terpene responsible for the smell of oranges and other citrus fruits.

(a) Given that (R)-limonene rotates plane-polarized light in the clockwise direction, should it be referred to as (d) or (l)?

(b) Is it (+) or (-)?

Textbook Question

Which of the following diastereomers of hexane-2,5-diol is optically inactive? Why?

Textbook Question

A solution of 0.50 g of (−)-epinephrine (see Figure 5-16) dissolved in 10.0 mL of dilute aqueous HCl was placed in a 20-cm polarimeter tube. Using the sodium D line, the rotation was found to be −5.1° at 25 °C. Determine the specific rotation of epinephrine.

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Textbook Question

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

a. 1.00 g of sample is dissolved in 20.0 mL of ethanol. Then 5.00 mL of this solution is placed in a 20.0-cm polarimeter tube. The observed rotation is 1.25° counterclockwise.

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°.

Textbook Question

If you had the two enantiomers of carvone in unmarked bottles, could you use just your nose and a polarimeter to determine

a. whether it is the (+) or (−) enantiomer that smells like spearmint

b. whether it is the (R) or (S) enantiomer that smells like spearmint?

c. With the information given in the drawings of carvone above, what can you add to your answers to (a) and (b)?

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Draw the structure for a compound with molecular formula C2H2I2F2c. that is optically active.

Key Concepts

Optical Activity

Chirality and Stereocenters

Molecular Structure and Isomerism

Watch next

Draw the structure for a compound with molecular formula C₂H₂I₂F₂
c. that is optically active.