Draw Lewis structures for the following free radicals. a. The ethyl radical, CH 3 —ĊH 2 b. The tert -butyl radical, (CH 3 ) 3 C•

Hey everyone, let's solve this problem. It says write the appropriate loose structures for these free radicals. Ch three with a radical and CH c H 32 with a radical. So don't worry about the radicals, we still treat this pretty much the exact same as any other compound. So when we're drawing a lewis structure we need to know all the bonding and non bonding electrons. So we're going to count the total number of valence electrons. Do we include that free radical? No. Okay, so don't don't count the radical as an extra. Okay, so you wouldn't total for CH three and then say, oh, it has a radical. So let me add one more. No, the radical will be taken into account just by totaling the valence electrons from the periodic table. Okay, and then next we'll draw our structure like a skeleton. Rough draft with our sigma bonds and then we can add any lone pairs or are radical. And remember we don't add a radical on hydrogen. Okay, cool. So let's do this. Our first compound is CH three with a radical. So how many valence electrons carbon has four hydrogen has one times three is three which is seven. Ah so notice how it's an odd number. That's sort of a hint that it's going to have a radical, right, Because usually bonds and lone pairs are all come in twos. Right, so having an odd number, sort of gives us a hint about that. So let's draw our carbon Central carbon with three hydrogen that uses 246 valence electrons. Ah so we need one more and that's going to be our radical which will go on the carbon, right? We said we don't put them on hydrogen and that will give carbon a little bit closer to a satisfied octet. Although not quite All right. That's as good as we can get. But radicals are known to not be stable. Right? That's kind of their whole deal. So it's okay. Don't freak out. Okay, now we have C H and C H 32 so two methyl groups. So if we have carbon is four plus one plus four plus three times one is three times two. So this is 19 villains electrons. Again, you notice it's an odd number. So the structure is similar except instead of three H. S. We have one H group And then two methyl groups. So that's already using 2468, 10, 12, 14, 16, 18, valence electrons. We only need one more which is going to be our free radical. I feel like it's kind of easy to catch your eye where that radical should go. Right, this carbon is the only one that is not that does not have its full octet. So we'll pop that radical right there. Okay, these are your correct lewis structures. I hope this video is helpful and thanks for watching

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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

1. A Review of General Chemistry

Lewis Structure

Organic Chemistry

1. A Review of General Chemistry

Lewis Structure

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4:04 minutes

Problem 1

Textbook Question

Draw Lewis structures for the following free radicals.
a. The ethyl radical, CH₃—ĊH₂
b. The tert-butyl radical, (CH₃)₃C•

Verified step by step guidance

Step 1: Understand that a free radical is a molecule that contains an unpaired electron. This unpaired electron is typically represented as a dot (•) in Lewis structures.

Step 2: For the ethyl radical (CH3—CH2•), start by drawing the Lewis structure for ethane (CH3—CH3). Ethane consists of two carbon atoms single-bonded to each other, with each carbon atom also bonded to three hydrogen atoms.

Step 3: To convert ethane to the ethyl radical, remove one hydrogen atom from the CH2 group, leaving an unpaired electron on the second carbon atom. The structure should now be CH3—CH2•, with the dot representing the unpaired electron on the second carbon.

Step 4: For the tert-butyl radical ((CH3)3C•), begin by drawing the Lewis structure for tert-butyl cation ((CH3)3C+). This involves a central carbon atom bonded to three methyl groups (CH3).

Step 5: To form the tert-butyl radical, replace the positive charge with an unpaired electron. The central carbon atom will have three single bonds to the methyl groups and one unpaired electron, represented as (CH3)3C•.

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Key Concepts

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

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms of a molecule and the lone pairs of electrons that may exist. They are essential for visualizing the arrangement of electrons around atoms, which helps in understanding the molecule's geometry, reactivity, and properties. In the case of free radicals, the Lewis structure will show an unpaired electron, which is crucial for identifying the radical nature of the molecule.

Free Radicals

Free radicals are atoms or molecules that contain an unpaired electron, making them highly reactive. In organic chemistry, radicals are often intermediates in reactions and can be identified by a single dot representing the unpaired electron in their Lewis structures. Understanding free radicals is important for predicting reaction mechanisms and the stability of different radical species.

Hybridization and Molecular Geometry

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals, which influences the geometry and bonding properties of molecules. For example, the carbon atoms in ethyl and tert-butyl radicals are typically sp3 hybridized, leading to a tetrahedral geometry. Recognizing the hybridization helps in accurately drawing the Lewis structures and understanding the spatial arrangement of atoms and electrons in radicals.

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