Draw three Lewis structures for C 3 H 8 O.

Hello everyone. So in this video, we're gonna go ahead and draw the three appropriate Lewis structures for the given species below so that it's going to be C four H 80. So first we need to go ahead and calculate for the degree of saturation. So we have a formula for this. So we have half times two C plus two plus N minus H minus X. So let's kind of break this down. So for our see here, that's going to be our carbons are end here represents our nitrogen, our age represents our protons or hide Jen's. And then lastly, our X is going to be all of our religions. So using the degree of saturation formula that we have just ran out and applying it to the given molecular formula of C four H 80, we can go ahead and find the degree oven situation for this molecule. So again, we have half times, let's see here. So for C again, as carbons, we see here in this formula, we have four carbons. So add that in plus two plus N, we don't have any nitrogen. So that's zero. For ages, we have eight ages. So we'll put the number eight here and then for all of our highlights, we have zero. All right. So now we can go ahead and put everything into the calculator once we do, so we'll get this to be equal to one. So the given formula or molecular formula corresponds to a structure with one element of unsaturated in which can be ring or a double bond. So to draw three appropriate lower structures for the given molecular formula of C four H 80 or o for the auction, we firstly determine the total number of valence electrons that corresponds to the given formula. So let's go ahead and calculate that as well. So starting off with our carbons here, we see that we have four atoms of carbon, each atom of carbon will give us four valence electrons. So four times four is equal to 16. So we'll get a total of 16 valence electrons coming off of just our carbons alone. Moving on to hydrogen. We see here that we have eight atoms of hydrogen, each gives us one valence electron. So eight times one is equal to eight valence electrons. Next, we have auction here, auction, we just have one atom of auction. Each atom will give us six valence electrons. So that equals 21 times six is always just going to be itself and that is six here in our case. So we have 60 coming off from the carbons eight coming off from our approaches are high regions and then six coming off from our auction. All right. So second step is going to go ahead and draw a appropriate Sigma bond framework. So three possible Sigma bond frameworks with one element of saturation is going to be possible. So let's go ahead and draw this out. So we'll draw all the bonds here. So we have A siege three at the end, we can see here again that this molecule needs to have four total carbons. Each carbon likes to have four bonds. So it is usually sort of a central atom. And lastly, we have our auction still and one carbon left as well as one proton. So we'll have carbon double bond auction, single bond proton. So that's one of our Sigma frameworks here. Next, we're gonna go ahead and actually make this molecule into a cycle al King. So we have a cyclic single bond molecule here. Again, we have four carbons total. All right. And because we need to use up all eight protons as well as having carbon liking for bonds. We'll go ahead and add two protons per carbon. Alright, so we've completed the cyclic version of this molecule. And lastly, we're just gonna go ahead and sort of do the one that we did in the first. We're gonna go ahead and add or put our carbon, you'll group into a different area. So we have each, well, we have a carbon at the end connected to three hydrogen is here. And of course, like we said, carbon is always just going to be central atom. So we can do a carbon double bound auction here in the second carbon. Now, of course, we have four carbons. I'll put the other two here and of course, just fulfilling its octet with all the leftover protons that we still have left. Alright. And third is just to basically complete our octet using lone pairs. So all of our hydrants are happy because of the duet rule. All of our carbons are happy because of the octet rule. But the auction's here in our case, still are not satisfied yet. So auctions like to have two bonds and two lone pairs. So far Carbonell group, which already has a double bond, which is going to go ahead and add two more lone pairs. Now, for our cyclic auction here, this one has two bonds as well instead of double bonds, it has two separate single bonds. And again, recently go ahead and add two lone pairs. So that right there, these three charters are going to be my final answer for this problem.

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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
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14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
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18. Aromaticity2h 34m
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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
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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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5:51 minutes

Problem 18b

Textbook Question

Draw three Lewis structures for C₃H₈O.

Verified step by step guidance

Step 1: Understand the molecular formula C3H8O. This indicates the molecule contains 3 carbon (C) atoms, 8 hydrogen (H) atoms, and 1 oxygen (O) atom. The task is to draw three possible Lewis structures that satisfy this formula.

Step 2: Recall that carbon typically forms 4 bonds, hydrogen forms 1 bond, and oxygen typically forms 2 bonds. Use this information to arrange the atoms in different configurations while ensuring all atoms satisfy their valency.

Step 3: Start with the first structure: Arrange the three carbon atoms in a straight chain (linear structure). Attach the oxygen atom to one of the carbons, and distribute the hydrogen atoms around the carbons and oxygen to satisfy their valency.

Step 4: For the second structure, consider a branched configuration. Place two carbons in a chain and attach the third carbon as a branch to the middle carbon. Attach the oxygen atom to one of the carbons and distribute the hydrogen atoms accordingly.

Step 5: For the third structure, consider a functional group variation. Place the oxygen atom between two carbons (as an ether group), and arrange the remaining carbon and hydrogen atoms to satisfy the molecular formula and valency rules.

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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 in a molecule and the lone pairs of electrons that may exist. They use dots to represent electrons and lines to represent bonds between atoms. Understanding how to draw Lewis structures is essential for visualizing molecular geometry and predicting reactivity.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are crucial for forming chemical bonds. In the case of C3H8O, knowing the number of valence electrons for carbon, hydrogen, and oxygen helps in constructing accurate Lewis structures. Each atom's valence electrons determine how they will bond with each other.

Resonance Structures

Resonance structures are different ways of drawing the same molecule that illustrate the delocalization of electrons. For C3H8O, multiple Lewis structures can be drawn to represent different arrangements of electrons while maintaining the same connectivity. Recognizing resonance is important for understanding the stability and reactivity of the molecule.

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