Draw two Lewis structures for C 2 H 6 O.

Hello, everyone in this video, we're gonna go ahead and draw to appropriate lower structures for the species given below the species here is C two H 40. So first we need to go ahead and calculate for the degree of saturation who will simplify that to B D O U. So we have a formula for this. This goes half times two C plus two plus N minus H minus X. So let's break this equation down a little bit. So see here represents our carbons are end, here represents our nitrogen, H represents our protons or hide Jen's. And lastly, our X here represents are hella Jen's. So using this equation here, we can go ahead and apply us to our molecule. So we have again half times to see. So according to our formula here, the number of carbons that we have is two plus two, then our nitrogen is here. We don't have any nitrogen. So that's just plus zero. And then our protons, we have four protons. So that's -4. And lastly for our intelligence here that we have zero. So that's just zero. Alright. So if I put everything into my calculator. The total that I get here is just one. So we have one degree of saturation. So they given molecular formula corresponds to a structure which is one element of saturation which can be ring or a double bond. So they draw so to draw to appropriate lower structures for the given molecular formula again of C two H four oh This is going to be consisting of two or three total steps. First being to determine the total number of valence electrons that corresponds to a given formula. So we can go ahead and sort of calculate this here. So starting off with my carbon atoms here, we see that again, we have two atoms of carbon, each carbon atom will contribute for total fans electrons. So two times four is equal to eight. So we get a total of eight valence electrons coming off of just from the carbons. Next is going to be our Hae Jin's. We again see four atoms of hydrogen, each will give us one valence electrons. So four times one is just equal to itself in their case, it's four. So four valence electrons. And lastly, we have auction, we just have one atom of this. Each one will give us six total valence electrons. So one time six of course, that is just six, just adding everything up eight plus four plus six that gives us a total of 18 valence electrons in our molecule. So the second step is to draw an appropriate sigma bond framework to possible sigma bond frameworks with one element of saturation are going to be possible. So we can go ahead and draw that out. Starting off with RCH three at the end of our molecule. Then we have, let's see. So we do have two carbons in our structure. So the only way that we have this auction here is a double bond auction and of course, adding the last proton there and the next one would be a cyclic ring. So we have a three member ring, one being auction and two being the carbons. So auction does like to have two bonds and carbon likes to have four bonds. So we can go ahead and fill up those other bonds with just protons or the hydrants. Alright. So last step here is to complete the octet using lone pairs. We can see here that all the carbons and all the hydrogen are already fulfilled for their octet or their duet role. But for the auctions here, we see that they do each have two bonds for the carbon Neil group. Here we have a double bound. And then for cyclic ring here we have two separate single bonds. So auctions do like to have two single or two bonds and two lone pairs. So we need to add two lone pairs to both of our lower structures here. Once we put in those lone pairs, we have completed our structures for our species of C two H 40 Go and Hide. That is my final answer for this question.

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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
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12. Alcohols, Ethers, Epoxides and Thiols2h 42m
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14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
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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
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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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4:20 minutes

Problem 18a

Textbook Question

Draw two Lewis structures for C₂H₆O.

Verified step by step guidance

Step 1: Identify the molecular formula C₂H₆O and determine the total number of valence electrons. Carbon (C) has 4 valence electrons, hydrogen (H) has 1 valence electron, and oxygen (O) has 6 valence electrons. Calculate the total valence electrons: (2 × 4) + (6 × 1) + (1 × 6) = 20 valence electrons.

Step 2: Recognize that C₂H₆O can form two distinct isomers: ethanol (an alcohol) and dimethyl ether (an ether). These isomers differ in the connectivity of their atoms.

Step 3: For ethanol, arrange the atoms such that one carbon is bonded to three hydrogens, the second carbon is bonded to two hydrogens and an oxygen, and the oxygen is bonded to a hydrogen. Ensure all atoms satisfy the octet rule (or duet rule for hydrogen).

Step 4: For dimethyl ether, arrange the atoms such that the oxygen is bonded to two carbons, and each carbon is bonded to three hydrogens. Again, ensure all atoms satisfy the octet rule (or duet rule for hydrogen).

Step 5: Draw the Lewis structures for both isomers. For ethanol, the structure will show a C-C-O-H backbone with appropriate single bonds and lone pairs on oxygen. For dimethyl ether, the structure will show a C-O-C backbone with appropriate single bonds and lone pairs on oxygen.

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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 in determining how atoms bond with each other. In the case of C2H6O, counting the total number of valence electrons helps in constructing accurate Lewis structures. Carbon has four, hydrogen has one, and oxygen has six, leading to a total of 14 valence electrons for C2H6O.

Resonance Structures

Resonance structures are different ways of drawing the same molecule that show the delocalization of electrons. For C2H6O, drawing multiple Lewis structures can illustrate how electrons are shared or distributed differently among atoms. Recognizing resonance is important for understanding the stability and reactivity of molecules.

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