Draw a Lewis structure for each of the following: b. CH 3 OCH 3

Hi, everyone. And welcome back. Our next problem says, give the Lewis structure of CH three C H2O, CH two CH three. So we're given a condensed structure where we see kind of the order in which our atoms are. So we know how they're connected to each other, but we don't know what type of bonds connect them if there's lone pairs or anything like that. And we need to go to a Lewis structure which shows bonds, the lone pairs and formal charges on all of our atoms. So let's think about what our molecule looks like. We have two carbons and then oxygen and then two more carbons. We have an ether here. So we know we have carbon, carbon, oxygen, carbon, carbon, all connected by single bonds. And then we have our going left to right, our terminal carbon has three oxygens. So we put them on there. Next one is CH two. So we add two hydrogens there. Ro doesn't have anything connected to it. Then ch two has two hydrogens, NCH three with those three hydrogens on there. So now we have how everything will go together. We need to look at how many valence electrons overall do we have and does our structure fit this? So total valence electrons? So we have 1234 carbons. So before carbons multiplied by four valance electrons plus we have three plus two is five plus two, is seven plus three, is 1010 hydrogens. So 10 hydrogens multiplied by one valence electron. And then we have our one oxygen that would be six valence electrons for that. So we'd have four, multiplied by four, is 16 plus 10 plus six equals 32 total valence electrons. So let's look at our structure and count up the bonds that we have here. So we have 123456789, 1011, 1213, 14. So total balance electrons, we have 14 bonds multiply by two, which would equal 28. So we know we're missing two as this is drawn or missing four, excuse me, four electrons as this is drawn. So let's look at where those missing electrons might go. Who has a completed octet? And who doesn't hear? Well, the hydrogens all have their single bond. So they've got all the electrons they need our carbons, each of them has four bonds. So for carbon, we've got four bonds multiplied by two electrons equals eight electrons. So they've got a complete octet, but our oxygen only has two bonds. So two bonds multiplied by two electrons equals four electrons. So oxygen must have two lone pairs since there are four more electrons. Well, there's the missing four electrons that we need. So put two lone pairs on our oxygen. And now the only thing left is to check if we need to add a formal charge anywhere. So for carbon hydrogens, all of them are single bonds. So hydrogen of course, has one valence electron, one bond, one minus one is zero. So our hydrogens is drawn, all have a former charge of zero carbon formal charge would equal the number of valence electrons which is four minus the number of non bonding electrons, which is zero minus the number of bonds, which is four, which equals zero. So no formal charge. In this case, the oxygen, its formal charge, we would have the number of valence electrons as six minus we have two lone pairs. So four non bonding electrons minus two bonds, six minus four minus two is zero. So the oxygen has no formal charge. So none of these atoms here has a formal charge. So we're done with our, with this structure just highlight that we've got all our carbons with their accompanying hydrogens. We know how this is connected to form ether and our oxygen has two lone pairs added to it in the structure. See you next video.

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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

Condensed Structural Formula

Organic Chemistry

1. A Review of General Chemistry

Condensed Structural Formula

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5:24 minutes

Problem 58b

Textbook Question

Draw a Lewis structure for each of the following:
b. CH₃OCH₃

Verified step by step guidance

Step 1: Count the total number of valence electrons in the molecule. Carbon (C) has 4 valence electrons, hydrogen (H) has 1 valence electron, and oxygen (O) has 6 valence electrons. Add these together for all the atoms in CH₃OCH₃.

Step 2: Determine the central atoms and the general structure. In CH₃OCH₃, the oxygen (O) atom is the central atom, with two methyl groups (CH₃) attached to it. Arrange the atoms so that the oxygen is in the center, and each carbon is bonded to the oxygen.

Step 3: Place single bonds between the central oxygen atom and the two carbon atoms. Also, place single bonds between each carbon atom and its three hydrogen atoms. Each single bond represents two electrons.

Step 4: Distribute the remaining valence electrons around the oxygen atom to satisfy its octet rule. Oxygen typically needs 8 electrons in its valence shell, so place lone pairs on the oxygen atom as needed.

Step 5: Verify that all atoms satisfy the octet rule (or duet rule for hydrogen) and that the total number of valence electrons used matches the total calculated in Step 1. Adjust the structure if necessary to ensure all requirements are met.

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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 organic molecules, the arrangement of valence electrons influences the structure and stability of the compound. For CH3OCH3, knowing the number of valence electrons for carbon, hydrogen, and oxygen is key to constructing its Lewis structure.

Functional Groups

Functional groups are specific groups of atoms within molecules that are responsible for the characteristic chemical reactions of those molecules. In the case of CH3OCH3, the presence of the ether functional group (–O–) is significant. Recognizing functional groups helps in understanding the properties and reactivity of organic compounds.

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