Draw a Lewis structure for each of the following: 4. CH 3 CONH 2

Hello everyone. In this video, we want to draw the correct Lewis structure for this molecule here. So the Lewis dot Structure is a representation of a molecule including all the atoms bonds, lone pairs and formal charges. It basically represents all the valence electrons and the molecule. So to draw the correct Lewis dot Structure, we have to first calculate the total number of valence electrons. So carbon has four valence electrons, Hydrogen has one valence electron Nitrogen has five valence electrons. Then we have auction and this has six valence electrons. So the total number of valence electrons, we can go ahead and calculate it with this provided information here. So give it to were calculated for the total number of valence electrons in our molecule. So if you count the number of carbons that we have, so we see 123 and four, so four times four. So one of these Ford's represents that we have four carbon atoms in this molecule and then we have four valence electrons coming off one atom of carbon. So we do four times four. Then next, we're gonna go ahead and count our hydrogen. I see a total of nine hydrogen and each will provide one valence electron. Then we can go ahead and count next one being nitrogen. I can see we have just one nitrogen atom right over here. So that's just five times one. And lastly for our auction, I see only one as well. So that's going to be six times one. So putting this into my calculator, I get a total sum of 36 valence electrons. So now next step is to draw the single bonding structure of this molecule. So we are giving this compact formula here. So it makes it a little bit easier for us. So starting off our molecule, we have our terminal ch three group. So that's one carbon with three hydrogen is connected to it that we have a central carbon with two hydrogen followed by another carbon. So we see we just have a carbon and auction. So that will give us a double bond auction there. So a carbon Neil functional group is connected to a nitrogen. I apologize. We are doing the single bonded version first. So we have a single bound auction, single bound nitrogen with a proton there. And then following the rest of the molecule, we C ch three group. So that's another terminal siege three group, same thing on the other side of the molecule. Alright. So now that we have the single bond drive the molecule, we can go ahead and now complete the octaves by adding the long pairs and double bonds. So carbon usually likes to have four bonds. So it's usually a central atom auction likes to have two long pairs and two bonds. So we see here, if we add those, now, this central carbon or the carbon carbon will have fulfilled its octet as well as this carbon auction there. As for our nitrogen, this has three bonds and it does like to have one lone pair. And we see now that the nitrogen does have its octet being filled. We, as we recall, hydrogen only needs to have two ways electrons. So it requires only just one bond because of the duet role. So we can go ahead and add formal charges two items. If the total number of valence electrons are not equal to the sum of the bonded pair of electrons and lone pairs and to calculate formal charges, go ahead and write this formula apps. So we can sure formula charge or formal charge just F C. So the equation for that is we take the number of valence electrons of the atom minus the number of electrons in lone pairs plus the number of bonds that is forming. And that gives us our equation to calculate our formal charge. But ask the question asking us is to, is to draw the correct Lewis structure for this molecule here. But I've highlighted just now in flu is going to be my final answer for this Lewis structure for this compact molecule there.

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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
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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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4:57 minutes

Problem 77a(4)

Textbook Question

Draw a Lewis structure for each of the following:
4. CH₃CONH₂

Verified step by step guidance

Step 1: Identify the total number of valence electrons in the molecule. Carbon (C) has 4 valence electrons, hydrogen (H) has 1 valence electron each, oxygen (O) has 6 valence electrons, and nitrogen (N) has 5 valence electrons. Add these together for all the atoms in CH3CONH2.

Step 2: Determine the connectivity of the atoms. The molecule consists of a methyl group (CH3) attached to a carbonyl group (C=O), which is further attached to an amide group (NH2). The central structure is CH3-C(=O)-NH2.

Step 3: Place single bonds between the atoms to connect them as described in Step 2. Use a single bond between the CH3 group and the carbonyl carbon, a double bond between the carbonyl carbon and oxygen, and a single bond between the carbonyl carbon and the nitrogen atom of the NH2 group.

Step 4: Distribute the remaining valence electrons to satisfy the octet rule for each atom. Start by placing lone pairs on the oxygen atom to complete its octet, then place lone pairs on the nitrogen atom. Ensure that the carbon atoms have a full octet by sharing electrons through bonds.

Step 5: Verify the structure by counting the total number of valence electrons used and ensuring it matches the total calculated in Step 1. Also, confirm that all atoms (except hydrogen) satisfy the octet rule and that the formal charges are minimized.

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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.

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 CH3CONH2, the amide functional group (–CONH2) is crucial for understanding its properties and reactivity. Recognizing functional groups helps in predicting the behavior of organic compounds.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are involved in forming bonds with other atoms. The number of valence electrons determines how atoms will bond and the overall structure of the molecule. For CH3CONH2, counting the valence electrons is necessary to accurately draw its Lewis structure and ensure that all atoms achieve a stable electron configuration.

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