Draw a Lewis Structure for each species. i. CH 3 OSO 2 OCH 3 j. CH 3 C(NH)CH 3 k. (CH 3 ) 3 CNO

Hello everyone. Today, we have the following problem provide the lowest structures of the molecules listed below. So for our first structure, we have three meal groups and we know that because we have a subscript of three. So what that means is that if we were to draw their Lewis structures, which involves drawing out bond lines and the connectivity of various atoms, we would note that the three methyl groups are also bound to a central carbon in the following manner. Now, as for that center carbon, we see that it is also bound to two additional methylene groups. And that just means that those car that those groups are carbons that have two hydrogens. And then last but not least the last methylene group to the right, is bound to an eo group or a nitrogen and oxygen group. So if we connect nitrogen and oxygen, we now had to take into account their bond preferences and how many bonds they like to, to, to make. So nitrogen prefers to make three, while oxygen prefers to make two, as we see, neither of them have fulfilled that preference. But if we simply add one bond line, we can fulfill nitrogen's preference. But also we have to make nope that each bottom line is two electrons. So, so far nitrogen has six electrons and it needs eight to fulfill octet rule. So we just add a lone pair on nitrogen. And then as for oxygen, we can add two lone pairs. Now, moving on to our second structure, we have to the right, but to the left, we have a methyl group bound to a methylene carbon. And that methylene carbon is bound to an additional carbon which has a, an NCH three or an N methyl group. So we will draw that in the foley manner so that nitrogen is bound to a methyl group as follows. And then that carbon that branched off with that nitrogen group is just bound to one other methyl group. Now, as we mentioned with octet rules, nitrogen prefers to have three bonds and it only has two in this case. Well, since the carbon it's bound to also needs an additional bo line, we can just draw one in between. And because nitrogen now only has six and it needs eight electrons, we just draw a loan pair on the nitrogen. And then last but not least, we have our third and final structure. We have a methyl group from the far left which is bound to a methylene carbon. And that methylene carbon is bound to an oxygen which is bound to a sulfur that has two oxygens. And an additional oxygen to the right of it. And then that oxygen. So the right of it is bound to a methylene carbon, which is bound to a methyl. Now, sulfur is an interesting atom in that it is in third row of the periodic table and therefore can expand its octet. So if we wanted to fulfill the oxygen's rule of not only needing two bonds, but eight electrons total, we can just draw two in two additional bonds. So from the sulfur to the oxygen, and then each oxygen to the side will get two pairs of lone pairs. And then the oxygens at the top of the double bond will get two oxygen as well. And with that, we have solved the problem overall, I hope this helped. And until next time.

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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
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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

Condensed Structural Formula

Organic Chemistry

1. A Review of General Chemistry

Condensed Structural Formula

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Problem 31i-k

Textbook Question

Draw a Lewis Structure for each species.
i. CH₃OSO₂OCH₃
j. CH₃C(NH)CH₃
k. (CH₃)₃CNO

Verified step by step guidance

Identify the total number of valence electrons for each species. For example, in CH3OSO2OCH3, count the valence electrons from each atom: C (4), H (1), O (6), and S (6).

Arrange the atoms to reflect the molecular structure. Typically, the least electronegative atom is central, except for hydrogen, which is always terminal. For CH3OSO2OCH3, consider the connectivity: CH3-O-SO2-O-CH3.

Distribute the valence electrons around the atoms, starting with the outer atoms, to satisfy the octet rule (or duet for hydrogen). For example, in CH3OSO2OCH3, ensure each oxygen has 8 electrons around it.

Form double or triple bonds if necessary to satisfy the octet rule for each atom. For instance, in CH3OSO2OCH3, you may need to form a double bond between sulfur and one of the oxygens.

Check the formal charges on each atom to ensure the most stable structure. Adjust the bonding if needed to minimize formal charges, ensuring the overall charge of the molecule is correct.

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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 help visualize the arrangement of electrons in molecules, showing how atoms are connected and the distribution of valence electrons. Understanding Lewis structures is crucial for predicting molecular geometry, reactivity, and properties.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are involved in forming chemical bonds. The number of valence electrons determines an atom's bonding behavior and is essential for drawing Lewis structures. For example, carbon has four valence electrons, allowing it to form four covalent bonds, which is key in organic molecules.

Formal Charge

Formal charge is a concept used to determine the charge distribution within a molecule. It is calculated by comparing the number of valence electrons in an isolated atom to the number assigned in a Lewis structure. Minimizing formal charges in a Lewis structure leads to the most stable configuration, which is important for accurately representing molecules like CH3OSO2OCH3, CH3C(NH)CH3, and (CH3)3CNO.

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