Predict the approximate bond angles: c. the C—N—H bond angle in (CH 3 ) 2 NH d. the C—N—C bond angle in (CH 3 ) 2 NH

All right. Hello everyone. So this question is asking us to estimate the values for the CNC and the CNH bond angles in the secondary amine diprop amine. Here we have four different choices labeled A through D proposing different values for both bond angles. These values include 109 or 109.5 degrees, 120 degrees or 107.3 degrees. So let's go ahead and get started by drawing out the lowest structure of diprop amine diprop amine is a secondary amine which if you recall has an initial Lewis structure in which nitrogen is bound to two R groups as well as one hydrogen atom and it has a lone pair of electrons. This is because nitrogen in order to fulfill its octet prefers to have three covalent bonds and one lone pair of electrons. Now because we're given a specific structure for this amine, we're going to replace the R groups with the carbon chains provided in the text of the question, it just so happens that the R groups are the same because we have two propel groups or three carbon chains on either side of nitrogen here. So when drawing out the lowest structure of an R group, the atom with the highest bond preference should go in the center carbon has a bond preference of four. Whereas hydrogen has a bond preference of only one because of this carbon should be in the center of the lowest structure of the art group. So carp in here is in the center and hydrogen atoms are connecting to each of the carbons. This allows for hydrogen to fulfill its one bond requirement and for carbon to fulfill its four bond requirement. So once again, because the R groups are the same, I'm going to go ahead and substitute the R on the right side with the same R group. So with this in mind, let's go ahead and talk about bond angles recall that bond angles are created between three atoms that make two bonds to determine the bond angle. We're going to look at the hybridization of the atom in the center of the bond angle or simply known as the point of reference. In both cases here, the atom in the center and therefore the point of reference is nitrogen. To understand the hybridization on nitrogen, we have to recall the number of quote unquote groups surrounding our target atom. In this case, nitrogen. When I refer to groups in this context, I am referring to two different categories, either loan pairs of electrons or and or I should say other atoms connecting to my atom of interest. So focusing our attention back on nitrogen, nitrogen has one lone pair as well as three atoms connecting directly to it. This makes a total of four groups, which if you recall corresponds to sp three hybridization. Now the bond angles are not only determined by hybridization but also the number of lone pairs surrounding a given atom. When an atom is sp three hybridized with zero lon pairs of electrons, then that bond angle is generally 109.5 degrees. But when you increase the number of lone pairs on that sp three hybridized central atom, the angle tends to go below 109.5 degrees to be specific. When one electron pair is present on an sp three hybridized atom, then the angle decreases somewhat to about 107.3 degrees instead. So this means that our answer is going to be option d in the multiple choice because nitrogen is in the center of both bond angles and nitrogen is sp three hybridized with one electron pair. Their bond angles are going to be 107.3 degrees for both the C and C bond and the C and H bond. And there you have it. So with that being said, thank you so very much for watching. And I hope you found this helpful.

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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

Molecular Geometry

Organic Chemistry

1. A Review of General Chemistry

Molecular Geometry

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6:00 minutes

Problem 53c,d

Textbook Question

Predict the approximate bond angles:
c. the C—N—H bond angle in (CH₃)₂NH
d. the C—N—C bond angle in (CH₃)₂NH

Verified step by step guidance

Step 1: Understand the molecular geometry around the nitrogen atom in (CH3)2NH. The nitrogen atom in this molecule is sp³ hybridized because it is bonded to two methyl groups (CH3), one hydrogen atom, and has one lone pair of electrons.

Step 2: Recall that sp³ hybridization typically results in a tetrahedral geometry with ideal bond angles of approximately 109.5°. However, the presence of a lone pair on nitrogen will cause slight deviations due to lone pair-bond pair repulsion being greater than bond pair-bond pair repulsion.

Step 3: For part (c), the C—N—H bond angle will be slightly less than 109.5° because the lone pair on nitrogen pushes the bonded atoms closer together, reducing the bond angle.

Step 4: For part (d), the C—N—C bond angle will also be slightly less than 109.5° for the same reason as in part (c). The lone pair on nitrogen exerts repulsion, compressing the bond angles between the bonded groups.

Step 5: Conclude that both the C—N—H and C—N—C bond angles in (CH3)2NH will be slightly less than the ideal tetrahedral angle of 109.5°, but the exact values depend on the specific molecular environment and steric effects.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) Theory is a model used to predict the geometry of individual molecules based on the repulsion between electron pairs in the valence shell of the central atom. According to VSEPR, electron pairs will arrange themselves to minimize repulsion, leading to specific bond angles characteristic of different molecular shapes.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate bonding. In the case of (CH3)2NH, the nitrogen atom undergoes sp3 hybridization, resulting in four equivalent hybrid orbitals that influence the bond angles around the nitrogen atom, typically leading to angles close to 109.5 degrees.

Steric Effects

Steric effects refer to the influence of the spatial arrangement of atoms in a molecule on its reactivity and geometry. In (CH3)2NH, the presence of two methyl groups (CH3) attached to the nitrogen creates steric hindrance, which can slightly alter the ideal bond angles due to the repulsion between the bulky groups, affecting the C—N—H and C—N—C angles.

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