Table of contents

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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8:26 minutes

Problem 54a,b,c

Textbook Question

Predict the hybridization and geometry of the carbon and nitrogen atoms in the following molecules and ions. (Hint: Resonance.)
a.
b.
c.

Verified step by step guidance

For molecule (a) [CH3C(=O)CH2]-, identify the hybridization of each carbon atom. The first carbon (CH3) is sp3 hybridized due to its four single bonds. The second carbon (C=O) is sp2 hybridized because it forms a double bond with oxygen and a single bond with the adjacent carbon. The third carbon (CH2-) is sp3 hybridized as it has three single bonds and a lone pair, considering the negative charge.

For molecule (a), determine the geometry around each carbon atom. The first carbon (CH3) has a tetrahedral geometry due to its sp3 hybridization. The second carbon (C=O) has a trigonal planar geometry because of its sp2 hybridization. The third carbon (CH2-) also has a tetrahedral geometry due to its sp3 hybridization.

For molecule (b) [H2NCH=CHCH2]+, identify the hybridization of the nitrogen and carbon atoms. The nitrogen atom is sp3 hybridized as it forms three sigma bonds and has a lone pair. The first carbon (CH) is sp2 hybridized due to its double bond with the second carbon. The second carbon (CH) is also sp2 hybridized because of its double bond with the first carbon and a single bond with the third carbon. The third carbon (CH2)+ is sp2 hybridized due to the positive charge and its bonding.

For molecule (b), determine the geometry around each atom. The nitrogen atom has a trigonal pyramidal geometry due to its sp3 hybridization. The first and second carbons (CH=CH) have a trigonal planar geometry because of their sp2 hybridization. The third carbon (CH2)+ also has a trigonal planar geometry due to its sp2 hybridization.

For molecule (c) [CH2C≡N]-, identify the hybridization of the carbon and nitrogen atoms. The first carbon (CH2) is sp3 hybridized as it forms three sigma bonds and has a lone pair. The second carbon (C≡N) is sp hybridized due to its triple bond with nitrogen. The nitrogen atom is also sp hybridized because of its triple bond with carbon.

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Key Concepts

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

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that can accommodate the bonding requirements of atoms in a molecule. For carbon, common hybridizations include sp3, sp2, and sp, which correspond to different geometries: tetrahedral, trigonal planar, and linear, respectively. Understanding hybridization helps predict the shape and bond angles in molecules.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It is determined by the number of bonding pairs and lone pairs of electrons around the central atom, which influences the overall shape. Common geometries include linear, trigonal planar, tetrahedral, and trigonal bipyramidal, each with specific bond angles that can be predicted based on hybridization.

Resonance

Resonance is a phenomenon in which a molecule can be represented by two or more valid Lewis structures, known as resonance structures. These structures differ only in the placement of electrons, not in the arrangement of atoms. Resonance helps explain the delocalization of electrons in molecules, affecting their stability, reactivity, and hybridization, which is crucial for accurately predicting molecular geometry.

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Textbook Question

Predict the hybridization, geometry, and bond angles for the central atoms in

a. but-2-ene, CH₃CH=CHCH₃

Textbook Question

a. Which of the species have bond angles of 109.5°?

b. Which of the species have bond angles of 120°?

H₂O H₃O⁺ ⁺CH₃ BF₃

Textbook Question

Do the sp² carbons and the indicated sp³ carbons lie in the same plane?

Textbook Question

For each of the following molecules, indicate the hybridization of each carbon and give the approximate values of all the bond angles:

a. CH₃C≡CH

Textbook Question

In most amines, the nitrogen atom is sp³ hybridized, with a pyramidal structure and bond angles close to 109°. In urea, both nitrogen atoms are found to be planar, with bond angles close to 120°. Explain this surprising finding. (Hint: Consider resonance forms and the overlap needed in them.)

Textbook Question

Predict the hybridization, geometry, and bond angles for the carbon and nitrogen atoms in acetonitrile (CH₃–C≡N:).

Textbook Question

Describe the orbitals used in bonding and the bond angles in the following compounds:

e. BF₃

Textbook Question

Predict the approximate bond angles in

a. the methyl cation.

b. the methyl radical.

c. the methyl anion.

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