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

Electronegativity

Organic Chemistry

1. A Review of General Chemistry

Electronegativity

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5:31 minutes

Problem 29e

Textbook Question

For each of the following compounds,
1. draw the Lewis structure.
2. show how the bond dipole moments (and those of any nonbonding pairs of electrons) contribute to the molecular dipole moment.
3. estimate whether the compound will have a large, small, or zero dipole moment.
e.

Verified step by step guidance

Step 1: Draw the Lewis structure for the compound. Begin by identifying the central atoms and their connectivity. In this case, the central atoms are two carbon atoms connected by a double bond. Each carbon atom is bonded to two cyano groups (NC). Ensure that each atom satisfies the octet rule, with carbon having four bonds and nitrogen having three bonds.

Step 2: Identify the bond dipole moments. The cyano group (NC) is polar due to the difference in electronegativity between nitrogen and carbon. The dipole moment points from carbon to nitrogen. Since there are two cyano groups attached to each carbon, consider the vector sum of these dipole moments.

Step 3: Consider the molecular geometry. The compound is symmetrical with respect to the central C=C bond. The dipole moments of the cyano groups on each carbon are equal in magnitude but opposite in direction, which can lead to cancellation.

Step 4: Evaluate the overall molecular dipole moment. Due to the symmetrical arrangement of the cyano groups, the dipole moments may cancel each other out, resulting in a small or zero net dipole moment for the molecule.

Step 5: Estimate the dipole moment. Given the symmetry and the cancellation of dipole moments, the compound is likely to have a zero or very small dipole moment. This is typical for symmetrical molecules where dipole moments are oriented in opposite directions.

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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 are essential for visualizing the arrangement of atoms, the distribution of electrons, and the connectivity within a molecule. In the given compound, each carbon atom is bonded to two nitrogen atoms and one other carbon atom, with a double bond between the carbon atoms.

Bond Dipole Moments

Bond dipole moments occur due to differences in electronegativity between bonded atoms, resulting in a partial charge distribution. The direction of the dipole is from the less electronegative atom to the more electronegative atom. In the compound, the C≡N bonds have significant dipole moments due to the high electronegativity of nitrogen compared to carbon, influencing the overall molecular dipole moment.

Molecular Dipole Moment

The molecular dipole moment is the vector sum of all individual bond dipoles in a molecule. It determines the molecule's polarity, affecting its physical properties and interactions. For the given compound, the symmetry and arrangement of the C≡N bonds suggest that the individual dipole moments may cancel out, potentially resulting in a small or zero overall dipole moment.

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Textbook Question

For each alkene, indicate the direction of the dipole moment. For each pair, determine which compound has the larger dipole moment

b. cis-1,2-dibromoethene or trans-2,3-dibromobut-2-ene

Textbook Question

For each alkene, indicate the direction of the dipole moment. For each pair, determine which compound has the larger dipole moment.

c. cis-1,2-dibromo-1,2-dichloroethene or cis-1,2-dichloroethene

Textbook Question

The N—F bond is more polar than the N—H bond, but NF₃ has a smaller dipole moment than NH₃. Explain this curious result.

NF₃

μ= 0.2 D

NH₃

μ = 1.5 D

Textbook Question

For each of the following compounds,

1. draw the Lewis structure.

2. show how the bond dipole moments (and those of any nonbonding pairs of electrons) contribute to the molecular dipole moment.

3. estimate whether the compound will have a large, small, or zero dipole moment.

Textbook Question

The electrostatic potential maps for ammonia and water are shown here. The structure of ammonia is shown within its EPM. Note how the lone pair creates a region of high electron potential (red), and the hydrogens are in regions of low electron potential (blue). Show how your three-dimensional structure of water corresponds with its EPM.

Textbook Question

Why are Na⁺ and K⁺ unable to form covalent bonds?

Textbook Question

Show the ionic compound that you would expect to form between the given metal and nonmetal. Label the charges on each species.

(b) Mg and Br

Textbook Question

Give the charge most likely to result from ionization of the following metals.

(b) Aluminum

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