Ch. 2 - Acids and Bases; Functional Groups

Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook

Wade 9th Edition

Ch. 2 - Acids and Bases; Functional Groups

Problem 29e

Chapter 2, Problem 29e

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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Related Practice

Textbook Question

Which of the following pure compounds can form hydrogen bonds? Which can form hydrogen bonds with water? Which ones do you expect to be soluble in water?

a. (CH₃CH₂)₂NH

b. (CH₃CH₂)₃N

c. CH₃CH₂CH₂OH

d. (CH₃CH₂CH₂)₂OH

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Textbook Question

The C≡N triple bond in acetonitrile has a dipole moment of about 3.6 D and a bond length of about 1.16 Å. Calculate the amount of charge separation in this bond. How important is the charge-separated resonance form in the structure of acetonitrile?

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Textbook Question

Which of the following pure compounds can form hydrogen bonds? Which can form hydrogen bonds with water? Which ones do you expect to be soluble in water?

e. CH₃(CH₂)₃CH₃

f. CH₂=CH—CH₂CH₃

g. CH₃COCH₃

h. CH₃CH₂COOH

2805

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Textbook Question

Sulfur dioxide has a dipole moment of 1.60 D. Carbon dioxide has a dipole moment of zero, even though C―O bonds are more polar than S―O bonds. Explain this apparent contradiction.

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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.

995

views

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.

a. CH₃CH=NCH₃

b. CH₃CH₂OH

c. CBr₄

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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 video answer for a similar problem:

Key Concepts

Lewis Structures

Bond Dipole Moments

Molecular Dipole Moment

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.