Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question
26
![**Question: Calculate the degree of unsaturation (DU) for the molecule with the formula of C₉H₁₀O₂.**
**Explanation:**
The degree of unsaturation provides insight into the structure of a molecule by indicating the total number of π bonds and rings. It is calculated using the formula:
\[
DU = \frac{2C + 2 + N - H - X}{2}
\]
Where:
- \( C \) = number of carbon atoms
- \( N \) = number of nitrogen atoms
- \( H \) = number of hydrogen atoms
- \( X \) = number of halogens (F, Cl, Br, I)
For the given formula C₉H₁₀O₂:
- \( C = 9 \)
- \( H = 10 \)
- \( N = 0 \)
- \( X = 0 \)
Oxygen does not affect the degree of unsaturation; thus, it is not included in the formula.
Plugging in the values:
\[
DU = \frac{2(9) + 2 - 10}{2}
\]
\[
DU = \frac{18 + 2 - 10}{2}
\]
\[
DU = \frac{10}{2} = 5
\]
Therefore, the degree of unsaturation for the molecule C₉H₁₀O₂ is 5. This indicates that the compound could have a combination of five rings and/or double bonds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ae82246-2ba7-4804-abb5-6328a4c1ee5a%2F1a0094cd-665d-421e-b0b4-a112ad8fd4ba%2F3zxd6ll_processed.png&w=3840&q=75)
Transcribed Image Text:**Question: Calculate the degree of unsaturation (DU) for the molecule with the formula of C₉H₁₀O₂.**
**Explanation:**
The degree of unsaturation provides insight into the structure of a molecule by indicating the total number of π bonds and rings. It is calculated using the formula:
\[
DU = \frac{2C + 2 + N - H - X}{2}
\]
Where:
- \( C \) = number of carbon atoms
- \( N \) = number of nitrogen atoms
- \( H \) = number of hydrogen atoms
- \( X \) = number of halogens (F, Cl, Br, I)
For the given formula C₉H₁₀O₂:
- \( C = 9 \)
- \( H = 10 \)
- \( N = 0 \)
- \( X = 0 \)
Oxygen does not affect the degree of unsaturation; thus, it is not included in the formula.
Plugging in the values:
\[
DU = \frac{2(9) + 2 - 10}{2}
\]
\[
DU = \frac{18 + 2 - 10}{2}
\]
\[
DU = \frac{10}{2} = 5
\]
Therefore, the degree of unsaturation for the molecule C₉H₁₀O₂ is 5. This indicates that the compound could have a combination of five rings and/or double bonds.
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