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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![**Calculate the degree of unsaturation (DU) for a molecule with the formula of C₅H₁₁F.**
The degree of unsaturation (DU) is a calculation used in organic chemistry to determine the total number of rings and/or multiple bonds (double bonds and triple bonds) present in the molecular structure.
To calculate DU for a molecule with the formula C₅H₁₁F, use the following formula:
\[ \text{DU} = \frac{2C + 2 + N - H - X}{2} \]
where:
- \( C \) is the number of carbon atoms,
- \( N \) is the number of nitrogen atoms,
- \( H \) is the number of hydrogen atoms,
- \( X \) is the number of halogens (F, Cl, Br, I).
For C₅H₁₁F:
- \( C = 5 \)
- \( H = 11 \)
- \( X = 1 \)
- \( N = 0 \) (since there are no nitrogen atoms)
Substituting these values into the formula:
\[ \text{DU} = \frac{2(5) + 2 - 11 - 1}{2} = \frac{10 + 2 - 11 - 1}{2} = \frac{0}{2} = 0 \]
Therefore, the degree of unsaturation for C₅H₁₁F is 0, indicating that the molecule is fully saturated with no rings or double/triple bonds.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6987f833-598b-481a-acf8-6172d851b160%2F66585c74-88d5-4c78-a62c-464aefe603e8%2Fwjv2gk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calculate the degree of unsaturation (DU) for a molecule with the formula of C₅H₁₁F.**
The degree of unsaturation (DU) is a calculation used in organic chemistry to determine the total number of rings and/or multiple bonds (double bonds and triple bonds) present in the molecular structure.
To calculate DU for a molecule with the formula C₅H₁₁F, use the following formula:
\[ \text{DU} = \frac{2C + 2 + N - H - X}{2} \]
where:
- \( C \) is the number of carbon atoms,
- \( N \) is the number of nitrogen atoms,
- \( H \) is the number of hydrogen atoms,
- \( X \) is the number of halogens (F, Cl, Br, I).
For C₅H₁₁F:
- \( C = 5 \)
- \( H = 11 \)
- \( X = 1 \)
- \( N = 0 \) (since there are no nitrogen atoms)
Substituting these values into the formula:
\[ \text{DU} = \frac{2(5) + 2 - 11 - 1}{2} = \frac{10 + 2 - 11 - 1}{2} = \frac{0}{2} = 0 \]
Therefore, the degree of unsaturation for C₅H₁₁F is 0, indicating that the molecule is fully saturated with no rings or double/triple bonds.
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