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

15. Analytical Techniques:IR, NMR, Mass Spect

Infrared Spectroscopy

Organic Chemistry

15. Analytical Techniques:IR, NMR, Mass Spect

Infrared Spectroscopy

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6:01 minutes

Problem 46f

Textbook Question

Based on Hooke's law, choose the bond in each pair that you expect to vibrate at a higher wavenumber.
(f) C―S vs C=O

Verified step by step guidance

Understand Hooke's Law in the context of molecular vibrations: Hooke's Law relates the vibrational frequency of a bond to the masses of the atoms and the stiffness of the bond. The formula for the vibrational frequency (ν) is given by:

ν = \frac{1}{2 π} \sqrt{\frac{k}{μ}}

, where k is the force constant (bond stiffness) and μ is the reduced mass of the bonded atoms.

Compare the bond types: The C=O bond is a double bond, while the C―S bond is a single bond. Double bonds generally have higher force constants (k) compared to single bonds, meaning they are stiffer and vibrate at higher frequencies.

Consider the reduced mass (μ): The reduced mass is calculated using the formula:

μ = m

₁m₂m₁+m₂, where m₁ and m₂ are the masses of the bonded atoms. Oxygen is lighter than sulfur, which affects the reduced mass and thus the vibrational frequency.

Analyze the effect of bond stiffness and reduced mass: Given that the C=O bond is stiffer (higher k) and involves a lighter atom (lower μ), it will vibrate at a higher frequency compared to the C―S bond.

Conclude which bond vibrates at a higher wavenumber: Based on the analysis, the C=O bond is expected to vibrate at a higher wavenumber than the C―S bond due to its higher stiffness and lower reduced mass.

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Key Concepts

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

Hooke's Law in Vibrational Spectroscopy

Hooke's Law describes the relationship between the force needed to extend or compress a spring and the distance it is stretched or compressed. In vibrational spectroscopy, it is used to model the vibrational frequencies of chemical bonds, where the bond acts like a spring. The frequency of vibration is proportional to the square root of the bond strength and inversely proportional to the square root of the reduced mass of the bonded atoms.

Bond Strength and Wavenumber

The wavenumber, which is the reciprocal of the wavelength, is directly related to the vibrational frequency of a bond. Stronger bonds, such as double bonds, typically vibrate at higher frequencies and thus have higher wavenumbers compared to weaker, single bonds. This is because stronger bonds have higher force constants, leading to higher vibrational frequencies according to Hooke's Law.

Reduced Mass in Vibrational Frequency

Reduced mass is a concept used in the calculation of vibrational frequencies, representing the effective inertial mass appearing in the two-body problem of two atoms bonded together. It is calculated using the formula μ = (m1 * m2) / (m1 + m2), where m1 and m2 are the masses of the two atoms. A lower reduced mass results in a higher vibrational frequency, and thus a higher wavenumber, for a given bond strength.

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