Ch. 14 - Structural Identification 1: Infrared Spectroscopy and Mass Spectrometry

Mullins - Organic Chemistry: A Learner Centered Approach 1st Edition

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Mullins 1st Edition

Ch. 14 - Structural Identification 1: Infrared Spectroscopy and Mass Spectrometry

문제 46f

13장, 문제 46f

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

검증된 단계별 안내

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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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.

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