The compound \(\mathrm{CH}_{3} \mathrm{CH}=(\mathrm{CHCHO}\) has a strong absorption in the ultraviolet at \(46950 \mathrm{~cm}^{-1}\) and a weak absorption at \(30000 \mathrm{~cm}^{-1}\) Justify these features in terms of the structure of the molecule.

Short Answer

Expert verified
The strong absorption at 46950 cm^{-1} is justified by π→π* transitions due to the double bond in the molecule, while the weak absorption at 30000 cm^{-1} is likely due to n→π* transitions from the non-bonding electrons of the oxygen in the CHO group.

Step by step solution

01

Understanding Electronic Transitions

Consider the molecular structure of the compound and possible electronic transitions that can occur due to ultraviolet (UV) light absorption. Strong absorption is often due to π→π* transitions, associated with double bonds, while weak absorption may be attributed to n→π* transitions in parts of the molecule where non-bonding electrons can be excited.
02

Identifying Transitions in Given Molecule

The compound has a double bond (C=C) as indicated by CH=(CHCHO. This double bond allows π→π* transitions. Additionally, oxygen in the CHO group has lone pairs that can undergo n→π* transitions. These are the likely transitions for the absorptions at 46950 cm^{-1} (strong) and 30000 cm^{-1} (weak).
03

Justifying the Absorption Wavelengths

The strong absorption at 46950 cm^{-1} corresponds to the energy difference between the ground state and the excited state of a π→π* transition, which are typically more energy-intensive due to the bonding π electrons. The weak absorption at 30000 cm^{-1} is likely an n→π* transition, which involves non-bonding electrons and typically requires less energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

π→π* Transition
When we talk about π→π* transitions, we are discussing a specific type of electronic transition that occurs in molecules containing pi (π) bonds — these are the bonds formed by the sideways overlap of p orbitals. During UV light absorption, electrons in the π bonding orbitals can be promoted (excited) to π* antibonding orbitals, which require a particular amount of energy.

In our molecule, CH3CH=(CHCHO, the presence of the C=C double bond signifies a locus for such a transition. The strong absorption peak at 46950 cm−1 in the UV spectrum is the fingerprint of a π→π* transition. This high-energy transition reflects the fact that π electrons are initially in a bonding state, and the jump to an antibonding (π*) state represents a significant alteration in electron density and energy within the molecule.

This type of transition is typically responsible for the vivid colors seen in organic compounds and is a crucial indicator while analyzing UV-Vis spectroscopy data for conjugated systems.
n→π* Transition
On the other side, we have the n→π* transitions. These occur when a nonbonding electron (n) absorbs energy and transitions to an antibonding pi (π*) orbital. Nonbonding electrons are found in atoms with lone pairs, such as the oxygen in the aldehyde group (CHO) of our example molecule.

The weaker absorption peak at 30000 cm−1 corresponds to this lower energy transition. Why is it lower energy? Unlike π→π* transitions, where the electron departs from a bonding molecular orbital, n→π* transitions involve nonbonding electrons. These electrons are not involved in bonding and are already at a higher energy state compared to bonding electrons. Thus, the energy required to promote them to the antibonding π* state is less.

Understanding n→π* transitions is especially important for compounds that contain atoms with spare lone pairs subject to excitation. These absorbances are pivotal for elucidating the structure of organic compounds with oxygen, nitrogen, or sulfur atoms.
UV Light Absorption
Diving deeper into the concept of UV light absorption, we discover that it's a fundamental process used in spectroscopy to study the electronic structure of molecules. When UV light is shone upon a sample, electrons within the molecule may absorb the energy and move to higher energy states. The specific wavelengths at which these absorptions occur can give us detailed insight concerning the molecule’s structure.

The energy of UV light absorbed is quantified by its wavelength or frequency. As seen in our problem, two distinct peaks at 46950 cm−1 and 30000 cm−1 are observed, each indicative of different transitions for electrons. The intensity of the peaks — strong for π→π* and weak for n→π* — also informs us about the probability and nature of these transitions.

Practically speaking, UV spectroscopy is an analytical method widely used for characterizing the presence of conjugated systems, functional groups, and overall molecular architecture, making it a cornerstone technique in both academic and industrial chemistry settings.

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Most popular questions from this chapter

A Birge-Sponer extrapolation yields \(7760 \mathrm{~cm}^{-1}\) as the area under the curve for the \(\mathrm{B}\) state of the oxygen molecule described in Problem \(14.1\). Given that the B state dissociates to ground-state atoms (at zero energy, \({ }^{3} \mathrm{P}\) ) and \(15870 \mathrm{~cm}^{-1}\left(\frac{55}{1} \mathrm{D}\right)\) and the lowest vibrational state of the \(\mathrm{B}\) state is 49363 \(\mathrm{cm}^{-1}\) above the lowest vibrational state of the ground electronic state, calculate the dissociation energy of the molecular ground state to the groundstate atoms.

The vibrational wavenumber of the oxygen molecule in its electronic ground state is \(1580 \mathrm{~cm}^{-1}\), whereas that in the first excited state \({ }^{\left(B^{\prime} \Sigma_{j}\right),}\), to which there is an allowed electronic transition, is \(700 \mathrm{~cm}^{-1}\) Given that the separation in energy between the minima in their respective potential energy curves of these two electronic states is \(6.175 \mathrm{eV}\), what is the wavenumber of the lowest energy transition in the band of transitions originating from the \(\mathrm{v}=\) 0 vibrational state of the electronic ground state to this excited state? Ignore any rotational structure or anharmonicity.

Consider some of the precautions that must be taken when conducting single- molecule spectroscopy experiments. (a) What is the molar concentration of a solution in which there is, on average, one solute molecule in \(1.0 \mu \mathrm{m}^{3}(1.0 \mathrm{fl} .)\) of solution? (b) It is important to use pure solvents in single-molecule spectroscopy because optical signals from fluorescent impurities in the solvent may mask optical signals from the solute. Suppose that water containing a fluorescent impurity of molar mass \(100 \mathrm{~g} \mathrm{~mol}^{-1}\) is used as solvent and that analysis indicates the presence of 0. \(10 \mathrm{mg}\) of impurity per \(1.0 \mathrm{~kg}\) of solvent. On average, how many impurity molecules will be present in \(1.0 \mu \mathrm{m}^{3}\) of solution? You may take the density of water as \(1.0 \mathrm{~g} \mathrm{~cm}^{-3} .\) Comment on the suitability of this solvent for single-molecule spectroscopy experiments.

The Beer-Lambert law states that the absorbance of a sample at a wavenumber \(\$ is proportional to the molar concentration \)[1]\( of the absorbing species \)\mathrm{J}\( and to the length 1 of the sample (eqn \)13.4)\(. In this problem you will show that the intensity of fluorescence emission from a sample of \)\mathrm{J}\( is also proportional to \)[\mathrm{J}]\( and \)l\(. Consider a sample of \)\mathrm{J}\( that is illuminated with a beam of intensity \){ }^{I_{0}(0)}\( at the wavenumber \){ }^{0 .}\( Before fluorescence can occur, a fraction of \)\mathrm{I}_{0}\left({ }^{0)}\right.\( must be absorbed and an intensity \){ }^{\prime(0)}\( 'will be transmitted. However, not all of the absorbed intensity is emitted and the intensity of fluorescence depends on the fluorescence quantum yield, \)\varphi_{\mathrm{f}}\(, the efficiency of photon emission. The fluorescence quantum yield ranges from 0 to 1 and is proportional to the ratio of the integral of the fluorescence spectrum over the integrated absorption coefficient. Because of a Stokes shift of magnitude \)\Delta \mathrm{v}_{\text {Slokes }}\(, fluorescence occurs at a wavenumber \)\mathrm{v}_{\mathrm{f}}\(, with \)\mathrm{v}_{\mathrm{f}}+\Delta \mathrm{v}_{\text {stokes }}=\mathrm{v} .\( It follows that the fluorescence intensity at \)\mathrm{v}_{\mathrm{f}}, \mathrm{I}_{\mathrm{f}}\left(\mathrm{v}_{\mathrm{f}}\right)\(, is proportional to \)\varphi_{\mathrm{f}}\( and to the intensity of exciting radiation that is absorbed by \)\mathrm{J}, \mathrm{I}_{\mathrm{abs}}(\mathrm{v})=\mathrm{I}_{0}(\mathrm{v})-\mathrm{I}(\mathrm{v})\(. (a) Use the Beer-Lambert law to express \)\mathrm{I}_{\mathrm{abs}}(\mathrm{v})\( in terms of \)\mathrm{I}_{0}(\mathrm{v}),[\mathrm{J}], \mathrm{I}\(, and \)\varepsilon(\mathrm{v})\(, the molar absorption coefficient of \)\mathrm{J}\( at \)\mathrm{V}\(. (b) Use your result from part (a) to show that \)\left.\mathrm{I}_{\mathrm{f}}\left(\mathrm{v}_{\mathrm{f}}\right) \infty \mathrm{I}_{0}(\mathrm{v}) \varepsilon(\mathrm{v}) \varphi_{\mathrm{f}} \mathrm{f}\right] \mathrm{I}$.

Use a group theoretical argument to decide which of the following transitions are electric-dipole allowed: (a) the \(\pi^{*} \leftarrow \pi\) transition in ethene, (b) the \(\pi^{*} \leftarrow \mathrm{n}\) transition in a carbonyl group in a \(\mathrm{C}_{2 \mathrm{v}}\) environment.

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