Structure-Property Relationships - Revision history

Cmditradmin at 22:22, 3 November 2011

2011-11-03T22:22:39Z

← Older revision		Revision as of 15:22, 3 November 2011
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	In the bottom structure we replace the C=O with a C=N+ (an iminium group) which is a stronger charge acceptor. The electrons move from one side to the other. The structure on the right is an identical to the structure on the left. The charge will be equally distributed between the two nitrogens resulting in 1.5 bonds between all carbons. The two forms have the same energy. When the two resonance forms contribute equally to the ground-state structure, the molecule exhibits essentially no BLA. This zero BLA limit is the so-called cyanine limit. The position of the electron will be more sensitive to the electric field in the cyanine case; this is also known as polarizability.		In the bottom structure we replace the C=O with a C=N+ (an iminium group) which is a stronger charge acceptor. The electrons move from one side to the other. The structure on the right is an identical to the structure on the left. The charge will be equally distributed between the two nitrogens resulting in 1.5 bonds between all carbons. The two forms have the same energy. When the two resonance forms contribute equally to the ground-state structure, the molecule exhibits essentially no BLA. This zero BLA limit is the so-called cyanine limit. The position of the electron will be more sensitive to the electric field in the cyanine case; this is also known as polarizability.

	Second order optical materials ~~requires~~ asymmetrical polarizability. Neither the first case with total transfer (polyene limit) and or the last case with equal transfer (cyanine limit) will have asymmetry; they will not have no β. The middle molecule will has some asymmetrical polarizability. Starting at the polyene limit or the cyanine limit you can add weak acceptors and thereby increase the asymmetry, and the β increases. The curve of β vs bond length alternation will start at zero, go up to a peak and then go back down to zero. For example, with stilbene and paranitroaniline and add stronger donors and acceptor groups, or make the molecules longer, you can increase the β. However, past a certain point stronger donors and acceptors simply results in a zwitterion and the polarizability decreases. The optimal point can be described and predicted this using quantum mechanics and four orbitals.		Second order optical materials require asymmetrical polarizability. Neither the first case with total transfer (polyene limit) and or the last case with equal transfer (cyanine limit) will have asymmetry; they will not have no β. The middle molecule will has some asymmetrical polarizability. Starting at the polyene limit or the cyanine limit you can add weak acceptors and thereby increase the asymmetry, and the β increases. The curve of β vs bond length alternation will start at zero, go up to a peak and then go back down to zero. For example, with stilbene and paranitroaniline and add stronger donors and acceptor groups, or make the molecules longer, you can increase the β. However, past a certain point stronger donors and acceptors simply results in a zwitterion and the polarizability decreases. The optimal point can be described and predicted this using quantum mechanics and four orbitals.

	=== Electric Field Perturbation of Structure ===		=== Electric Field Perturbation of Structure ===
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	:<math>\alpha \propto \left ( \frac {\mu^{2}_{ge}}{E_{ge}}\right )\,\!</math>		:<math>\alpha \propto \left ( \frac {\mu^{2}_{ge}}{E_{ge}}\right )\,\!</math>

	A linear ~~polarizablity~~ α(alpha) vs BOA curve starts small, increases to a maximum and then goes down. A graph of polarizability vs charge is a sigmoidal curve so the maximum slope of such a curve occurs near the inflection point where BOA is zero. The polarizability is highest at the cyanine limit, and the hyperpolarizability is zero at the cyanine limit. Rate of change of polarizability is zero at the cyanine limit and from a particle- in-a-box perspective the molecule does not have an asymmetrical polarizability. Thus without any sum-over-states calculations you can draw conclusions about the linear polarizability and hyperpolarizability using only ground state dipole moment calculations.		A linear polarizability α(alpha) vs BOA curve starts small, increases to a maximum and then goes down. A graph of polarizability vs charge is a sigmoidal curve so the maximum slope of such a curve occurs near the inflection point where BOA is zero. The polarizability is highest at the cyanine limit, and the hyperpolarizability is zero at the cyanine limit. Rate of change of polarizability is zero at the cyanine limit and from a particle- in-a-box perspective the molecule does not have an asymmetrical polarizability. Thus without any sum-over-states calculations you can draw conclusions about the linear polarizability and hyperpolarizability using only ground state dipole moment calculations.

	In a sum-over-states calculation the linear polarizability is defined as the sum over the square of all of the transition dipole moments, between the ground and any of the excited states, divided by the energy gap. The result is the red curve which peaks at the cyanine limit. Spectroscopically molecules like these usually have a single strong peak which has a transition dipole moment that is larger than others and that peak tends to be the lowest energy, resulting in a larger energy gap.		In a sum-over-states calculation the linear polarizability is defined as the sum over the square of all of the transition dipole moments, between the ground and any of the excited states, divided by the energy gap. The result is the red curve which peaks at the cyanine limit. Spectroscopically molecules like these usually have a single strong peak which has a transition dipole moment that is larger than others and that peak tends to be the lowest energy, resulting in a larger energy gap.

	In a two state model we choose to ignore the contribution from all terms except the one that give rise to the strong peak. α is the transition dipole moment squared between the ground and one excited state, over the energy gap between these two states results in the blue open circles. This matches the red curve very well so the two state model seems to capture the essence of the relationship. The question is which term is responsible for the peak; is it the transition dipole moment (blue triangle) or one over the energy gap (blue square)? Calculating and plotting these two terms shows that both peak near the cyanine limit. As a result the polarizabilty peaks at the cyanine limit.		In a two state model we choose to ignore the contribution from all terms except the one that give rise to the strong peak. α is the transition dipole moment squared between the ground and one excited state, over the energy gap between these two states results in the blue open circles. This matches the red curve very well so the two state model seems to capture the essence of the relationship. The question is which term is responsible for the peak; is it the transition dipole moment (blue triangle) or one over the energy gap (blue square)? Calculating and plotting these two terms shows that both peak near the cyanine limit. As a result the polarizabilty peaks at the cyanine limit.
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	First, there is a Coulombic term that is destabilizing when charge is separated. Adding donors and acceptors does encourage separation of charge but this also costs energy. Increasing the strength of the donor and acceptor end groups, and/or placing the molecule in a more polar solvent to the molecule can each lead to stabilization of the charge separated form.		First, there is a Coulombic term that is destabilizing when charge is separated. Adding donors and acceptors does encourage separation of charge but this also costs energy. Increasing the strength of the donor and acceptor end groups, and/or placing the molecule in a more polar solvent to the molecule can each lead to stabilization of the charge separated form.

	In addition, there is also an energy consideration associated with the topology of the molecule.If a molecule has 6 π-electrons in a ring the molecule has an additional resonance stabilization and is referred to as aromatic. In the case of stilbene with added donor and acceptors there is a loss of aromaticity, (a loss of 36kcal/mol). This favors the molecule that is aromatic.		In addition, there is also an energy consideration associated with the topology of the molecule. If a molecule has 6 π-electrons in a ring the molecule has an additional resonance stabilization and is referred to as aromatic. In the case of stilbene with added donor and acceptors there is a loss of aromaticity, (a loss of 36kcal/mol). This favors the molecule that is aromatic.

	For molecules whose neutral forms are aromatic, charge separation will interrupt the aromaticity and yield structure with a higher energy quinoidal resonance form.		For molecules whose neutral forms are aromatic, charge separation will interrupt the aromaticity and yield structure with a higher energy quinoidal resonance form.

Cmditradmin: /* Electric Field Perturbation of Structure */

2010-11-29T18:54:35Z