Why does benzene have 3 π–π* transitions in the UV range, but ethene, butadiene, and hexatriene each have...
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Are the π and π* orbitals of benzene much closer together in energy which allows for multiple transitions? Why is this not the case for hexatriene if they both contain the same number of pi bonds?
molecular-orbital-theory hydrocarbons symmetry group-theory uv-vis-spectroscopy
edited Nov 18 at 2:36
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Are the π and π* orbitals of benzene much closer together in energy which allows for multiple transitions? Why is this not the case for hexatriene if they both contain the same number of pi bonds?
molecular-orbital-theory hydrocarbons symmetry group-theory uv-vis-spectroscopy
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
asked Nov 18 at 2:19
ETS
574
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Are the π and π* orbitals of benzene much closer together in energy which allows for multiple transitions? Why is this not the case for hexatriene if they both contain the same number of pi bonds?
molecular-orbital-theory hydrocarbons symmetry group-theory uv-vis-spectroscopy
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
asked Nov 18 at 2:19
ETS
574
Are the π and π* orbitals of benzene much closer together in energy which allows for multiple transitions? Why is this not the case for hexatriene if they both contain the same number of pi bonds?
molecular-orbital-theory hydrocarbons symmetry group-theory uv-vis-spectroscopy
molecular-orbital-theory hydrocarbons symmetry group-theory uv-vis-spectroscopy
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
asked Nov 18 at 2:19
ETS
574
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
asked Nov 18 at 2:19
ETS
574
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
edited Nov 18 at 2:36
orthocresol♦
37.3k7109221
37.3k7109221
asked Nov 18 at 2:19
ETS
574
asked Nov 18 at 2:19
ETS
574
asked Nov 18 at 2:19
ETS
574
574
add a comment |
add a comment |
1 Answer
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It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.
If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:
The ground state has overall symmetry $mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(mathrm{e_{1g}})^3(mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:
$$mathrm{E_{1g} times E_{2u} = B_{1u} oplus B_{2u} oplus E_{1u}}$$
Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state
$$mathrm{A_{1g} to B_{1u}, quad A_{1g} to B_{2u}, quad A_{1g} to E_{1u}}$$
should fall within the UV range.
For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(pi)^3(pi^*)^1$ configuration will only correspond to one electronic state.
This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $langle i | hat{mu} | j rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.
If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:
The ground state has overall symmetry $mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(mathrm{e_{1g}})^3(mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:
$$mathrm{E_{1g} times E_{2u} = B_{1u} oplus B_{2u} oplus E_{1u}}$$
Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state
$$mathrm{A_{1g} to B_{1u}, quad A_{1g} to B_{2u}, quad A_{1g} to E_{1u}}$$
should fall within the UV range.
For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(pi)^3(pi^*)^1$ configuration will only correspond to one electronic state.
This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $langle i | hat{mu} | j rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
add a comment |
up vote
4
down vote
It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.
If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:
The ground state has overall symmetry $mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(mathrm{e_{1g}})^3(mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:
$$mathrm{E_{1g} times E_{2u} = B_{1u} oplus B_{2u} oplus E_{1u}}$$
Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state
$$mathrm{A_{1g} to B_{1u}, quad A_{1g} to B_{2u}, quad A_{1g} to E_{1u}}$$
should fall within the UV range.
For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(pi)^3(pi^*)^1$ configuration will only correspond to one electronic state.
This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $langle i | hat{mu} | j rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
add a comment |
up vote
4
down vote
up vote
4
down vote
It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.
If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:
The ground state has overall symmetry $mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(mathrm{e_{1g}})^3(mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:
$$mathrm{E_{1g} times E_{2u} = B_{1u} oplus B_{2u} oplus E_{1u}}$$
Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state
$$mathrm{A_{1g} to B_{1u}, quad A_{1g} to B_{2u}, quad A_{1g} to E_{1u}}$$
should fall within the UV range.
For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(pi)^3(pi^*)^1$ configuration will only correspond to one electronic state.
This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $langle i | hat{mu} | j rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
It seems that here we only need to consider the lowest energy transitions, i.e. HOMO–LUMO transitions. The difference between benzene and the linear polyenes is that both the HOMO and LUMO of benzene are doubly degenerate, whereas the linear polyenes do not have degeneracy.
If we think about this HOMO–LUMO excitation in the case of benzene, we can schematically represent it like this:
The ground state has overall symmetry $mathrm{A_{1g}}$ because all electrons are paired, but the first excited electronic configuration above $(mathrm{e_{1g}})^3(mathrm{e_{2u}})^1$ corresponds to three different electronic states, each with a slightly different spatial distribution of electrons. The symmetries of these three states are obtained by taking a direct product of the relevant irreps:
$$mathrm{E_{1g} times E_{2u} = B_{1u} oplus B_{2u} oplus E_{1u}}$$
Because these three states arise from the same electronic configuration, their energies are going to be fairly close to each other, and all three electronic transitions from the ground state
$$mathrm{A_{1g} to B_{1u}, quad A_{1g} to B_{2u}, quad A_{1g} to E_{1u}}$$
should fall within the UV range.
For any linear polyene (including ethene, butadiene, and hexatriene), both HOMO and LUMO are nondegenerate, so the $(pi)^3(pi^*)^1$ configuration will only correspond to one electronic state.
This is a rather simplified analysis, because we are only considering transitions which conserve spin, i.e. singlet–singlet transitions. These are so-called "spin-allowed" transitions. The singlet–triplet transitions also occur within the UV range, but are technically "forbidden". On top of that, not all of the three singlet–singlet transitions in benzene are equal: two of them are also "forbidden", as the transition dipole moment $langle i | hat{mu} | j rangle$ vanishes. But that is a story for another day. The keen reader may wish to check out J. Mol. Struct. 1973, 19, 143–166 for an assignment of the electronic transitions of benzene.
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
edited Nov 18 at 3:19
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
answered Nov 18 at 2:50
orthocresol♦
37.3k7109221
37.3k7109221
add a comment |
add a comment |
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