Explanation for bond lengths in trans-hexatriene
$begingroup$
Hexatriene is an unsaturated hydrocarbon with six carbon atoms and five carbon-carbon bonds, three of which are double bonds.
However, the bond lengths of the $ce{C=C}$ bonds are not the same. The middle $ce{C=C}$ bond has a length of 137 pm while the $ce{C=C}$ bonds at the end of the molecule have lengths of 134 pm, the length of a standard $ce{C=C}$ bond. The two carbon-carbon single bonds are 146 pm long, also off from the standard 154 pm length of carbon-carbon single bonds.
Clayden's organic chemistry hints that the explanation has to do with the molecular orbits formed and the conjugation system in the molecule. However, I do not fully understand this explanation.
Why do these carbon-carbon bonds show this unusual bond length behavior? A thourough explanation using MO theory would be appreciated.
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.
organic-chemistry bond molecular-orbital-theory
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
$endgroup$
add a comment |
$begingroup$
Hexatriene is an unsaturated hydrocarbon with six carbon atoms and five carbon-carbon bonds, three of which are double bonds.
However, the bond lengths of the $ce{C=C}$ bonds are not the same. The middle $ce{C=C}$ bond has a length of 137 pm while the $ce{C=C}$ bonds at the end of the molecule have lengths of 134 pm, the length of a standard $ce{C=C}$ bond. The two carbon-carbon single bonds are 146 pm long, also off from the standard 154 pm length of carbon-carbon single bonds.
Clayden's organic chemistry hints that the explanation has to do with the molecular orbits formed and the conjugation system in the molecule. However, I do not fully understand this explanation.
Why do these carbon-carbon bonds show this unusual bond length behavior? A thourough explanation using MO theory would be appreciated.
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.
organic-chemistry bond molecular-orbital-theory
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
$endgroup$
add a comment |
$begingroup$
Hexatriene is an unsaturated hydrocarbon with six carbon atoms and five carbon-carbon bonds, three of which are double bonds.
However, the bond lengths of the $ce{C=C}$ bonds are not the same. The middle $ce{C=C}$ bond has a length of 137 pm while the $ce{C=C}$ bonds at the end of the molecule have lengths of 134 pm, the length of a standard $ce{C=C}$ bond. The two carbon-carbon single bonds are 146 pm long, also off from the standard 154 pm length of carbon-carbon single bonds.
Clayden's organic chemistry hints that the explanation has to do with the molecular orbits formed and the conjugation system in the molecule. However, I do not fully understand this explanation.
Why do these carbon-carbon bonds show this unusual bond length behavior? A thourough explanation using MO theory would be appreciated.
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.
organic-chemistry bond molecular-orbital-theory
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
$endgroup$
Hexatriene is an unsaturated hydrocarbon with six carbon atoms and five carbon-carbon bonds, three of which are double bonds.
However, the bond lengths of the $ce{C=C}$ bonds are not the same. The middle $ce{C=C}$ bond has a length of 137 pm while the $ce{C=C}$ bonds at the end of the molecule have lengths of 134 pm, the length of a standard $ce{C=C}$ bond. The two carbon-carbon single bonds are 146 pm long, also off from the standard 154 pm length of carbon-carbon single bonds.
Clayden's organic chemistry hints that the explanation has to do with the molecular orbits formed and the conjugation system in the molecule. However, I do not fully understand this explanation.
Why do these carbon-carbon bonds show this unusual bond length behavior? A thourough explanation using MO theory would be appreciated.
References
Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.
organic-chemistry bond molecular-orbital-theory
organic-chemistry bond molecular-orbital-theory
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
edited Dec 27 '18 at 0:23
orthocresol♦
39.4k7114241
39.4k7114241
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
asked Dec 26 '18 at 23:57
EthiopiusEthiopius
3671113
3671113
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):
I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
$endgroup$
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
2
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
3
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
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votes
$begingroup$
If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):
I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
$endgroup$
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
2
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
3
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
add a comment |
$begingroup$
If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):
I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
$endgroup$
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
2
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
3
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
add a comment |
$begingroup$
If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):
I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
$endgroup$
If you derive the π-type molecular orbitals of hexatriene, the three lower-energy MOs which are filled would look something like this (image from p 33 of Fleming's Molecular Orbitals and Organic Chemical Reactions, Reference Edition):
I suspect what Clayden is getting at is that in the second MO, there is some antibonding character between C3 and C4, whereas the C1/C2 and C5/C6 interaction is purely bonding.
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
answered Dec 27 '18 at 1:37
orthocresol♦orthocresol
39.4k7114241
39.4k7114241
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
2
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
3
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
add a comment |
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
2
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
3
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
$begingroup$
And this antibonding interaction between C3 and C4 would cause the double bond to have slight single bond character, which would explain the longer than usual bond length for the C3=C4 bond?
$endgroup$
– Ethiopius
Dec 27 '18 at 1:41
2
2
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
$begingroup$
Yes, pretty much. So it is something like a 1.99-bond, if that makes any sense. (That number's made up, of course.)
$endgroup$
– orthocresol♦
Dec 27 '18 at 1:43
3
3
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
$begingroup$
You do not have to make it up. You can work out net pi bonding in each linkage by multiplying the coefficients on each pair of bonded atoms, doubling, and adding up the results for all orbital. This gives 0.87 pi bond between C-1 and C-2, 0.48 between C-2 and C-3, and 0.78 between C-3 and C-4. Other bonds are determined by symmetry.
$endgroup$
– Oscar Lanzi
Dec 27 '18 at 12:09
add a comment |
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