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Is hybridization of the cyclopropyl anion sp5?
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Neil Farbstein
Is hybridization of the cyclopropyl anion sp5?
Background
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated.
The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.
See these earlier answers for interesting examples where the theorem is applied:
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom. In other words, we have increased the electron density in this carbon orbital. Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in. In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond. Hence, $$\lambda^2_{\ce{C-C}}>3.74$$$$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$
Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated.The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.
See these earlier answers for interesting examples where the theorem is applied:
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom. In other words, we have increased the electron density in this carbon orbital. Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in. In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond. Hence, $$\lambda^2_{\ce{C-C}}>3.74$$$$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.
I dont get the leap that s character will be removed from the bonds. I would assume the lone pair to be in a p like orbital, increasing s character in the bonds. Could you clarify that please.More
@Martin-マーチン Just like in ammonia where the molecule is approximately tetrahedral with the lone pair in an approximately $\ce{sp^3}$ orbital; this being a lower energy geometry than the molecule being planar with the lone pair in a p orbital.More
Background
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated. The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.
See these earlier answers for interesting examples where the theorem is applied:
The Question
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom. In other words, we have increased the electron density in this carbon orbital. Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in. In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond. Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$ Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.
Background
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist. It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated.The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.
See these earlier answers for interesting examples where the theorem is applied:
The Question
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°. From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom. In other words, we have increased the electron density in this carbon orbital. Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in. In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond. Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.
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