It is well known that double bonds are "stiff" and make particles less likely to rotate around the bond. But why? What makes it so that the bond has to be orientated a specific way?
$\begingroup$
$\endgroup$
3
asked Jul 4, 2016 at 22:02
-
1$\begingroup$ Reduce overlap. $\endgroup$– RodriguezCommented Jul 4, 2016 at 22:05
-
5$\begingroup$ What do you know about how the second bond of a double-bond forms? $\endgroup$– NijCommented Jul 4, 2016 at 23:06
-
$\begingroup$ youtu.be/nxVO2cz4OCM $\endgroup$– KaushikCommented Sep 21, 2019 at 8:31
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Besides the thermodynamic barrier with respect to double bond rotation, we could also look at it from a symmetry perspective.
Rotating a double bond, without breaking the $\pi$ bond first, requires that you rotate the left or right orbitals that participate in the $\sigma$ and $\pi$ bonds (but not both atoms' orbitals, because that would be rotating the molecule, not the bond).
Suppose that we are looking at a $2p_x-2p_x$ $\pi$ overlap for a $\text{C}=\text{C}$ bond. If we rotate the $2p_x$ orbital $90^o$ about the internuclear ($z$) axis, we can transform the $2p_x$ orbital into a $2p_y$ orbital. Ethene, for example, is a molecule of $D_{2h}$ symmetry, and contains a $\text{C}=\text{C}$ bond.
According to this D2h character table, in ethene, the $2p_x$ orbital transforms under the $B_{3u}$ IRREP, but the $2p_y$ orbital transforms under the $B_{2u}$ IRREP.
Two orbitals transforming under different IRREPs cannot overlap, so the $\pi$ bond can no longer be made if one of the orbitals is rotated.
Beyond that, if you rotate one of the $2p_x$ orbitals about the internuclear axis by some small angle (instead of specifically $90^o$), you would change the orientation of the x-axis for only that orbital, and they would technically not be the same orbital anymore because each would be following a different x- (and y-)axis convention.
-
$\begingroup$ What you say about symmetry is true, but in essence the lack of rotation is about the stability of the double bond vs. thermal energy available. The double bond's strength is far greater than this and so is stable wrt rotation. The bond has to have two similarly oriented p orbitals as these otherwise do not overlap, i.e. we say that px, py , pz etc are orthogonal to one another. $\endgroup$ Commented Jul 6, 2016 at 7:17
-
2$\begingroup$ It is quite easy to break a double bond with a photon of the appropriate energy. This promotes an electron into an 'antibonding' LUMO from which rotation is far easier as $\pi$ electron overlap is reduced. $\endgroup$ Commented Jul 6, 2016 at 11:26
-
$\begingroup$ @porphyrin therefore can we say that cis-trans forms are interconvertible at high temperatures $\endgroup$– LalitCommented Jul 28, 2021 at 15:32
-
$\begingroup$ Yes, technically they are interconvertible any any temperature but both the rate constants will become bigger as the temperature increases, as per Arrhenius for example. $\endgroup$ Commented Aug 2, 2021 at 9:11