2 replaced http://chemistry.stackexchange.com/ with https://chemistry.stackexchange.com/ edited Apr 13 '17 at 12:57 Generally, all single bonds are said to permit free rotation while all double bonds are said to inhibit free rotation. This can actually be explained numerically as well: the rotation around the $$\ce{O-O}$$ bond in $$\ce{H2O2}$$ has an approximate rate constant of $$2.0 \times 10^{8}~\mathrm{s^{-1}}$$ while the rotation around the $$\ce{C=C}$$ bond in ethene has $$1.3 \times 10^{-31}~\mathrm{s^{-1}}$$This can actually be explained numerically as well: the rotation around the $$\ce{O-O}$$ bond in $$\ce{H2O2}$$ has an approximate rate constant of $$2.0 \times 10^{8}~\mathrm{s^{-1}}$$ while the rotation around the $$\ce{C=C}$$ bond in ethene has $$1.3 \times 10^{-31}~\mathrm{s^{-1}}$$. This translates to an equilibration half-life of ethene which is greater than the age of the universe. Generally, all single bonds are said to permit free rotation while all double bonds are said to inhibit free rotation. This can actually be explained numerically as well: the rotation around the $$\ce{O-O}$$ bond in $$\ce{H2O2}$$ has an approximate rate constant of $$2.0 \times 10^{8}~\mathrm{s^{-1}}$$ while the rotation around the $$\ce{C=C}$$ bond in ethene has $$1.3 \times 10^{-31}~\mathrm{s^{-1}}$$. This translates to an equilibration half-life of ethene which is greater than the age of the universe. Generally, all single bonds are said to permit free rotation while all double bonds are said to inhibit free rotation. This can actually be explained numerically as well: the rotation around the $$\ce{O-O}$$ bond in $$\ce{H2O2}$$ has an approximate rate constant of $$2.0 \times 10^{8}~\mathrm{s^{-1}}$$ while the rotation around the $$\ce{C=C}$$ bond in ethene has $$1.3 \times 10^{-31}~\mathrm{s^{-1}}$$. This translates to an equilibration half-life of ethene which is greater than the age of the universe. 1 answered Dec 10 '16 at 0:59 Jan 51k77 gold badges128128 silver badges272272 bronze badges Generally, all single bonds are said to permit free rotation while all double bonds are said to inhibit free rotation. This can actually be explained numerically as well: the rotation around the $$\ce{O-O}$$ bond in $$\ce{H2O2}$$ has an approximate rate constant of $$2.0 \times 10^{8}~\mathrm{s^{-1}}$$ while the rotation around the $$\ce{C=C}$$ bond in ethene has $$1.3 \times 10^{-31}~\mathrm{s^{-1}}$$. This translates to an equilibration half-life of ethene which is greater than the age of the universe. It is, of course, helpful to understand why such a large barrier exists. This is due to the way double bonds are formed. ‘Normal’ single bonds are σ-symmetric overlaps of orbitals, i.e. they are necessarily totally symmetric with respect to rotation of the bond axis. Thus, any mechanisms that inhibit single bond rotations must derive from outside these bonds. The ‘second’ bon, however, is a π-symmetric bond, meaning that a plane of symmetry exists which is parallel to the bond axis (i.e. the bond axis is in said plane of symmetry). This automatically means that the bond is everything but symmetric with respect to rotation around the bond axis; if the p-orbitals do not align parallelly the overlap gets smaller and the bond starts to get broken. Note that triple bonds, which contain two perpendicular π bonds, are freely rotatable again since what is lost in one direction (e.g. vertical) is mathematically strictly gained in the other (e.g. horizontal). You can see the three types of bonds in the figure below. Figure 1: Single, double and triple $$\ce{C-C}$$ bonds. The single bond is formed by σ overlap of $$\mathrm{sp}^n$$ hybrid orbitals, double and triple bonds by π-symmetric p-orbital overlap. It may be immediately obvious that the overlap of a σ bond is generally much better than that of a π bond. Hence, the difference in orbital energies between σ and σ* is much greater than that between π and π*: π orbitals are less stabilised than σ orbitals. In turn, that means that the HOMO will generally be a π orbital (least stabilisation), while the LUMO will typically be a π* orbital (least destabilisation). Thus, the energy difference between π and π* is the smallest energy difference between an occupied and an unoccupied orbital. If a photon with the appropriate wavelength comes along, it can excite the π bonding electron into the antibonding π* orbital. (It also needs to come along from the appropriate direction — parallel to the $$\ce{C=C}$$ bond axis in ethene if I did my group theory correctly, but that is a minor issue.) If we excite $$1\unicode[Times]{x3c0}^2\,2\unicode[Times]{x3c0}^0 \ce{->} 1\unicode[Times]{x3c0}^1\,2\unicode[Times]{x3c0}^1$$, we have effectively reduced the bond order from $$1$$ to $$0$$, creating a ‘no bond’ where there used to be a ‘second single’ bond. If there is no overall bonding interaction, the rotation restrictions no longer apply. As long as one electron is excited into π*, the molecule is freely rotatable. When the electron finally relaxes, chances are that the geometry of the double bond has inverted. You can’t observe that in (symmetric) ethene, but other systems such as stilbene or hemithioindigo have been studied extensively. If you don’t wish to invoke molecular orbital theory, you can also explain the entire idea with Lewis structures. Under the Lewis formalism, an incoming photon ($$h\cdot \nu$$) will cleave the $$\ce{C=C}$$ double bond while leaving the underlying $$\ce{C-C}$$ single bond intact. It can be written as: $$\ce{-C=C- ->[h\cdot \nu] -C^.-C^.-{}}$$ Here too, you are making a double bond a single bond. The former cannot rotate, the latter is freely rotatable. Upon relaxation, whichever rotamere the molecule was in will be fixed again. Since the trans rotamer is more stable then the cis rotamer in almost all linear cases, that one will be accessed by irradiation. Tl;dr: Yes, double bonds are not freely rotatable, but by irradiation you are breaking the double bond into a formal single bond diradicalic structure.