# When is rotation around double bonds to form cis-trans isomers allowed?

Textbook Question:

The rhodopsin molecules in the retina contain a derivative of vitamin A called cis-retinal (Figure 6). On exposure to light, cis-retinal becomes an isomer called trans-retinal.

This process helps cause an electrical impulse to be transferred to the brain. Around which numbered bonds does the isomerism rotation occur?

The answer is bond 11 (shown in red).

What I don't understand is: I thought rotation is not allowed around double bonds.

So when light hits the structure in Figure $$6$$, how does it cause rotation around bond $$11$$ to form the structure in Figure $$7$$?

## 2 Answers

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.

• I really appreciate the thorough explanation. It made me aware of other things I need to read up on such as triple bonds. I assumed they were just the same as double bonds and couldn't rotate. Dec 10, 2016 at 21:46
• @K-Feldspar You’re welcome. Yeah, the thing about triple bonds is interesting; you don’t really get to appreciate it with carbon because a $\ce{C#C}$ triple bond is always flanked by two single bonds. But it can be observed with metals, e.g. with $\ce{[Re2Cl8]}$ where a quadruple bond is present between the rhenium atoms. If an electron is excited from $\delta$ to $\delta^*$, the quadruple bond is broken and the resulting triple bond rotates by $45^\circ$ to prevent eclipsed chlorine atoms.
– Jan
Dec 11, 2016 at 0:43
• Do you have any thoughts on this animation demonstrations.wolfram.com/RotationAboutCarbonCarbonBonds You won't be able to see it as it requires the wolfram player download but here are some frames. i.imgur.com/b4PAxmI.png The double bond makes sense, as one carbon rotates the bond gets twisted and stuck. Dec 11, 2016 at 1:31
• But from the looks of things the triple bond should do the same. The issue is: it seems almost as if both carbons on either side of the triple bond are rotating (which suggests the bond itself isn't rotating but the single bond adjacent to it is?) Wouldn't the 3 lines from the triple bond get twisted and stuck the same way as in the double bond? In the animation they don't, and they freely rotate (i.e. one side of each line forming the triple bond rotates at the same rate the other side does so they never get twisted). Dec 11, 2016 at 1:31
• @K-Feldspar The issue is that you should draw it with orbitals.
– Jan
Dec 11, 2016 at 1:39

In rhodopsin this normally occurs only after the 11 cis in the chromophore (a retinylidene schiff base) is placed into an electronically excited singlet state after absorbing a visible photon. This excited state has one electron promoted from the HOMO orbital into the LUMO (antibonding) orbital of the conjugated $\pi$ system. Most data has been obtained from the protein bactoeriorhodposin, which acts a proton pump, as it is easier to use and has a similar chromophore to rhodopsin.

In the protein the rhodopsin molecule is somewhat strained by adjacent (tryptophan) and other amino acids. The isomerisation takes about 200fs so is extremely fast; among the fastest reaction ever studied. After that there is a series of at least 6 intermediates that slowly returns the trans molecule to the cis form and these thermal reactions take about a second or so.

The initial reaction is thought to proceed as follows; just after the electronically excited state is formed calculations indicate that the cis bond is rapidly stretched (~20 femtoseconds) and experiments show that a small barrier exists between the minimum energy of the excited state and the pathway to isomerisation. At or close to this barrier rotation & oscillation of the groups either side of the 'double' bond occur during isomerisation. (Kobayashi et.al Nature, 2001, v414,p531).

Many experiments have ben performed on trans/cis stilbene isomerisation which shows similar behaviour. Trans stilbene is the most stable isomer and can be converted into the cis also by photo excitation. The cis ground state has a relatively low barrier to the trans at room temperature (the opposite is not true) and so reverts to the trans spontaneously. This does not happen in rhodopsin so the protein has to go through a complex set of reactions.