(E)-stilbene may be separated by crystallization from (Z)-stilbene and the side product of the Wittig reaction, triphenylphospine oxide (TPPO), by using absolute ethanol. My question is why? (E)-stilbene and (Z)-stilbene are isomers of each other, why (and how) do they differ in polarity? If both are of low polarity, why is (Z)-stilbene soluble in ethanol, but not (E)-stilbene? Also why is (Z)-stilbene the more polar isomer of the two and has a lower $R_f$ value in TLC?

An initial reasoning of mine was that (Z)-stilbene were more hindered than the (E)-isomer, and thus less stable. Yet, this doesn't make sense to me.

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    $\begingroup$ Please pay attention to your spelling: it’s stilbene, not stillbene. You get it correct half the time. Stereochemical descriptors like E and Z should be italicised, placed in parentheses, and connected to the main name by a hyphen: so the correct names are (E)-stilbene and (Z)-stilbene. $\endgroup$
    – orthocresol
    May 6 '21 at 10:53
  • $\begingroup$ Neither E nor Z stilbene are very polar. But their polarities must differ if their geometries are different and they are very different. $\endgroup$
    – matt_black
    May 6 '21 at 12:22

With the $\ce{C=C}$ double bond of the stilbenes as geometric reference, you may think of the phenyl rings as substitutents increasing the electron density of the former. This creates a small, locally constraint, dipole moment. Dipoles like dipoles differ from a scalar (like e.g., temperature) that they have a direction, which is why I drew little arrows in the simplified representation of either isomer in the illustration below:

enter image description here

In the case of (E)-stilbene, these vectors oppose each other; for symmetry reasons, they may cancel out each other as if there were no vector present. This contrasts to the case of (Z)-stilbene where the sum of the two vectors does not vanish; symbolized (the doodle is a concept, not at exact scale) by the dashed arrow: there is a non-zero molecular dipole moment.

The drawing of (Z)-stilbene's configuration however does not state (much) about the molecule's energetically preferred conformation(s). To lessen intramolecular steric hindrance of the hydrogens of the two phenyl rings, the following spatial arrangement is more likely to happen:

enter image description here

Note, regardless how the phenyl rings rotate around the bond between $\ce{C_{ar}-C_{sp^2}}$, the overall (molecular) dipole moment will not be affected from these local variations.

Equally note that to achieve an efficient recrystallization of the three constituents of your mixture, the isomers of stilbene and TPPO, you use a mixture of ethanol/water heated to reflux then allowed to cool slowly to room temperature. The addition of water is to attenuate the solution capacity of ethanol for each of the three compounds so that one preferentially crystallizes out. This is why this purification method is called fractional crystallization.

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    $\begingroup$ Also worth noting that the factors influencing propensity to crystallise easily are far more complex that just the dipole moment if the isomer. In this case one isomer is much flatter than the other and likely to have far larger pi-pi interactions between neighboring molecules' phenyl rings. Explanations are not going to be simple! $\endgroup$
    – matt_black
    May 7 '21 at 17:01
  • $\begingroup$ @matt_black True, crystallization of (organic) matter depends on many factors in addition to «just» dipole moments. Rigidity and shape of the molecules equally contribute to this, too. One of the well-known in the field is Kitaigorodski (e.g.,«Molecular Crystals and Molecules», one of the references mentioned here). One may argue about the wording to describe the complexity known and unknown parameters bring to this, yet there may be many ... $\endgroup$
    – Buttonwood
    May 8 '21 at 8:23
  • $\begingroup$ Thank you very much @Buttonwood, if you allow me though, in a specific step of this lab, we get trans/cis isomers and the side product TPPO, so basically two solids and a liquid, to get the cis isomer, can we just used filtration to get the liquid which is the cis isomer, and then we are left with two solids, one is polar and the second is not which is easy to seperate because of the difference in polarity? $\endgroup$ May 8 '21 at 11:18
  • $\begingroup$ I assume the specific step you describe is after performing the reaction, extraction against dichloro methane and water, drying the organic phase, and gentle evaporation of dichloro methane. If so, I would not simply filter off the liquid, even if pure (Z)-stilbene is liquid (at room temperature) while the other two chemicals are solids, because (Z)-stilbene may equally act as a solvent for them. Recrystallization of this mixture from a bit of ethanol, from boiling down to an ice bath (slow cooling, not crushing precipitation) helps you to concentrate (E)-stilbene. $\endgroup$
    – Buttonwood
    May 8 '21 at 12:22
  • $\begingroup$ By means of atom economy, assuming your Wittig is like benzaldehyde + benzyltriphenylphosphonium chloride + a base, it may be more efficient to create a stilbene, e.g., by reaction of benzaldehyde + the Grignard of benzylbromide, and a subsequent dehydration. (Perhaps refluxing the raw product in toluene with a grain of iodine increases the percentage of (E)-stilbene (if necessary).) Depends, of course, of what is at your disposition, but basically, in [PPh3-CH2-Ph+][Cl-], there is only one Ph in your wanted, but three Ph in the by-product. $\endgroup$
    – Buttonwood
    May 8 '21 at 12:30

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