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This is a "homework" question.

I'm writing up a lab in which we synthesized 1,4-diphenyl-1,3-butadiene. We analyzed the products using TLC with 4:1 Hexane:Ethyl Acetate as the solvent.

cis,trans isomer trans,trans isomer

One of the isomers has a larger Rf value than the other (0.83 and 0.95).

TLC plate

I'm having trouble understanding how to figure out which of the isomers is which on the TLC. I know more polar compounds are going to have a lower Rf, like cinnamaldehyde.

My best guess is that, although the all the double bonds are conjugated, the phenyl rings have a higher electron density than the rest of the molecule. Since the trans,trans isomer is essentially symmetrical, it has little or no molecular dipole. The cis,trans isomer has a small molecular dipole because it is asymmetric. Therefore the cis-trans isomer has more affinity for the stationary phase (giving it the Rf value of 0.83).

Am I on the right track here?

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    $\begingroup$ Your analysis seems logical. $\endgroup$ – jerepierre Mar 2 '15 at 17:52
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Well, apparently I was wrong.

In fact, the trans,trans isomer is adsorbed more strongly by the stationary phase because it is entirely planar. The cis,trans-isomer cannot adopt a planar conformation because of steric hindrance between the two phenyl rings. The non-coplanar geometry lowers the efficiency of interactions with the stationary phase, making it more mobile.

So the cis,trans isomer has a higher Rf, and the trans,trans isomer has a lower Rf.

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