2
$\begingroup$

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?

$\endgroup$
1
  • 1
    $\begingroup$ Your analysis seems logical. $\endgroup$
    – jerepierre
    Mar 2, 2015 at 17:52

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.