TLC Chromatography Question

[IH8ColdWeath3r]

So, I know that with TLC, the molecule of interest travels up the plate due to the solvent that is present. USUALLY (I know some passages specify, some don't, and hopefully the MCAT will) the plate is made of of silicon oxide and is polar.

I've read that the more polar the compound is, the more it will be attracted to the plate and thus travel LESS farther up the plate. This would lead to a lower Rf value, correct?

well here is a question from a passage in TPR. Here is a brief description of the passage:

A chemist has a mixture of compounds that she wants to separate. A TLC of the mixtures shows four distinct spots with Rf values of 0.9, 0.7, 0.2, 0.0 when eluted with 20% ethyl acetate in hexanes. She follows the extraction procedure below: (long, separation with several steps, none relevant)

Q: The Rf value for compound 3 is 0.7 when eluted with ethyl acetate in hexanes. Which of the following is the most likely Rf value for compound 3 eluted with straight hexanes.

A) 0.2
B) 0.7
C) 0.8
D) Cannot be determined

I chose C but the answer is A. I thought that if you elute with straight hexane as opposed to ethyl acetate in hexanes, that this would make the compound travel further up the plate and thus the Rf value would increase. (The effluent is like the solvent, right?)

Heres their explanation:
The Rf in 20% ethyl acetate in hexanes is 0.7. Pure hexane is a much less polar elutent, so the Rf value should decline substantially. Therefore, the answer must be A.

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[MT Headed]

Silicon oxide is polar. Ethyl acetate is polar. With an original Rf of 0.7, i'm guessing compound 3 is slightly polar.

Now you remove the ethyl acetate. The polar portion of compound 3 is looking for somebody else polar to play with, and the only available playmate is the silicon oxide itself. So it sticks to the silicon oxide more, and gets a smaller Rf.

Hmm, using this logic, all of the compounds would have a smaller Rf except for a purely nonpolar compound, which would have an Rf of 1.0 in both cases.

 

[IH8ColdWeath3r]

Silicon oxide is polar. Ethyl acetate is polar. With an original Rf of 0.7, i'm guessing compound 3 is slightly polar.

Now you remove the ethyl acetate. The polar portion of compound 3 is looking for somebody else polar to play with, and the only available playmate is the silicon oxide itself. So it sticks to the silicon oxide more, and gets a smaller Rf.

Hmm, using this logic, all of the compounds would have a smaller Rf except for a purely nonpolar compound, which would have an Rf of 1.0 in both cases.

but I thought since the only thing there now is hexane which is nonpolar, it would help it travel farther up. Since ethyl acetate was removed, it would help the compound of interest move further up, as opposed to staying further down?

I see what your saying in that the ethyl acetate would compete with the plate, but the logic explained above is how I interpreted the question and my answer. Be interesting to see what others think.

 

[mcloaf]

For TLC always assume the plate is highly polar unless they tell you otherwise. I suspect there must be something hidden in the rest of the passage that indicates the compound in question is polar--otherwise you really couldn't say (unless you follow MT's logic that all non-polar compounds have an Rf of 1 regardless of solvent, which experience says is not the case). Either that or it's just a bad question. I guess it just depends on what assumptions you think you can make about how the observed Rf in a specified solvent indicates the nature of the compound polarity-wise, though to be honest I'd be very surprised to see the MCAT make a logical leap like that.