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Topic: Eluition order of some molecules in HPLC.  (Read 282 times)

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Offline BluntTheKnives

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Eluition order of some molecules in HPLC.
« on: July 27, 2019, 05:34:39 AM »
Hello, everyone!

I have a question that I really hope someone can help me with, I need to answer some question about Environmental Analitical Chemistry and I'm having troubles with these two:

I need to incidcate the correct expected eluition order of four molecules analized HPLC with normal phase: the molecules are benzene (PhH), phenol (PhOH), chlorobenzene (PhCl) and acetophenone (C6H5-CO-CH3).

Now, I know that in normal phase I do expect to see exiting first the more neutral molecule, and last the more polar one, problem is that I haven't been given any tools to precisely evaluate and match the polarity of a molecule respect to others so similar.

I'm pretty sure the first one, the more neutral one, is for sure the benzene (PhH).
Then, trying to compare them considering polarity as charge/mass ratio, I'd go with acetophenone as the second molecule expected, since its low polarity is spreaded on a bigger surface.
Third I'd go with chlorobenzene (PhCl), since its charge is spreaded on a smaller surface.
Lastly I'd say phenol, since it's ability to from hydrogen bonds makes it the most polar of the group.

So my answer would be:
1. Benzene
2. Acetophenone
3. Chlorobenzene
4. Phenol

Is my reasoning correct? Can anyone provide me with a reliable tool/strategy to correctly evaluate the polarity of such similar molecules?

The second question is practically tha same, but in HPLC in reverse phase, so I expect first the most polar one. The molecules in this one are: Naphtalene, Nitrobenzene (PhNO2), Salicylic acid, Chloroform.

Iintially I've put them in this order:
1.Chloroform (since I thought that its small charge against such a small surface gave it a greater polarity)
2. Salicylic acid
3. Nitrobenzene
4. Naphtalene

But I've seen that the correct one is:
1. Salicylic acid
2. Nitrobenzene
3. Chloroform
4. Naphtalene
Because Chloroform is poorly soluble in water.

So it's clear that I have quite a bit of confusion about this topic, but I wasn't able to find any answer elsewhere.

Can someone help me out?
Thank you so much in advance!

Offline BluntTheKnives

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Re: Eluition order of some molecules in HPLC.
« Reply #1 on: July 27, 2019, 01:59:28 PM »

Trying to come up with an objective base to evaluate the relative polarity (which, I'd like to underline, hasn't been give to me during the lectures) I thought it could be appropriate to see the individual values of Topological Polar Surface Area and solubility in water, hoping that they reflect, somehow, the eluition order in HPLC.

This is what I've found on pubchem:

Reverse phase:
1.   Salicylic acid
•   Topological Polar Surface Area   57.5 A^2
•   Solubility in water   2 g/l a 20 °C
2.   Nitrobenzene
•   Topological Polar Surface Area   45.8 A^2
•   Solubility in water   1,90 g/l a 293 K
3.   Chloroform
•   Topological Polar Surface Area   0 A^2
•   Solubility in water   8 g/l a 293 K
4.   Naphtalene
•   Topological Polar Surface Area   0 A^2
•   Solubility in water   0,03 g/l a 293 K

Normal phase:

1.   Benzene
•   Topological Polar Surface Area   0 A^2
•   Solubility in water   1,770 g/L a 293 K
2.   Chlorobenzene
•   Topological Polar Surface Area   0 A^2
•   Solubility in water   0,5 g/L (20 °C)
3.   Acetophenone
•   Topological Polar Surface Area   17.1 A^2
•   Solubility in water   6,9 g/L a 25 °C
4.   Phenol
•   Topological Polar Surface Area   20.2 A^2
•   Solubility in water   84 g/l a 293 K

I've put them in this order giving priority to the TPSA value and, where it was equal, seeing the value of solubility. Is this a correct way to proceed? Does it make sense?

In normal phase I've put benzene as first anyway, even if its solubility is higher than chlorobenzene's just because one of the sources quoted on PubChem quoted an absolute nonsolubility in water, even if the others were giving values around 1,7g/L (which is way higher than the one of chlorobenzene, thing that I find quite unexpected).

Thank you again to anyone who will help me sorting this all out.


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