How to take a peak out of chromatogram?

[Fraish]

Hi, apologies if this not right forum for this question.

I am separating four peaks using LC-UV but I usually find a fifth unknown peak eluted before them. I am using methanol water mobile phase C18 column , isocratically using 50 micron injection volume.

How can I take this peak out for further chracterisation by NMR, etc to know what structure of this peak, mass etc.

After long reading, comes these idea:

I thought I must use ( semi- preparative LC ( wide column and do large volume injection then fractionation to collect this peak).

Then I said what I know that LC-UV is passive detector( not destructive) why not delay time of the four peaks and just take the waste of the detector before unknown peak eluted then make sure it is eluted in the test tube so the uknown will be in it.

Then I said why not just put flow splitter between column and detector and collect the peak( but i dont have flow splitter).

I think the only option left is to collect the unknown from the detector waste , what do you think?

Thanks in advance

[lmh]

Yes, you can take the outflow of the detector and collect it. There is nothing wrong with this approach. In fact it's a very good approach wherever possible because it simplifies finding the delay time(*).

The reason that people resort to flow-splitting is usually either that they have a destructive detector (e.g. MS), or because their flow-rate isn't compatible with their detector. For example, if you typically use your PDA or fluorescence detector for analytical work, its flow-cell may be optimised for flows between 0.1-1.0mL/min; anything higher will create more back-pressure, and more back-pressure will increase the risk of the cell leaking or breaking.

If you can get enough of your peak for NMR using a 4.6mm column, then don't feel obliged to use a 10mm column just because other people do!

(* delay time: the peak will take a little while to travel down tubing from the detector to wherever you are collecting fractions. You can find this time by experiment or estimate it based on the volume of the tubing between the two, and the flow-rate. It is very hard to do when flow-splitting because anything that affects the split-ratio will affect the rate of flow, and hence the delay time. Splitters are basically just a pair of resistances, so any additional resistance in either line will affect the ratio)

(Flow splitters: if you don't have one, you can also use a simple tee-piece and adjust the ratio by adding lengths of fine tubing to either side as desired).

[Fraish]

Thanks lmh for your rich answer

"If you can get enough of your peak for NMR using a 4.6mm column, then don't feel obliged to use a 10mm column just because other people do!"

Yes my column is 250 mm, 4.6 mm C18

The size of unknown peak is same as the size of ( 1 mg/L analyte peak) so i guess its concentration is usually 1 ppm always whatever is the standard concentration.

Now how can I obtain enough unknown peak to further do NMR etc :

What I think that :

if i inject 50 micron from 1 ppm pure standards ( of four compounds) thats may give me ( 50 ng of unknown peak by weight)

If inject 50 micron from 100 mg/L pure standard this may give me (also 50 ng)

How can I get higher weight of the unknown peak? Please not whether i use 1ppm or 100 ppm standard , the unknown peak has same size.

Or

Do I repeat the injection and collect and re-run again and collect in same test tube? But this is require repeating injection in the same test tube maybe time consuming

Thanks

[tom jupille]

Please not whether i use 1ppm or 100 ppm standard , the unknown peak has same size.

That raises a very large red flag! Are you sure that your "unknown" peak is not simply t0 upset?

[bunnahabhain]

My first guess is dissolved air:
http://www.shimadzu.com/an/hplc/support ... 37lab.html

Try to purge your blank with argon or nitrogen and inject immediately to see if the additional peak disappears.

[Fraish]

Tom I am sure it is not t0 peak. 

bunnahabhain I read this article long long time ago In my case I don't thing what they suggest apply to my situation. I am working with catalysts.

Please let me ask the question in other way. That's, If I have 1 mg/L peak this remains usually same and not related to the standard concentration. Now:

How I can get enough weight from this peak in a test tube ( say I need a round 100 mg or 1000 mg)? the reason I need to do many characterization with many catalyst involve. One big professor said to me this peak is comes from catalyst not understand his talk very well as I met him in seconds.

[tom jupille]

The loading capacity of a typical HPLC column is on the order of 100 micrograms. If you gave enough resolution, you can overload, but getting to 100 milligrams is going to be tedious.

But, rather than speculate, you can establish the limits in a day. Just do a loading study: a series of runs where you increase the concentration in geometric progression (e.g., 1, 2, 4, 8, . . . or 1, 3, 10, 30, . . . ) until the band width gets so bad that the nearest other peak just touches the one you're interested in. That will give you a good idea how much you can get from a single injection.

Just out of curiosity, though, what is your flow rate and what is the approximate retention time of that unknown peak?

[lmh]

(1) To do nmr, you don't need a huge amount of material; depending on your instrument, somewhere in the region 0.2-1.0mg seems to keep most NMR operators happy.
(2) If the peak you wish to analyse isn't changing size when you inject more or less sample, then the peak is nothing to do with the sample. I wouldn't bother trying to purify it for identification until I'd worked out why it isn't changing linearly with concentration of sample. Is it something in the solvent you're using to prepare the sample? Does it happen in blank runs?

[Fraish]

"what is your flow rate and what is the approximate retention time of that unknown peak?"

1 mL/min

Retention time= 3 min

Thanks for the good suggestions.

[Fraish]

Thanks Imh I will work in all these suggestion. Or I might just ignore it as I have enough resolution and I will be able to quantify precisely.

[tom jupille]

The void volume of a 250 x 4.6 mm column is somewhere around 2.5 mL. At 1 mL min, t0 is somewhere around 2-3 minutes.

Or I might just ignore it as I have enough resolution and I will be able to quantify precisely.

That is the best choice.

[lmh]

I think what Tom means is that since your mystery peak is unretained, and since you also know that it's not dependent on the concentration of sample that you add, there is a strong chance that it is actually an injection peak caused by the solvent in which you dissolved the sample being slightly different from the running solvent. There is no point in trying to collect the injection solvent for further analysis.
(correct me if I'm putting words in your mouth, Tom)

[tom jupille]

there is a strong chance that it is actually an injection peak caused by the solvent in which you dissolved the sample being slightly different from the running solvent. There is no point in trying to collect the injection solvent for further analysis.
(correct me if I'm putting words in your mouth, Tom)

That's it exactly.