Chromatography Forum

I am new to ELS Detection. I need your help.

How can 'no co-elution' be proved in ELSD..?
in other word - How can be proved that a peak is pure in ELSD..?

Very many thanks in advace.

baz kar

You can't really do it as you do not have any spectral information.

ELSD is used for it's universal detection i.e. it can see molecules that do not contain chromophores, assuming they are non-volatile...

Assay a placebo matrix, and show no peaks there (assumes such a matrix is available).

Dear All,
very many thanks for your responses.

When we do forced degradation study with ELSD - how can we state that the method is capable to separate impurity peak from the active...when some one argue the active peak could be degradation peak…?

Any suggestion welcome…!

baz

I can't think of any way of doing this by ELSD. UV is better but it assumes that the UV profile is sufficiently different to be able to identify a co-eluting peak. MS is better, but this assumes that there is no ion suppression of any co-eluters. LC-IR would work especially as you really only need to prove the qulity of the peak but quantitation isn't that easy. Similarly LC-NMR is good, depending on the changes that occur during degradation, but it places restrictions on your chromatography and isn't as sensitive as the other techniques. A combination of one or more of the above with ELSD is usually how its done.

Hi Baz

Becuuse the ELSD is indiscriminate it its response to different compounds there is no way that you can use the detector itself to prove or disprove the presence of degradation products under (or instead of !) the active peak.

Since the detector cannot do it you have to use the separation. If the peak co-elutes with a standard of the active when you change the mobile phase composition by both small increments and more drastically (say plus and minus 10% of the organic component, and also substituting e.g. acetonitrile for methanol with appropriate changes to gradients etc) then the chances that the peak is due to a breakdown product get smaller and smaller with every set of conditions that still gives perfect co-elution.

This approach is horribly time consuming and it is far better to go with DAD or MS if you can.

Peter