Why does the same chromatogram give a different purity at 214 nm and at 280 nm?

Accepted answer

The two wavelengths are looking at two different chromophores, so they weight the same mixture differently. Neither is a sensitivity setting; they answer different questions, and 214 nm is the right one for a purity claim.

What each wavelength sees

214 nm (or 210, 215, 220 depending on the method) is the tail of the amide n-to-pi* absorption of the peptide bond itself. Every residue in the backbone contributes. The consequence is that response is roughly proportional to the number of peptide bonds, which is to say roughly proportional to peptide mass. That approximate proportionality to mass is exactly what a purity percentage needs, and it is why every pharmacopoeial peptide-purity method sits in the 210 to 220 nm region.

280 nm sees only the aromatic side chains, and effectively only three of them. Standard molar extinction coefficients in water, per residue:

So at 280 nm the detector is counting tryptophans, with a small contribution from tyrosines. For our two usual molecules:

• Semaglutide: one Trp (position 31) and one Tyr (19). 5500 + 1490 = 6990
• Tirzepatide: one Trp (25) and two Tyr (1 and 10). 5500 + 2 x 1490 = 8480

A single tryptophan is carrying 79% of semaglutide's 280 nm absorbance and 65% of tirzepatide's.

Worked: why an impurity's apparent level changes with wavelength

Take semaglutide and a truncation impurity comprising residues 7 to 30 — 24 residues, so it has Tyr19 but has lost Trp31. Suppose it is genuinely present at 2.00 mol% of the mixture.

At 214 nm, response scales with peptide bonds: 23 for the impurity, 30 for the parent. Relative response factor 23 / 30 = 0.767.

• Impurity area contribution: 2.00 x 0.767 = 1.533
• Parent: 98.00 x 1.000 = 98.000
• Reported impurity: 1.533 / (98.000 + 1.533) = 1.54%
• Reported purity: 98.46%

At 280 nm, response scales with the aromatic sum: the impurity has only Tyr, so 1490 against the parent's 6990. Relative response factor 1490 / 6990 = 0.213.

• Impurity area contribution: 2.00 x 0.213 = 0.426
• Reported impurity: 0.426 / (98.000 + 0.426) = 0.43%
• Reported purity: 99.57%

Same vial, same injection, 98.46% versus 99.57%. The impurity did not shrink; the detector stopped counting the part of the molecule that was missing. A truncation that removes the tryptophan is nearly invisible at 280 nm, and truncations at the C-terminus are among the most common synthesis by-products.

Your peak that appears at 280 and not at 214

Almost certainly not a peptide. A species with a strong aromatic or conjugated chromophore and little peptide backbone will show that pattern: residual protecting groups and their scavenger adducts, dibenzofulvene and its piperidine adduct from Fmoc deprotection, trityl-derived species, some plasticisers, and dye or leachable contamination from a stopper or a filter. Several of these absorb well at 280 to 300 nm.

It is worth chasing rather than dismissing, because a non-peptide organic impurity is real material in the vial that a 214 nm purity figure may under-weight, and its identity tells you something about the purification. The way to chase it is the MS trace at that retention time. A diode-array report that includes a full spectrum for each peak makes this a thirty-second check: read the peak's absorbance maximum. Peptide impurities have essentially no maximum above 230 except a shoulder at 275 to 280 if they contain Trp or Tyr; a species with a clean maximum at 265 or 300 is something else.

Which number to quote, and whether to be suspicious

Quote 214 nm. It is the compendial region, it approximates mass proportionality, and it is what any comparison you make against another report will assume.

On suspicion: 280 nm as the primary purity wavelength for a peptide is either a mistake or a choice, and it reliably flatters. It should be treated as a method red flag on the same level as a 20-minute gradient. But 280 nm as a second channel is genuinely useful and its presence is a sign of a better report, not a worse one:

• Ratio-based peak purity. If the 280/214 area ratio of an impurity differs from the parent's, the impurity has a different aromatic content — a direct structural clue that costs nothing. Truncations losing Trp show a low ratio; species that gained a chromophore show a high one.
• Trp oxidation detection. Oxidation of tryptophan to N-formylkynurenine destroys the 280 nm absorbance and creates absorbance near 320 nm. An impurity with a suppressed 280/214 ratio and a rise at 320 is an oxidised-Trp variant, and 320 nm is a channel worth asking for on any Trp-containing peptide.
• Detector linearity. A heavily loaded main peak can saturate the detector at 214 nm while remaining linear at 280, since the 280 absorbance is far weaker. Comparing the two channels catches saturation, which otherwise inflates purity by flattening the top of the main peak.

So: a report giving 214 nm as the purity figure and 280 nm alongside for structural information is doing it right. A report giving only 280 nm has chosen the number.

edited 19 Oct 2024 by coring_risk — tightened the wording; no substantive change