# How would you distinguish between ortho and para nitrophenol using infrared spectroscopy?

Jan 24, 2016

The thing is, anyone would have a hard time justifying which of these two compounds they had made from using just an IR spectrum, even if they had both.

STRUCTURES AND PREDICTED VIBRATIONAL MOTIONS

The structures are:

(o-nitrophenol)

(p-nitrophenol)

The peaks you should see based on the structure of the molecule (and based on what I have in my reference text, Techniques in Organic Chemistry, Mohrig) are:

• $\setminus m a t h b f \left(\text{C"="C}\right)$ aromatic (stretch): 1620~1440 ${\text{cm}}^{- 1}$; medium to weak; sharp
• $\setminus m a t h b f \left(\text{C"-"H}\right)$ aromatic (stretch, bend): 3100~3000, 900~680 ${\text{cm}}^{- 1}$, respectively; medium to weak; sharp
• $\setminus m a t h b f \left(\text{C"-"O}\right)$ alcohol (stretch): 1300~1000 ${\text{cm}}^{- 1}$; strong; sharp
• $\setminus m a t h b f \left(\text{O"-"H}\right)$ alcohol (stretch): 3650~3200 ${\text{cm}}^{- 1}$; medium to strong; broad
• $\setminus m a t h b f \left({\text{NO}}_{2}\right)$ nitro (stretch): 1570~1490, 1390~1300 ${\text{cm}}^{- 1}$; strong; sharp

THE IR SPECTRA

Here are two sample spectra run on a KBr pellet (I seem to recall them being a few thousand dollars each?):

You should notice that the spectra are very similar above ${\text{1700 cm}}^{- 1}$, except for the $\text{O"-"H}$ ALCOHOL stretches near ${\text{3200 cm}}^{- 1}$, which are STRONGER in the ortho spectrum.

Also, you may notice though is that the ortho spectrum is messier below ${\text{1700 cm}}^{- 1}$. Many of those peaks are still the same, such as at ${\text{1400~900 cm}}^{- 1}$, but some peaks, like the NITRO and $\text{C"-"O}$ ALCOHOL stretches at ${\text{1600~1300 cm}}^{- 1}$, have additional splitting and are STRONGER in the ortho spectrum.

That's important; why are they stronger for the ortho isomer?

PEAKS IN IR SPECTRA ARISE FROM A CHANGE IN DIPOLE MOMENT

One thing you should remember is that peaks in IR spectra arise from a change in dipole moment during the sample analysis. You can probably imagine that changes in dipole moment are lessened when there is more symmetry in the molecule.

Not surprisingly, the main difference between the two compounds is thus the amount of symmetry they have---notice how the para isomer has a (${C}_{2}$) symmetry axis right down carbons 1 and 4. The ortho isomer doesn't have that.

MORE SYMMETRY REDUCES THE CHANGE IN DIPOLE MOMENT

The vector direction of the nitro group stretches on the para isomer is more symmetrical with respect to the vector direction of the hydroxyl group stretches than in the ortho isomer, which reduces the overall change in dipole moment during those stretches, thus weakening any peaks pertaining to those vibrational motions that are predicted to show up in the IR spectrum.

As a result, we see weaker peaks in the more symmetrical para isomer for the $\text{C"-"O}$ and $\text{O"-"H}$ alcohol stretches near $1300$ and ${\text{3200 cm}}^{- 1}$ respectively, and for the ${\text{NO}}_{2}$ nitro stretches near ${\text{1600~1300 cm}}^{- 1}$.

ALTERNATIVE APPROACH?

Of course, had you not had BOTH spectra in front of you, you might not have been able to notice such subtleties. So, I wouldn't say this is a practical way to approach the determination of which product you actually made more of during your synthesis.

You should also consider your $\setminus m a t h b f \left(\text{^13 "C}\right)$ NMR spectra, as that would tell you more. The more symmetry you have on the molecule, the less number of carbons would appear to show up on the $\text{^13 "C}$ spectrum, and that should be a much clearer indicator than your IR spectra as to whether you made the ortho or the para isomer.