Why is the ester carbonyl peak in benzocaine IR spectrum so low?

Question

Having looked at multiple IR spectra for benzocaine, I am struggling to understand why the carbonyl C=O peak is so low in the spectra compared to where an ester carbonyl group usually is on other IR spectra

Answer 1

Let's first have a look at the structure of benzocaine, to see how it is different from an usual aliphatic ester:

One thing that should be immediately obvious is that the ester is attached to an aromatic ring (which has an amine group). The aromatic ring would participate in conjugation with the ester oxygen, particularly because the amine group is strongly electron-donating (via conjugation).

This is the IR spectrum of benzocaine (from Chemical book):

The peak at $\mathrm{1682\;cm^{-1}}$ can be assigned to the ester C=O stretch. Usually aliphatic esters have IR stretches around $\mathrm{1730-1750\;cm^{-1}}$. Esters attached to unsaturated systems or aromatic groups have lower stretches, as the conjugation weakens the C=O bond.

Compare the $\ce{C=O}$ IR stretches of some esters:

Ethyl acetate : $\mathrm{1752 cm^{-1}}$

Ethyl benzoate : $\mathrm{1726 cm^{-1}}$

The presence of the benzene group has reduced the stretch wavenumber by $\mathrm{\sim30 cm^{-1}}$.

methylparaben : $\mathrm{1682 cm^{-1}}$

Here the $\ce{-OH}$ group is quite strongly electron-donating, and it brings down the ester C=O stretch wavenumber even further, and now the IR stretch resembles that of benzocaine. $\ce{-NH2}$ is also a stronlgy electron-donating group like $\ce{-OH}$.