Functional groups like $\ce{OH}$, $\ce{NH2}$, attached to unsaturated compounds (chromophores) are called auxochromes. These groups tend to shift the absorption wavelength to the infrared region, which is called a bathochromic shift.

  • Why do these groups shift the light absorption wavelength?
  • Why exactly to the infrared region; that is, why is the absorption red-shifted or lowered in energy?

Generally speaking, the typical auxochromic groups possess (at least) one pair of non-bonded $n$-electrons and -- if taken alone -- do not absorb in the UV. If attached, for example to a benzene ring, their pair(s) of $n$-electron interact(s) with the $\pi$-electrons, and the delocalization of the $\pi$-system is extended.

Now the catch:

  1. Bathchromic auxochromic groups (like the mentioned $\ce{-OH}$ and $\ce{-NH2}$) increase the electron density in the $\pi$-system. This is typically caused by a strong $+M$-effect. As they lower the excited $\pi^*$-level, a longer wavelength is needed for the excitation along $\pi{} \rightarrow {} \pi^*$, hence the redshift.
  2. In contrast to the former, substitution equally may alter the $n \rightarrow{} \pi^*$ transition, inducing the opposite effect, moving the absorption towards higher absorption (blue) and a hypsochromic shift. One example is the to compare the absorption of indigo yielding "jeans blue" with 6,6'-dibromoindigo, main constitutent of tyrian purple. A second, that heteroatoms next to a $\ce{C=O}$ carbonyl group induce a hypsochromic shift like in the case of acetaldehyde example here in comparison to acetic acid here

Note the use of the terms of "bathochromic shift" and "hypsochromic shift" along solvatochromism.

  • $\begingroup$ I wasn't aware that electron density played a role. I thought the basic premise was that any extension of the conjugation path leads to a smaller HOMO-LUMO separation and consequently a bathochromic shift (particle in a box). Looking at the Woodward-Fieser rules even bromine and chlorine lead to a bathochromic shift since they extend the conjugation path. $\endgroup$ – ron Apr 28 '15 at 1:12
  • $\begingroup$ @ron I altered the second part of my answer and provided some examples to support this section. Provided unsaturated compounds often provide both $\pi \rightarrow{} \pi^*$ and $n \rightarrow{} \pi^*$ transitions, I now speculate the Woodward-Fieser rules empirically base on the former ($\pi \rightarrow{} \pi^*$). $\endgroup$ – Buttonwood Apr 28 '15 at 19:58
  • $\begingroup$ Thanks for the clarification! I think you're right about the W-F rules. $\endgroup$ – ron Apr 28 '15 at 21:33

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