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The ELISA test on HIV works by verifying the presence of peroxidase within samples. The verification process is done through the enzymatic reaction that said peroxidase has on TMB (3,3',5,5'-tetramethylbenzidine) in an environment that contains $\ce{H2O2}$. This reaction leaves the enzyme untouched and transforms the $\ce{H2O2}$ into $\ce{2H2O}$ by adding two hydrogen atoms taken from the TMB. This also transforms the TMB into another compound, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which has a blue color in that specific case.

Therefore my question is, how could one predict from the structure of the one product that it is going to have a blue color in that situation? I know that it probably comes from the conjugated double bonds but how do their count relate to the absorbed wavelength(s)?

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Your question: How could one predict from the structure of the one product that it is going to have a blue color in that situation?

This is a very broad question. OP has to go to chemistry fundamentals to learn, how one can predict the absorbance wavelength of a compound by analyzing the structure of the given compound. One easy subject for this study for OP is basic fundamentals of UV-visible spectroscopy in organic chemistry. For example, how absorbance wavelength increase with each additional conjugated double bond (Absorption in the ultraviolet and visible regions is related to the transition of electrons). Look following table:

$$ \begin{array}{l|cr} \text{Substance} & \lambda_\mathrm{max}, \ \pu{nm} & \epsilon, \ \pu{L mol-1 cm-1} \\\hline \ce{CH2=CH2} & \pu{180nm} & 10000 \\ \ce{CH2=CH-CH=CH2} & \pu{217nm} & 21000 \\ \text{Vitamin A (5 conjugate double bonds)} & \pu{328nm} & 51000 \\ \text{Benzene} & \pu{255nm} & 180 \\ \text{Naphthalene} & \pu{286nm} & 360 \\ \text{Anthracene} & \pu{325nm} & 7100 \\ \hline \end{array}\\ \epsilon = \text{Molar Absorption Coefficient}; \ \lambda_\mathrm{max} = \text{Wavelength of maximum absorption peak} $$

Thus, one can suggest diimine, 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine, which contains 5 directly conjugated double bonds (two of them are imines), may absorb wavelengths of lights way larger than $\pu{328 nm}$ given for Vitamin A. That's because it involves transition of $n \rightarrow \pi^*$ in addition to $\pi \rightarrow \pi^*$ in Vitamin A (e.g., $\pi \rightarrow \pi^*$ in acetone is $\pu{195 nm}$ in addition to $n \rightarrow \pi^*$ transition at $\pu{274 nm}$, when compared to $\pi \rightarrow \pi^*$ transition of ethylene at $\pu{180 nm}$). Experimental evidence has shown that 4-(4-imino-3,5-dimethylcyclohexa-2,5-dien-1-yl)-2,6-dimethylcyclohexa-2,5-dien-1-imine absorbs light at $\pu{450 nm}$, giving its characteristic complimentary color $\color{blue}{\text{Blue}}$ (Ref.1).

References:

  1. P. D. Josephy, T. Eling, R. P. Mason, “The horseradish peroxidase-catalyzed oxidation of 3,5,3',5'-tetramethylbenzidine. Free radical and charge-transfer complex intermediates,” The Journal of Biological Chemistry 1982, 257(7), 3669-3675 (http://www.jbc.org/content/257/7/3669)(PDF).
  2. For basic learning, read: Master Organic Chemistry.
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