2
$\begingroup$

Could anybody describe which product is formed in the following reaction (o-Toluidine oxidation)?

$$\ce{H2O2 + o-Toluidine_{uncolored} - Peroxidase -> H2O + 1/2 O2 + o-Toluidine\; Oxidate_{colored}} $$

Note: Hydrogen peroxide is one of the products of glucose oxidation to glucolactone by glucose oxydase.

$\endgroup$
  • $\begingroup$ is this homework? $\endgroup$ – user4076 Mar 3 '14 at 19:04
  • $\begingroup$ @Amaterasu Not really. If I remember correctly, this was/is a method for the photometric determination of glucose in blood. $\endgroup$ – Klaus-Dieter Warzecha Mar 3 '14 at 19:57
  • $\begingroup$ @KlausWarzecha ah, good to know! $\endgroup$ – user4076 Mar 3 '14 at 20:02
1
$\begingroup$

The product most likely is 2-methylnitrobenzene (3), which means that the amino group ($\ce{-NH2}$) in o-toluidine (1) is oxidized to a nitro group ($\ce{-NO2}$). The oxidation proceeds via 2-methylnitrosobenzene (2).

toluidine oxidation

The formation of 2-methylnitrobenzene can be detected photometrically at $\lambda$ = 400 nm, where the aniline does not absorb.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.