So, basically I can convert Benzene to Phenol using Dow's synthesis process and then use $\ce{NH3}$, $\ce{ZnCl2}$, $\ce{HNO2}$, $\ce{NaNO2}$ under 0-5 °C controlled temperature and convert the phenol to Diazonium Chloride Salt. Then with Coupling Reaction, I can convert the Diazonium Chloride to Aniline Yellow. But, Chrysoidine has another $\ce{-NH2}$ group attached to the Ortho position. So, an explanation would be greatly appreciated as to how to obtain Chrysoidine from there on.

  • $\begingroup$ why go via the phenol? Easier surely to nitrate benzene then reduce to aniline. $\endgroup$ – Waylander Sep 6 '18 at 14:52
  • $\begingroup$ @Waylander Yes, I know that pathway to Aniline as well. However, I wanted to know whether Aniline Yellow, upon reacting with Nitric Acid and then being reduced by Sn and HCl, yields Chrysoidine. Or will the -NO2 group just get attached to the Meta position? $\endgroup$ – Rifat Raiyan Sep 6 '18 at 15:00

I think aniline yellow is likely to nitrate ortho to the aniline giving you the wrong structure. Also the Sn/HCl reduction may well reduce the azo double bond.

Here is a route from benzene. From benzene a simple nitration gives nitrobenzene. Divide your nitrobenzene, reduce one portion to aniline by hydrogenation over Pd or Pt catalyst. Nitrate the second portion to 1,3-dinitrobenzene procedure here. Reduce the 1,3-dinitrobenzene by hydrogenation over Pd or Pt catalyst to give m-phenylenediamine. Diazotise your aniline to the diazonium and react this with the phenylenediamine. This route was used industrially to prepare Chrysoidine in the 19th century reference here.

| improve this answer | |
  • $\begingroup$ My gratitude, sir. This helped a lot. :D $\endgroup$ – Rifat Raiyan Sep 7 '18 at 13:21
  • $\begingroup$ Please accept the answer if you are happy with it. $\endgroup$ – Waylander Sep 7 '18 at 13:32

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.