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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.

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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