The Raschig process (according to the german wikipedia) is divided in the following steps:

  1. $\ce{2NaOH + Cl2 -> NaOCl + NaCl + H2O}$
  2. $\ce{NaOCl + H2O -> HOCl + NaOH}$
  3. $\ce{HClO + NH3 -> NH2Cl + H2O}$
  4. $\ce{NH2Cl + NH3 -> N2H4 + HCl}$
  5. $\ce{N2H4 + 2NH2Cl -> N2 + 2NH4Cl}$

But what happens now if one changes $\ce{NH3}$ into $\ce{R-NH2}$, with $\ce{R}=$enter image description here?

Starting from step 3 of the Raschig process I came up to the reaction of the produced hydrazine derivate to give an azo compound.

  1. $\ce{HClO + R-NH2 -> R-NHCl + H2O}$
  2. $\ce{R-NHCl + R-NH2 -> R-NH-NH-R + HCl}$
  3. $\ce{R-NH-NH-R + x~R-NHCl -> R-N=N-R~+~?}$

The last reaction step bothers me a little ... is it simply $$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$ but without the cleavage?

  • 2
    $\begingroup$ Aliphatic diazocompounds (say, $\ce{CH3-N=N-CH3}$) are very unstable and tend to fragment into dinitrogen molecule and two alkyl radicals after a funny look. $\endgroup$
    – permeakra
    Jul 12, 2015 at 8:15
  • $\begingroup$ You are right at this. I added the information on my R- to the question. And that compound is stable as it was synthesised - not only be me. $\endgroup$ Jul 13, 2015 at 11:38

1 Answer 1


It can't be

$$\ce{R-NH-NH-R + 2R-NHCl -> R-N=N-R + 2R-NH2Cl}$$

The reason is that while the atoms are balanced, the electrons aren't. Going from $\ce{R-NH-NH-R}$ to $\ce{R-N=N-R}$ involves the loss of two protons and two electrons, but the $\ce{R-NHCl}$ are really only picking up two protons. (Nitrogens with four substituents are positively charged.)

When thinking of mechanisms, it may help to explicitly split the final listed reaction into two steps.

5a. $\ce{H-NH-NH-H + NH2Cl -> H-N=N-H + NH4Cl}$

5b. $\ce{H-N=N-H + NH2Cl -> N2 + NH4Cl}$

That is, each chloramine is involved in pulling one dihydrogen (two protons and two electrons) off of the diazo. - The addition of a "hydride" (a proton and two electrons) to the nitrogen kicks the chlorine off as a negatively charged ion, and the remaining proton protonates the lone pair on the nitrogen.

The analogous last step for the $\ce{R-NH2}$ case might be:

$$\ce{R-NH-NH-R + R-NHCl -> R-N=N-R + R-NH3Cl}$$

This is equivalent to reducing the $\ce{N-Cl}$ covalent bond to $\ce{R-NH2 + HCl}$ followed by an acid/base reaction.


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