In my textbook, all that was told was that ethanol is a mild reducing agent and hence reduces BDC while oxidising itself. Is there any mechanism or am I just supposed to remember this like a redox reaction?

I did search for a mechanism but did not find anything. I have no idea as to how this reaction works. Can someone please point me in the right direction?

  • 1
    $\begingroup$ If you use CH3CD2OH, I would imagine that you get benzene-d1, CH3CDO and N2. The rest is up to you. $\endgroup$
    – user55119
    Sep 17 '20 at 2:01

In reality, benzene and acetaldehyde are the minor product of the reaction of benzenediazonium chloride with ethanol. The major product is ethoxybemzene (phenetole). Peter Griess, who had discovered diazonium salts in 1858, had reported benzenediazonium salt (nitrate or sulfate) with ethanol undergoes aforementioned redox reaction in 1864 (Ref.1). However, by 1901, it was confirmed that the products from the redox reaction is minor (about ~9%) when compared to major product, phenetole (about ~64%) when benzenediazonium chloride was used. By 1956, Kelley (Ref.2) found that the yield is even low as 3% and his student, Miller (Ref.1) later confirmed this result.

Now, it is believed that the reaction follows the scheme $\bf{A}$ and $\bf{B}$ (Ref.1):

$$\ce{Ar-^+N#NX- + RCH2-OH -> Ar-O-CH2R + N2 + HX} \tag{$\bf{A}$}$$ $$\ce{Ar-^+N#NX- + RCH2-OH <=> Ar-N=N-O-CH2R HX \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -> ArH + RCHO + N2 + HX} \tag{$\bf{B}$}$$

where process $\bf{A}$ proceeding through a $\mathrm{S_N2}$ displacement of the diazonium groups while process $\bf{B}$ involves a homolytic fission of the bonds to produce radicals which lead ultimately to aromatic hydrocarbon and aldehyde.

Proposed mechanism:

$$\ce{Ar-N=N-X -> Ar^. + N2 + X^.} $$ $$\ce{Ar^. + R-CH2-OH -> ArH + R-^.CH-OH} $$ $$\ce{AR-^.CH-OH + X^. -> R-CHO + HX} $$

It was found that when oxygen is added to the system rate of reaction has not changed significantly. The propagation steps with $\ce{O2}$ has been suggested as:

$$\ce{AR-^.CH-OH + O2 -> R-CHO + H-O-O^.} $$ $$\ce{H-O-O^. + AR-CH2-OH -> AR-^.CH-OH + H-O-O-H} $$

Ref.1 and Ref.2 have reported the first order kinetics of this reaction with $k = 0.96 \times 10^{-4}$ at $\pu{25 ^\circ C}$ and $k = 2.03 \times 10^{-4}$ at $\pu{30 ^\circ C}$, which are consistent with other published data.

Note: More insight on mechanism, read Ref.3.


  1. Robert Warren Miller, "The stoichiometry of the reaction of benzenediazonium chloride with ethanol," MS Thesis 1957, University of Arizona, Arizona (PDF).
  2. A. E. Kelley, PhD Dissertation 1956, Purdue University, Indiana.
  3. DeLos F. DeTar, Takuo Kosuge, "Mechanisms of Diazonium Salt Reactions. VI. The Reactions of Diazonium Salts with Alcohols under Acidic Conditions; Evidence for Hydride Transfer," J. Am. Chem. Soc. 1958, 80(22), 6072–6077 (https://doi.org/10.1021/ja01555a044).

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