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In this particular reaction firstly $\ce{NaNO2/HCl}$ at $\pu{0-5°C}$ was taken and after that $\ce{H2/Ni}$ .

But what I doubt is, was there any role of $\ce{NaNO2/HCl}$ at $\pu{0-5°C}$ ? Since only reduction occurred which can also be done by $\ce{H2/Ni}$enter image description here

Poutnik
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