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 $\ce{H2/Ni}$
upright vs italic // Use plain texts in CH SE titles. // For more, see Math SE MathJax tutorial. – Poutnik Jun 17 '22 at 08:14