Can nitrous dioxide develop when copper and dilute nitric acid react? When doing an experiment in the laboratory with copper, dilute nitric acid and finally hydrochloric acid, the first reaction between copper and dilute nitric acid should end in nitric monoxide, cupric nitrate and water:
$$\ce{3 Cu(s) + 8 HNO3(aq) -&gt; 3 Cu(NO3)2(aq) + 4 NO(g) + 2 H2O(l)},\tag{R1}$$
but somehow nitric dioxide, a reddish-brownish gas appears with a poignant smell, develops from this type of reaction:
$$\ce{Cu(s) + 4 HNO3(aq) -&gt; Cu(NO3)2(aq) + 2 NO2(g) + 2 H2O(l)}.\tag{R2}$$
There was 5 mL of nitric acid and as we added nitric acid we quickly covered it with a watch glass. Then we put it in a sand bath instead of heating it normally with a Bunsen burner. Unfortunately, I cannot as it was one of the experiments that we just wanted to know what was going to happen and write down our observations and conclusion.
There are two ways to interpret an observation nitrogen dioxide.

Reactions between a metal and nitric acid rarely proceed by a single path to generate a specific nitric acid reduction product; in all likelihood there will be a combination of multiple reactions and thus a mixture of products.  With copper you could get both $\ce{NO}$ and $\ce{NO2}$.  And with a more powerful reducing metal like magnesium even more products could get into the mix, including ammonium nitrate.

You may have conducted the reaction in the presence of air.  $\ce{NO}$ is readily oxidized by the air to make $\ce{NO2}$ &quot;after the fact&quot;.  This is in fact part of the natural production of nitric acid in thunderstorms.  Further complicating this is the fact that nitrogen dioxide in turn reacts with water, forming nitric oxide and regenerating some of the nitric acid (this is also indicated in the above link).  To get the true product mix out of the nitric acid you needed to use an environment free of air and unintended water.