Synthesis of azo dye

Question: Which of the following form a brilliant coloured dye on treatment with NaNO₂ in dil HCl followed by addition to an alkaline solution of β-naphthanol.

Options:

Solution: First read the question carefully. It is based on the syntheses of dye. Now notice the first step of the reaction in which a compound reacts with NaNO₂ and dil HCl Or we can say with nitrous acid (HNO₂). 

(Since nitrous acid is unstable, it is prepared in stu by the action of cold and dilute hydrochloric acid on sodium nitrite.)

NaNO₂ + HCl = NaCl + HNO₂

So the first step of reaction and final product i.e. dye, indicate that the question is all about the syntheses of azo dye.  

If you remember, azo dyes are prepared in a two steps reaction, the first being the synthesis of an aromatic diazonium ion from an aniline derivative in the presence of sodium nitrite and dil HCl. The next step is coupling of the diazonium salt with an aromatic compound.

So let’s check our options:

a)

                                                                      

b)

c)

 d)

 Here options c and d give dizonium salt on treatment with NaNO2 and dil HCl.

But when Benzylamine i.e.option d reacts with nitrous acid it form unstable diazonium salt which in turn gives alcohol with the evolution of nitrogen gas.

On the other hand, option c reacts with HNO2 at a low temperature to form stable diazonium salt. Thus nitrogen gas is not evolved. On treatment with β-naphthanol it forms azo dye.

Thus, option c is correct.