Question

In the following reaction the product B is
$$A\stackrel{Bromination}{\to }B\stackrel{NaN{O}_2/HCl}{\to }C\underset{C_2{H}_{5}OH}{\overset{Boiling}{\to }}tribromobenzene$$
A. Salicylic acid
B. Benzoic acid
C. Phenol
D. 2,4,6-tribromoaniline

Answer 1

Hint: In the given question, we have given that there are three sequences of organic reactions. First reaction is a bromination, second one is diazotization and the third one is boiling in presence of ethyl alcohol

Complete step by step answer:
The tribromobenzene can be obtained from diazonium salt by boiling with ethyl alcohol. Thus the compound C could be 2,4,6-tribromobenzene diazonium chloride. The reaction is given below;
                             

The compound 2, 4, -tribromobenzene diazonium chloride(C) could be prepared by diazotization if $$2,4,6-tribromoaniline\left(B\right)$$ with $$NaN{O}_2/HCl$$ at low temperature. The reaction is given below;
                      

The compound $$2,4,6-tribromoaniline$$ (B) could be obtained by bromination of aniline group. Hence, the compound A could be aniline.the reaction is given below
              

From these steps we got that the product B is $$2,4,6-tribromoaniline$$
The complete reaction sequence can be represented as follows
 

So, the correct answer is Option D.

Note:
Bromination of both phenol and aniline is very difficult to control, with di-, and tri-bromo products forming readily. Because of their high nucleophilic reactivity, aniline phenol undergoes substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. In the question the reactions occur in a step by step manner. Starting with bromination of aniline , then diazotization of tribromoaniline and ends with tribromobenzene

In the solution the sequence of reactions starts with aniline. Like aniline there are several benzene derivatives(functional groups attached in benzene) such as phenol, carboxylic acid which gives different products in respective steps.so, we have to be aware of such things