In the reaction,Aniline xrightarrow[text{Dil. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reaction sequence is as follows:$1$. Aniline reacts with $NaNO_2$ and $	ext{Dil. } HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $

Vedclass · Accepted Answer

$para$-hydroxyazobenzene

Vedclass · Answer

(C) The reaction sequence is as follows:$1$. Aniline reacts with $NaNO_2$ and $	ext{Dil. } HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride $(P)$: $C_6H_5NH_2 + NaNO_2 + 2HCl ightarrow C_6H_5N_2^+Cl^- + NaCl + 2H_2O$$2$. Benzenediazonium chloride $(P)$ then undergoes an electrophilic substitution (coupling reaction) with phenol in the presence of $NaOH$ ($pH$ $9-10$). The coupling occurs primarily at the $para$-position due to steric hindrance at the $ortho$-position,forming $p$-hydroxyazobenzene $(Q)$: $C_6H_5N_2^+Cl^- + C_6H_5OH \xrightarrow{NaOH} C_6H_5-N=N-C_6H_4-OH(p) + HCl$

In the reaction,Aniline $\xrightarrow[\text{Dil. } HCl]{NaNO_2} P$ $\xrightarrow[NaOH]{\text{Phenol}} Q$,identify $Q$.

Similar Questions

Ethyl amine undergoes oxidation in the presence of $KMnO_4$ to form

What are $X$ and $Y$ in the following reactions?

Which of the following statements is true for Benzenediazonium fluoroborate?

Write short notes on the following:
$(i)$ Carbylamine reaction
$(ii)$ Diazotisation
$(iii)$ Hofmann's bromamide reaction
$(iv)$ Coupling reaction
$(v)$ Ammonolysis
$(vi)$ Acetylation
$(vii)$ Gabriel phthalimide synthesis.

The product formed when aniline is treated with chloroform in the presence of $KOH$ (alcoholic) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module