Identify product B in the following sequence | vedclass

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

Question

(D) Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $A$,which is Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: When

Vedclass · Accepted Answer

Phenol

Vedclass · Answer

(D) Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $273-278 \ K$ to form $A$,which is Benzenediazonium chloride $(C_6H_5N_2^+Cl^-)$.
Step $2$: When Benzenediazonium chloride is heated with water $(\Delta, H_2O)$,it undergoes hydrolysis to form $B$,which is Phenol $(C_6H_5OH)$,along with the evolution of Nitrogen gas $(N_2 \uparrow)$ and Hydrochloric acid $(HCl)$.
Therefore,product $B$ is Phenol.

Identify product $B$ in the following sequence of reaction:
Aniline $\xrightarrow[273 \ K]{NaNO_2+HCl} A$ $\xrightarrow[\Delta]{H_2O} B + N_2 \uparrow$

Similar Questions

The major product of the following reaction is:

Phenol is more acidic than:

For the given compounds,the correct order of increasing $pK_{a}$ value is:

Which of the following compounds has maximum volatility?

Identify the product in the following reaction: $Phenol \xrightarrow{CrO_3} \text{Product}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Identify product $B$ in the following sequence of reaction: Aniline $\xrightarrow[273 \ K]{NaNO_2+HCl} A$ $\xrightarrow[\Delta]{H_2O} B + N_2 \uparrow$