The compound A with the following sequence of | vedclass

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

Question

(B) The reaction sequence is as follows:$1. 	ext{Aniline } (C_6H_5NH_2)$ $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C} 	ext{Benzenediazonium chloride } (C

Vedclass · Accepted Answer

Aniline

Vedclass · Answer

(B) The reaction sequence is as follows:$1. 	ext{Aniline } (C_6H_5NH_2)$ $\xrightarrow{NaNO_2/HCl, 0-5^{\circ}C} 	ext{Benzenediazonium chloride } (C_6H_5N_2Cl) 	ext{ (Compound } B	ext{)}$$2. 	ext{Benzenediazonium chloride } (C_6H_5N_2Cl)$ $\xrightarrow{KCN/CuCN} 	ext{Benzonitrile } (C_6H_5CN) 	ext{ (Compound } C	ext{)}$$3. 	ext{Benzonitrile } (C_6H_5CN) \xrightarrow{H_3O^{+}} 	ext{Benzoic acid } (C_6H_5COOH)$Therefore,the compound $A$ is Aniline.

The compound $A$ with the following sequence of reactions gives benzoic acid.
$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{KCN} C$ $\xrightarrow{H_3O^{+}} \text{benzoic acid}$
The compound $A$ is:

Similar Questions

The correct statements of the following are:
$A$. Aniline forms a stable benzene diazonium chloride at $285 \ K$.
$B$. $N$-Phenylethanamide is less reactive towards nitration than aniline.
$C$. $p-CH_3C_6H_4COCl$ is Hinsberg reagent.

For the transformation shown below,the reagent used is:

Identify the amine that does not form alcohol by the action of $NaNO_2$ and $HCl$.

Nitrobenzene on reduction using zinc in alkaline medium results in $X$. The number of $\sigma$ and $\pi$ bonds in $X$ is

Identify $(a)$ and $(b)$ in the following reaction:
$CH_3CH_2NH_2 + CHCl_3 + 3KOH \rightarrow (a) + (b) + 3H_2O$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The compound $A$ with the following sequence of reactions gives benzoic acid.$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{KCN} C$ $\xrightarrow{H_3O^{+}} \text{benzoic acid}$The compound $A$ is: