The following is a confirmatory test for arom | vedclass

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Question

(D) The reaction shown is the diazotization of aniline followed by a coupling reaction to form an azo dye. Step $1$: Aniline reacts with $NaNO_2 + HCl

Vedclass · Accepted Answer

$A = NaNO_2 + HCl, 0-5^{\circ}C$; $B = \beta	ext{-Naphthol}$

Vedclass · Answer

(D) The reaction shown is the diazotization of aniline followed by a coupling reaction to form an azo dye. Step $1$: Aniline reacts with $NaNO_2 + HCl$ at $0-5^{\circ}C$ to form benzenediazonium chloride. Thus,reagent $(A)$ is $NaNO_2 + HCl$ at $0-5^{\circ}C$. Step $2$: Benzenediazonium chloride reacts with $\beta	ext{-naphthol}$ in the presence of $NaOH$ to form a scarlet red dye (an azo dye). Thus,reagent $(B)$ is $\beta	ext{-naphthol}$. Therefore,the correct option is $(D)$.

The following is a confirmatory test for aromatic primary amines. Identify reagent $(A)$ and $(B)$.

Similar Questions

Identify $R$ in the following sequence of reaction:

In the reactions given below,
$R-Cl \xrightarrow{(i)\, KCN, (ii)\, LiAlH_4}$ Product $A$
$R-Cl \xrightarrow{(i)\, AgCN, (ii)\, LiAlH_4}$ Product $B$
the compounds $A$ and $B$ are

The major product in the following reaction is:

$A$ compound of molecular weight $180$ is acetylated to give a compound of molecular weight $390$. The number of amino groups in the initial compound is:

Which one of the following gives the most stable diazonium salt?

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