Match the following reactions of benzene diaz | vedclass

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

Question

(A) The reactions of benzene diazonium chloride $(ArN_2^{+} Cl^{-})$ are as follows:$(A)$ Hydrolysis with $H_2O$ yields phenol $(ArOH)$. Thus,$(A) - (

Vedclass · Accepted Answer

$A-q, B-p, C-s, D-r$

Vedclass · Answer

(A) The reactions of benzene diazonium chloride $(ArN_2^{+} Cl^{-})$ are as follows:$(A)$ Hydrolysis with $H_2O$ yields phenol $(ArOH)$. Thus,$(A) - (q)$.$(B)$ Reaction with $HBF_4$ followed by $NaNO_2$ and heating yields nitrobenzene $(ArNO_2)$. Thus,$(B) - (p)$.$(C)$ Reduction with ethanol $(CH_3CH_2OH)$ yields benzene $(ArH)$. Thus,$(C) - (s)$.$(D)$ Balz-Schiemann reaction with $HBF_4$ followed by heating yields fluorobenzene $(ArF)$. Thus,$(D) - (r)$.Therefore,the correct matching is $A-q, B-p, C-s, D-r$.

Column-$I$	Column-$II$
$(A)$ $ArN_2^{+} Cl^{-} \rightarrow ArOH$	$(p)$ $HBF_4 / NaNO_2, \Delta$
$(B)$ $ArN_2^{+} Cl^{-} \rightarrow ArNO_2$	$(q)$ $H_2 O$
$(C)$ $ArN_2^{+} Cl^{-} \rightarrow ArH$	$(r)$ $HBF_4, \Delta$
$(D)$ $ArN_2^{+} Cl^{-} \rightarrow ArF$	$(s)$ $CH_3CH_2OH$

Similar Questions

Alkylamine dissolves in hydrochloric acid to form alkylammonium chloride. The nitrogen in the latter salt is:

Consider the following reaction:
$C_6H_5NO_2$ $\xrightarrow{Sn / HCl} X$ $\xrightarrow{C_6H_5COCl} Y + HCl$
What is $Y$?

In the following reaction,the product $(A)$ is

When aniline is treated with sodium nitrite and hydrochloric acid at $0\,^{\circ}C$,it gives:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module