Correct sequence of reagents for the conversi | vedclass

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(D) The direct nitration of aniline with $HNO_3 / H_2SO_4$ leads to the formation of $m$-nitroaniline due to the protonation of the $-NH_2$ group to $

Vedclass · Accepted Answer

$(CH_3CO)_2O / 	ext{Pyridine}$,$HNO_3 / H_2SO_4$ at $288 \ K$,$OH^-$ or $H^+$

Vedclass · Answer

(D) The direct nitration of aniline with $HNO_3 / H_2SO_4$ leads to the formation of $m$-nitroaniline due to the protonation of the $-NH_2$ group to $-NH_3^+$,which is meta-directing. To obtain $p$-nitroaniline,the $-NH_2$ group must first be protected by acetylation using $(CH_3CO)_2O / 	ext{Pyridine}$ to form acetanilide. The acetanilide group $(-NHCOCH_3)$ is ortho/para-directing and less activating,which allows for controlled nitration at the para position using $HNO_3 / H_2SO_4$ at $288 \ K$. Finally,the acetyl group is removed by hydrolysis using $OH^-$ or $H^+$ to yield $p$-nitroaniline. Thus,the correct sequence is: $(CH_3CO)_2O / 	ext{Pyridine}$,$HNO_3 / H_2SO_4$ at $288 \ K$,$OH^-$ or $H^+$.

Correct sequence of reagents for the conversion shown in the image is $-$

Similar Questions

Primary nitro alkanes are obtained in good yield by oxidising aldoximes with the help of $-$

The decreasing order of basicity is:
$(1) \ CH_3CONH_2$
$(2) \ CH_3CH_2NH_2$
$(3) \ PhCH_2CONH_2$

Assertion : Nitration of aniline can be conveniently done by protecting the amino group by acetylation.
Reason : Acetylation increases the electron density in the benzene ring.

Identify the product $C$ in the following reaction sequence:
$C_6H_5COOH$ $\xrightarrow{NH_3, \Delta} A$ $\xrightarrow{Br_2, KOH} B$ $\xrightarrow{CHCl_3, KOH} C$

Find the correct order of acidity of the following compounds:
$(I)$ $C_6H_5NH_2$
$(II)$ $C_6H_5CONH_2$
$(III)$ Phthalimide
$(IV)$ $CH_3CH_2NH_2$

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