What is the correct order of basicity for the | vedclass

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

Question

(D) The basicity of substituted anilines depends on the electron-donating or electron-withdrawing nature of the substituent attached to the benzene ri

Vedclass · Accepted Answer

$III < IV < I < II$

Vedclass · Answer

(D) The basicity of substituted anilines depends on the electron-donating or electron-withdrawing nature of the substituent attached to the benzene ring.Electron-donating groups $(EDG)$ increase the electron density on the nitrogen atom,thereby increasing basicity.Electron-withdrawing groups $(EWG)$ decrease the electron density on the nitrogen atom,thereby decreasing basicity.In the given structures:$(I)$ $p$-toluidine: $-CH_3$ is an $EDG$ (+$I$ effect).$(II)$ $p$-anisidine: $-OCH_3$ is a strong $EDG$ (+$M$ effect).$(III)$ $p$-nitroaniline: $-NO_2$ is a strong $EWG$ (-$M$ and -$I$ effect).$(IV)$ $p$-chloroaniline: $-Cl$ is an $EWG$ (-$I$ effect,but +$M$ effect is weak).Comparing the effects: The strongest $EDG$ is $-OCH_3$ $(II)$,followed by $-CH_3$ $(I)$. The EWGs are $-Cl$ $(IV)$ and $-NO_2$ $(III)$. Since $-NO_2$ is a much stronger $EWG$ than $-Cl$,it will make the aniline the least basic.Thus,the order of basicity is: $III < IV < I < II$.

What is the correct order of basicity for the substituted anilines ($I$ to $IV$)?

Similar Questions

$m$-Bromotoluene is prepared by :

Which of the following reaction/s will not give $p-$aminoazobenzene?

Identify the products $y$ and $x$ in the following reaction sequence:
$C_6H_5CONH_2 \xrightarrow{NaOBr} y$
$C_6H_5CONH_2 \xrightarrow[(ii) H_3O^+]{(i) C_6H_5SO_2Cl / py, \Delta} x$

Melting points are normally the highest for

Primary nitro compounds react with $HNO_2$ to form crystalline solids which,on treatment with $NaOH$,give:

Vedclass Test Series

Exam Paper Generator

Online Exam Module