Match the following:List-I (Amine)List-II (pK | vedclass

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(A) The basicity of amines is inversely proportional to their $pK_b$ values. $A$ lower $pK_b$ value indicates a stronger base.$1$. $N$-Ethylethanamine

Vedclass · Accepted Answer

$A-II, B-III, C-IV, D-I$

Vedclass · Answer

(A) The basicity of amines is inversely proportional to their $pK_b$ values. $A$ lower $pK_b$ value indicates a stronger base.$1$. $N$-Ethylethanamine (diethylamine) is an aliphatic secondary amine,which is much more basic than aromatic amines due to the absence of resonance stabilization of the lone pair on nitrogen. Thus,it has the lowest $pK_b$ value: $C-IV$ $(3.00)$.$2$. Among the aromatic amines,$N,N$-Dimethylaniline $(A)$ is more basic than $N$-Methylaniline $(D)$ and Aniline $(B)$ due to the electron-donating $+I$ effect of the two methyl groups,which increases electron density on the nitrogen atom. Thus,$A$ has the lowest $pK_b$ among the aromatics: $A-II$ $(8.92)$.$3$. $N$-Methylaniline $(D)$ is more basic than Aniline $(B)$ due to the $+I$ effect of one methyl group. Thus,$D-I$ $(9.30)$.$4$. Aniline $(B)$ is the least basic among these due to the resonance of the lone pair with the benzene ring. Thus,$B-III$ $(9.38)$.Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

List-$I$ (Amine)	List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline	$I$. $9.30$
$B$. Aniline	$II$. $8.92$
$C$. $N$-Ethylethanamine	$III$. $9.38$
$D$. $N$-Methylaniline	$IV$. $3.00$

List-$I$	List-$II$
$1.$ Aniline	$a.$ Used in the preparation of azo dyes
$2.$ Nitrobenzene	$b.$ Sulpha drug
$3.$ Sulphanilamide	$c.$ Solvent in Friedel-Crafts reaction
$4.$ Trinitrotoluene	$d.$ Used as an explosive

Match the following:
List-$I$ (Amine) List-$II$ ($pK_b$ value)
$A$. $N,N$-Dimethylaniline $I$. $9.30$
$B$. Aniline $II$. $8.92$
$C$. $N$-Ethylethanamine $III$. $9.38$
$D$. $N$-Methylaniline $IV$. $3.00$

The correct answer is

Similar Questions

$p$-Toluidine reacts with benzene diazonium chloride to form a compound,which on boiling with $aq. H_2SO_4$ gives ........ products.

Product $(E)$ is

The correct order of basic strengths of the given amines is:

In a reaction of aniline,a coloured product $C$ was obtained as shown below:
$A$ $\xrightarrow{NaNO_2/HCl} B$ $\xrightarrow{N,N-dimethylaniline, \text{cold}} C$
The structure of $C$ would be:

Match List-$I$ with List-$II$ and select the correct answer.

List-$I$ List-$II$

$1.$ Aniline $a.$ Used in the preparation of azo dyes

$2.$ Nitrobenzene $b.$ Sulpha drug

$3.$ Sulphanilamide $c.$ Solvent in Friedel-Crafts reaction

$4.$ Trinitrotoluene $d.$ Used as an explosive

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Match the following:List-$I$ (Amine)List-$II$ ($pK_b$ value)$A$. $N,N$-Dimethylaniline$I$. $9.30$$B$. Aniline$II$. $8.92$$C$. $N$-Ethylethanamine$III$. $9.38$$D$. $N$-Methylaniline$IV$. $3.00$The correct answer is