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

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

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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 (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	$IV$. $3.00$

Column $I$	Column $II$
$A$. $CH_3CH_2CH_2CN$	$p$. Reduction with $Pd-C/H_2$
$B$. $CH_3CH_2OCOCH_3$	$q$. Reduction with $SnCl_2/HCl$
$C$. $CH_3-CH=CH-CH_2OH$	$r$. Development of foul smell on treatment with chloroform and alcoholic $KOH$
$D$. $CH_3CH_2CH_2CH_2NH_2$	$s$. Reduction with diisobutylaluminium hydride $(DIBAL-H)$
	$t$. Alkaline hydrolysis

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

Number of structural isomers of $C_4H_{11}N$.

Benzyl isocyanide can be obtained from which of the following reactions? Choose the correct answer from the options given below:
$A$. Benzyl bromide + $AgCN$
$B$. Benzylamine + $CHCl_3$ + $Aq. NaOH$
$C$. Bromobenzene + $AgCN$
$D$. Aniline + $CHCl_3$ + $Aq. NaOH$
$E$. $2-$Phenylethyl bromide + $KCN$

In the above conversion,the correct sequence of reagents to be added is:

The correct sequence of reactions involved in the following conversion is:

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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