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Question

(A) The conversion of $p$-nitrotoluene to $2$-bromo-$4$-methylbenzonitrile involves the following steps:$1$. Bromination: $p$-Nitrotoluene reacts with

Vedclass · Accepted Answer

Bromination,reduction,carbylamine reaction

Vedclass · Answer

(A) The conversion of $p$-nitrotoluene to $2$-bromo-$4$-methylbenzonitrile involves the following steps:$1$. Bromination: $p$-Nitrotoluene reacts with $Br_2$ to form $2$-bromo-$4$-nitrotoluene.$2$. Reduction: The nitro group $(-NO_2)$ is reduced to an amino group $(-NH_2)$ using $Sn + HCl$ to form $2$-bromo-$4$-methylaniline.$3$. Carbylamine reaction: The primary amine $(-NH_2)$ is converted to an isocyanide $(-NC)$ using $CHCl_3 + KOH$ (Note: The final product in the provided image is an isocyanide,not a nitrile,which is characteristic of the carbylamine reaction).Thus,the correct sequence is Bromination,reduction,carbylamine reaction.

List-$I$ (Mixture of Compounds)	List-$II$ (Reagent used to distinguish)
$A$. Diethyl amine + Ethyl amine	$I$. Bromine water
$B$. Acetaldehyde + Acetone	$II$. $CHCl_3 + KOH, \Delta$
$C$. Ethanol + Phenol	$III$. Neutral $FeCl_3$
$D$. Benzoic acid + Cinnamic acid	$IV$. Ammonical silver nitrate

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 sequence of reactions involved in the following conversion is:

Similar Questions

Product $(E)$ is

Consider the above sequence of reactions. The number of bromine atom$(s)$ in the final product $(P)$ will be:

The major product $Z$ formed in the following sequence of reactions is:

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

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