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Question

(A) $1$. The starting material is $p$-nitrotoluene. Bromination in the presence of $FeBr_3$ occurs at the ortho position relative to the methyl group

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$2-$bromotoluene

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(A) $1$. The starting material is $p$-nitrotoluene. Bromination in the presence of $FeBr_3$ occurs at the ortho position relative to the methyl group (since $-CH_3$ is ortho/para directing and $-NO_2$ is meta directing),yielding $2$-bromo-$4$-nitrotoluene $(A)$.$2$. Reduction of the nitro group using $Sn/HCl$ converts the $-NO_2$ group to an $-NH_2$ group,yielding $2$-bromo-$4$-aminotoluene $(B)$.$3$. Treatment with $NaNO_2/HCl$ at $0-5^{\circ}C$ performs diazotization of the $-NH_2$ group,forming the diazonium salt,$2$-bromo-$4$-methylbenzenediazonium chloride $(C)$.$4$. Finally,treatment with $H_3PO_2$ and $H_2O$ reduces the diazonium group to a hydrogen atom,resulting in $2$-bromotoluene $(D)$.

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	$IV$. $3.00$

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Similar Questions

An alkyl halide reacts with alcoholic ammonia in a sealed tube,the product formed will be

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

Given below are some nitrogen-containing compounds. Each of them is treated with $HCl$ separately. $1.0 \ g$ of the most basic compound will consume $........$ $mg$ of $HCl$. ($Given$ molar mass in $g \ mol^{-1}$: $C:12, H:1, O:16, Cl:35.5$)

Ethanoic acid on heating with ammonia forms compound $A$ which on treatment with bromine and sodium hydroxide gives compound $B$. Compound $B$ on treatment with $NaNO_2/dil. HCl$ gives compound $C$. The compounds $A, B$ and $C$ respectively are

In the given reaction,the end product $(C)$ is:

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