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Question

(A) $1$. The starting material is $1$-methyl-$4$-nitrobenzene. Bromination in the presence of $Fe$ gives $2$-bromo-$1$-methyl-$4$-nitrobenzene $(A)$.$

Vedclass · Accepted Answer

$2-$bromobenzoic acid

Vedclass · Answer

(A) $1$. The starting material is $1$-methyl-$4$-nitrobenzene. Bromination in the presence of $Fe$ gives $2$-bromo-$1$-methyl-$4$-nitrobenzene $(A)$.$2$. Reduction of the nitro group with $Sn/HCl$ converts $(A)$ to $2$-bromo-$4$-methylaniline $(B)$.$3$. Treatment with $NaNO_2/HCl$ at $0-5 \ ^\circ C$ converts the amine group to a diazonium salt $(C)$.$4$. Reduction of the diazonium salt with $H_3PO_2$ removes the $-N_2^+$ group,yielding $2$-bromotoluene $(D)$.$5$. Oxidation of the methyl group with $KMnO_4/HO^-$ followed by acidification yields $2$-bromobenzoic acid $(E)$.

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

Product $(E)$ is

Similar Questions

$(a) \, CH_3CONH_2 + KOH + Br_2 \rightarrow$
$(b) \, CH_3COOH + \text{Soda lime} \rightarrow$
$(c) \, CH_3COOAg + Br_2 \rightarrow$
What is the common feature in the above three reactions?

An aromatic amine $(A)$ was treated with alcoholic potash and another compound $(Y)$ to form a foul-smelling gas with the formula $C_6H_5NC$. Compound $(Y)$ was formed by reacting a compound $(Z)$ with $Cl_2$ in the presence of slaked lime. The compound $(Z)$ is:

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

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 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:

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