How will you convert $4-$nitrotoluene to $2-$bromobenzoic acid?
(N/A) The conversion of $4-$nitrotoluene to $2-$bromobenzoic acid can be achieved through the following steps:
$1$. Bromination of $4-$nitrotoluene: $4-$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 give $2-$bromo-$4-$methylaniline.
$3$. Diazotization: $2-$bromo-$4-$methylaniline is treated with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt,$2-$bromo-$4-$methylbenzenediazonium chloride.
$4$. Deamination: The diazonium group is removed using $H_3PO_2$ and $H_2O$ to yield $2-$bromotoluene.
$5$. Oxidation: Finally,the methyl group $(-CH_3)$ is oxidized to a carboxylic acid group $(-COOH)$ using $KMnO_4/OH^-$ to obtain $2-$bromobenzoic acid.
$1$. Bromination of $4-$nitrotoluene: $4-$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 give $2-$bromo-$4-$methylaniline.
$3$. Diazotization: $2-$bromo-$4-$methylaniline is treated with $NaNO_2/HCl$ at $273-278 \ K$ to form the diazonium salt,$2-$bromo-$4-$methylbenzenediazonium chloride.
$4$. Deamination: The diazonium group is removed using $H_3PO_2$ and $H_2O$ to yield $2-$bromotoluene.
$5$. Oxidation: Finally,the methyl group $(-CH_3)$ is oxidized to a carboxylic acid group $(-COOH)$ using $KMnO_4/OH^-$ to obtain $2-$bromobenzoic acid.
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