How will you carry out the following conversions?
$(i)$ Aniline to $1,3$-dibromo-$5$-nitrobenzene
(ii) Aniline to $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene
$(i)$ Aniline to $1,3$-dibromo-$5$-nitrobenzene
(ii) Aniline to $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene
(N/A) The conversion steps are as follows:
$(i)$ Aniline to $1,3$-dibromo-$5$-nitrobenzene:
$1$. Aniline is treated with acetic anhydride in pyridine to form acetanilide.
$2$. Acetanilide is nitrated with conc. $HNO_3$ and conc. $H_2SO_4$ to give $p$-nitroacetanilide.
$3$. Hydrolysis with $H_2O/H^+$ yields $p$-nitroaniline.
$4$. Bromination with $Br_2/CH_3COOH$ gives $2,6$-dibromo-$4$-nitroaniline.
$5$. Diazotization with $HNO_2$ at $273-278 \ K$ gives the diazonium salt.
$6$. Reduction with $H_3PO_2$ removes the diazonium group to yield $1,3$-dibromo-$5$-nitrobenzene.
(ii) Aniline to $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene:
Follow the same steps as above until the formation of the diazonium salt ($2,6$-dibromo-$4$-nitrobenzenediazonium chloride).
Then,treat the diazonium salt with $KI$ to replace the diazonium group with an iodine atom,yielding $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene.
$(i)$ Aniline to $1,3$-dibromo-$5$-nitrobenzene:
$1$. Aniline is treated with acetic anhydride in pyridine to form acetanilide.
$2$. Acetanilide is nitrated with conc. $HNO_3$ and conc. $H_2SO_4$ to give $p$-nitroacetanilide.
$3$. Hydrolysis with $H_2O/H^+$ yields $p$-nitroaniline.
$4$. Bromination with $Br_2/CH_3COOH$ gives $2,6$-dibromo-$4$-nitroaniline.
$5$. Diazotization with $HNO_2$ at $273-278 \ K$ gives the diazonium salt.
$6$. Reduction with $H_3PO_2$ removes the diazonium group to yield $1,3$-dibromo-$5$-nitrobenzene.
(ii) Aniline to $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene:
Follow the same steps as above until the formation of the diazonium salt ($2,6$-dibromo-$4$-nitrobenzenediazonium chloride).
Then,treat the diazonium salt with $KI$ to replace the diazonium group with an iodine atom,yielding $1,3$-dibromo-$2$-iodo-$5$-nitrobenzene.
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