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(B) Step $i$: Aniline reacts with acetic anhydride/pyridine to form acetanilide $(C_6H_5NHCOCH_3)$. This protects the amino group and reduces its acti

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$1,4$-dibromobenzene

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(B) Step $i$: Aniline reacts with acetic anhydride/pyridine to form acetanilide $(C_6H_5NHCOCH_3)$. This protects the amino group and reduces its activating power.Step $ii$: Electrophilic aromatic substitution with $KBrO_3/HBr$ (which generates $Br_2$) occurs. The acetamido group is ortho/para directing,but due to steric hindrance,para-bromoacetanilide is the major product.Step $iii$: Hydrolysis with $H_3O^+/	ext{heat}$ removes the acetyl group to yield $p$-bromoaniline.Step $iv$: Diazotization with $NaNO_2/HCl$ at $273-278 \ K$ converts the amino group into a diazonium salt $(-N_2^+Cl^-)$.Step $v$: Sandmeyer reaction with $Cu/HBr$ replaces the diazonium group with a bromine atom,resulting in $1,4$-dibromobenzene ($p$-dibromobenzene).

The product$(s)$ of the following reaction sequence is(are):
Aniline $\xrightarrow[\text{ii) } KBrO_3/HBr]{\text{i) Acetic anhydride/pyridine}}$ $\xrightarrow[\text{iv) } NaNO_2/HCl, 273-278 \ K]{\text{iii) } H_3O^+, \text{heat}}$ $\xrightarrow{\text{v) } Cu/HBr} \text{Product}$

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Arrange the following in order of their decreasing acidic strength:
$(I)$ $CH_3NH_3^+$
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The product $(A)$ is:

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The product$(s)$ of the following reaction sequence is(are):Aniline $\xrightarrow[\text{ii) } KBrO_3/HBr]{\text{i) Acetic anhydride/pyridine}}$ $\xrightarrow[\text{iv) } NaNO_2/HCl, 273-278 \ K]{\text{iii) } H_3O^+, \text{heat}}$ $\xrightarrow{\text{v) } Cu/HBr} \text{Product}$