Consider the following sequence of reactions:
Chlorobenzene $\xrightarrow[ii) CO_2, H_3O^{+}]{i) Mg, \text{dry ether}} \text{Benzoic acid}$ $\xrightarrow{NH_3, \Delta} \text{Benzamide (A)}$ $\xrightarrow{Br_2, NaOH} \text{Aniline (B)}$
$11.25 \ mg$ of chlorobenzene will produce $.......... \times 10^{-1} \ mg$ of product $B$.
(Consider the reactions result in complete conversion.)
[Given molar mass of $C, H, O, N$ and $Cl$ as $12, 1, 16, 14$ and $35.5 \ g \ mol^{-1}$ respectively]

Given below are two statements:
Statement $I$: $o$-Phenylenediamine can be synthesized from $o$-xylene using simpler reagents in the order: $i)$ Acidic $KMnO_4$, $ii)$ Ammonia, $iii)$ Bromine and alkali.
Statement $II$: Aniline can be converted into benzene using reagents in the order: $(i)$ Bromine-$H_2O$, $(ii)$ $NaNO_2 / HCl$ $(0-5^{\circ} C)$, $(iii)$ aq. $H_3PO_2$.
In the light of the above statements, choose the correct answer from the options given below.

The product $(C)$ of the following reaction is:

Predict the major product $G$ of the following reaction sequence starting from $p$-nitrotoluene.

The ammonolysis of alcohol is represented by the reaction $xCH_3OH + yNH_3 \xrightarrow[Al_2O_3/ThO_2]{300-400^{\circ}C} \text{Products}$. The products formed are:

Consider the following reaction sequence starting from $p$-nitrotoluene:
$(i)$ $\text{Sn/HCl; OH}^-$
(ii) $(\text{CH}_3\text{CO})_2\text{O}$
(iii) $\text{Br}_2/\text{AlBr}_3$
(iv) $\text{H}_3\text{O}^+$
When the final product $(P)$ is subjected to Carius analysis using $\text{AgNO}_3$, $1.0 \text{ g}$ of the product $(P)$ will produce . . . . . . $\text{g}$ of the precipitate of $\text{AgBr}$. (Nearest Integer)
(Given: molar mass in $\text{g mol}^{-1}$: $\text{C} = 12, \text{H} = 1, \text{O} = 16, \text{N} = 14, \text{Br} = 80, \text{Ag} = 108$)