% of N in compound E is ____

The given reaction sequence starts with benzene and proceeds through the following steps to form compound E:

Step 1: Nitration of Benzene to form (A)
Benzene reacts with a nitrating mixture of concentrated ${\mathrm{HNO}}_3$ and concentrated $H_2{\mathrm{SO}}_4$ to undergo electrophilic aromatic substitution, yielding nitrobenzene (A):
$C_6{H}_6+{\mathrm{NO}}_2^{+}\to C_6{H}_5{\mathrm{NO}}_2\text{ (Nitrobenzene)}$

Step 2: Reduction of (A) to form (B)
Nitrobenzene (A) is reduced using tin and hydrochloric acid ($\mathrm{Sn}/\mathrm{HCl}$) to form aniline (B):
$C_6{H}_5{\mathrm{NO}}_2\stackrel{\mathrm{Sn}/\mathrm{HCl}}{\to }{C}_6{H}_5{\mathrm{NH}}_2\text{ (Aniline)}$

Step 3: Acetylation of (B) to form (C)
Aniline (B) reacts with acetic anhydride ($({\mathrm{CH}}_3\mathrm{CO}{)}_{2}O$) to protect the amino group, forming acetanilide (C):
$C_6{H}_5{\mathrm{NH}}_2+({\mathrm{CH}}_3\mathrm{CO}{)}_{2}O\to C_6{H}_5{\mathrm{NHCOCH}}_3\text{ (Acetanilide)}$

Step 4: Nitration of (C) to form (D)
Nitration of acetanilide (C) with concentrated ${\mathrm{HNO}}_3$ and concentrated $H_2{\mathrm{SO}}_4$ selectively introduces a nitro group at the para-position due to the steric hindrance of the acetamido group, yielding p-nitroacetanilide (D):
$C_6{H}_5{\mathrm{NHCOCH}}_3\stackrel{{\mathrm{HNO}}_3/H_2{\mathrm{SO}}_4}{\to }{O}_{2}N-C_6{H}_4-{\mathrm{NHCOCH}}_3\text{ (p-Nitroacetanilide)}$

Step 5: Hydrolysis of (D) to form (E)
Acidic hydrolysis of p-nitroacetanilide (D) regenerates the primary amine, yielding p-nitroaniline (E):
$O_{2}N-C_6{H}_4-{\mathrm{NHCOCH}}_3\stackrel{H^{+}/H_{2}O}{\to }{O}_{2}N-C_6{H}_4-{\mathrm{NH}}_2\text{ (p-Nitroaniline)}$

Calculation of Mass Percentage of Nitrogen in Compound E:
The chemical formula of p-nitroaniline (E) is $C_6{H}_6{N}_2{O}_2$.
Let us calculate its molar mass:
Molar Mass of Carbon (C) = $6\times 12=72\text{g/mol}$
Molar Mass of Hydrogen (H) = $6\times 1=6\text{g/mol}$
Molar Mass of Nitrogen (N) = $2\times 14=28\text{g/mol}$
Molar Mass of Oxygen (O) = $2\times 16=32\text{g/mol}$
Total Molar Mass of Compound E = $72+6+28+32=138\text{g/mol}$

Now, the percentage of Nitrogen (% of N) is calculated as:
$\text{\% of N}=\frac{\text{Mass of Nitrogen}}{\text{Molar Mass of E}}\times 100$

$\text{\% of N}=\frac{28}{138}\times 100\approx 20.29\%$

Rounding to the nearest integer, the percentage of Nitrogen in compound E is 20%.