Consider the following reaction sequence Benzene xrightarrowtextconc. HNO_3 + textconc. H_2SO_4, 333text K textP xrightarrow1.text Sn/HCl/Delta quad 2.text pH neutralised textQ xrightarrow(mathrmCH_3mathrmCO)_2mathrmO textR xrightarrow1.text conc. HNO_3 + text conc. H_2SO_4 quad 2.text pH neutralised (major product) textS xrightarrowmathrmHCl / mathrmEtOH / Delta textT The percentage of nitrogen in product ‘T’ formed is ____%. (Nearest integer) (Given molar mass in mathrmg\ mol^-1 H:1, C:12, N:14, O:16)

Numerical Answer Type:
Enter a numerical value Answer: 20 to 20 +4 marks

Solution & Explanation

### Core Logic Step 1: Nitration of benzene gives nitrobenzene (P). mathrmPh-H xrightarrowHNO_3/H_2SO_4 mathrmPh-NO_2 quad text(P)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Step 2: Reduction of nitrobenzene with Sn/HCl gives aniline (Q). mathrmPh-NO_2 xrightarrowSn/HCl mathrmPh-NH_2 quad text(Q)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Step 3: Acetylation of aniline with acetic anhydride gives acetanilide (R). This protects the amino group to prevent oxidation and polysubstitution in the next step. mathrmPh-NH_2 xrightarrow(CH_3CO)_2O mathrmPh-NH-CO-CH_3 quad text(R)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Step 4: Nitration of acetanilide gives predominantly p-nitroacetanilide (S) due to steric hindrance at ortho position. mathrmPh-NH-CO-CH_3 xrightarrowHNO_3/H_2SO_4 ptext-NO_2text-C_6textH_4text-NH-CO-CH_3 quad text(S) Step 5: Acidic hydrolysis of the amide linkage yields p-nitroaniline (T). ptext-NO_2text-C_6textH_4text-NH-CO-CH_3 xrightarrowHCl/EtOH/Delta ptext-NO_2text-C_6textH_4text-NH_2 quad text(T)
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Electrophilic Substitution Reactions diagram for Q74 - JEE Main 2026 Morning
Molecular formula of p-nitroaniline (T) is C_6H_6N_2O_2. Molar mass = (6 times 12) + (6 times 1) + (2 times 14) + (2 times 16) = 72 + 6 + 28 + 32 = 138text g/mol. Total mass of Nitrogen = 2 times 14 = 28text g. Percentage of Nitrogen = frac28138 times 100 approx 20.29\%. ### Step 1: Final Conclusion Nearest integer is 20. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Amines

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