Chemistry Q67 – JEE-MAIN 2026

Home Chemistry Basic Organic ChemistryQuestion #122

Given below are two statements : Statement I : Heating benzamide with bromine in an ethanolic solution of sodium hydroxide will give benzylamine. Statement II : Nitration of aniline with

H N O_{3} / H_{2} S O_{4}

at 288 K produces

m

-nitroaniline in higher amount than

o

-nitroaniline (pH adjusted). In the light of the above statements, choose the correct answer from the options given below :

Clue 1 of 3: Identify the reactions

For Statement I, recall the Hofmann bromamide degradation reaction. For Statement II, consider the nitration of aniline and the effect of acidic conditions on the amino group.

🧠 Full Solution Path

Step 1: Evaluate Statement I (Hofmann Bromamide Degradation)✦ Active

Statement I describes the reaction of benzamide ( $C_{6} H_{5} C O N H_{2}$ ) with bromine in an ethanolic solution of sodium hydroxide. This is the Hofmann bromamide degradation reaction. In this reaction, an amide is converted to a primary amine with one carbon atom less than the starting amide.

C_{6} H_{5} C O N H_{2} \overset{B r_{2} / N a O H}{\to} C_{6} H_{5} N H_{2} (Aniline)

The product formed is aniline ( $C_{6} H_{5} N H_{2}$ ), not benzylamine ( $C_{6} H_{5} C H_{2} N H_{2}$ ). Therefore, Statement I is false.

Step 2: Evaluate Statement II (Nitration of Aniline)○ Expand

Statement II concerns the nitration of aniline with $H N O_{3} / H_{2} S O_{4}$ at 288 K. Aniline ( $C_{6} H_{5} N H_{2}$ ) is an ortho/para directing group. However, under the strongly acidic conditions of nitration, aniline is protonated to form the anilinium ion ( $C_{6} H_{5} N H_{3}^{+}$ ).

C_{6} H_{5} N H_{2} + H^{+} ⇌ C_{6} H_{5} N H_{3}^{+}

The anilinium ion is a meta-directing and deactivating group. Consequently, nitration of aniline yields a significant amount of meta-product along with ortho and para products. Typical product distribution is approximately: $o$ -nitroaniline (~2%), $m$ -nitroaniline (~47%), and $p$ -nitroaniline (~51%).

Comparing the amounts, $m$ -nitroaniline (47%) is indeed produced in a higher amount than $o$ -nitroaniline (2%). Therefore, Statement II is true.

💡 Teacher's Secret Hint

Remember that the amino group in aniline gets protonated in acidic conditions, changing its directing nature.

Step 3: Conclusion○ Expand

Based on the analysis, Statement I is false, and Statement II is true. This corresponds to option 4.

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Chemistry Question 67 – JEE-MAIN 2026

Clue 1 of 3: Identify the reactions

🧠 Full Solution Path