Ethyl acetate reacts with $N H_{2} N H C O N H_{2}$ to form:

C H_{3} C O N H C O N H N H_{2}

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C H_{3} C O N (N H_{2}) C O N H_{2}

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C H_{3} C O N H N H C O N H_{2}

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C H_{3} C H_{2} N H N H C O N H_{2}

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Solution

The correct option is C $C H_{3} C O N H N H C O N H_{2}$
$C H_{3} C O O C_{2} H_{5} + H_{2} N N H C O N H_{2} \to C H_{3} C O N H N H C O N H_{2} + C_{2} H_{5} O H$

The lone pair on nitrogen will attack on $- C = O$ group and the alkoxy part ( $- O C_{2} H_{5}$ ) will leave.

Option C is correct.

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What happens when acetic acid reacts with ethanol in presence of conc. H₂SO₄?

H_{2} S O

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