Acetic anhydride with ammonia gives the product:

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

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

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C H_{3} C N

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C H_{3} C O O N H_{4}

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Solution

The correct option is B $C H_{3} C O N H_{2}$
$(C H_{3} C O)_{2} O + 2 N H_{3} \to C H_{3} C O N H_{2} + C H_{3} C O O^{-} N H_{4}^{+}$ .

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