$(a) {C H}_{3} C O {N H}_{2} + K O H ⟶$
$(b) {C H}_{3} C O O H + s o d a l i m e ⟶$
$(c) {C H}_{3} C O O A g + {B r}_{2} ⟶$
All the three reactions show similarity in :

Descending of carbon atoms

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Ascending of carbon atoms

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Both the above

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None of above

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Solution

The correct option is A Descending of carbon atoms
All the three reactions show similarity in descending of carbon atoms.
The number of carbon atoms in the product is lees than the number of carbon atoms in the reactant.
$(a) {C H}_{3} C O {N H}_{2} + B r_{2} + 4 K O H ⟶ X H_{3} N H_{2} + K_{2} C O_{3} + 2 K r + 2 H_{2} O$ Hoffmann's bromamide reaction.
$(b) {C H}_{3} C O O H + s o d a l i m e ⟶ C H_{4} + C O_{2}$ Decarboxylation.
$(c) {C H}_{3} C O O A g + {B r}_{2} ⟶ C H_{3} B r + C O_{2} + A g B r$

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