CH3MgBr→b) H3O+a) CO2(s)A→b) P2O5a) NH3/ΔB→ΔH3O+C The compound ‘C’ is

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${CH}_{3} MgBr \to_{{b) H}_{3} O^{+}}^{{a) CO}_{2} (s)} A \to_{{b) P}_{2} O_{5}}^{{a) NH}_{3} / Δ} B \to_{Δ}^{H_{3} O^{+}} C$ The compound ‘C’ is

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

${CH}_{3} - CO - {CH}_{3}$

${CH}_{3} COOH$

${({CH}_{3} CO)}_{2} O$

answer is C.

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A is $C H_{3} C O O H$ ; B is $C H_{3} C N$ . C is $C H_{3} C O O H$

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