Two moles of acetic acid are heated with , The product formed is — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

# Dehydration of Acetic Acid with $P_2O_5$ **Step 1: Identify the reactant and reagent** - Reactant: 2 moles of acetic acid ($CH_3COOH$) - Reagent: $P_2O_5$ (phosphorus pentoxide) — a strong dehydrating agent **Step 2: Recognize the reaction type** $P_2O_5$ causes dehydration of carboxylic acids. Two acetic acid molecules can undergo condensation with loss of water. **Step 3: Write the reaction** $$2\,CH_3COOH \xrightarrow{P_2O_5} CH_3-CO-O-CO-CH_3 + H_2O$$ **Step 4: Identify the product** The product is **acetic anhydride** ($CH_3CO)_2O$ or $(CH_3CO)_2O$ Two molecules of acetic acid lose one water molecule to form an anhydride, where the two acyl groups ($CH_3CO-$) are connected through an oxygen atom. **Why this is correct:** - $P_2O_5$ selectively dehydrates carboxylic acids to form their anhydrides - This is a characteristic reaction of acetic acid with $P_2O_5$ - No other products (like acetaldehyde or ketene as sole products) form under these conditions **Answer: Acetic anhydride** [$(CH_3CO)_2O$]