See image — Reaction Mechanism Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the final product and work backwards. The final product is CH3CH2CH2NH2 (propylamine, n-propylamine), a primary amine with 3 carbons. Step 2 - The last reaction is Br2/KOH (Hofmann bromamide degradation / Hofmann rearrangement). This reaction converts a primary amide (RCONH2) to a primary amine (RNH2) with one fewer carbon. So the intermediate C4H9ON must be a primary amide. If the product amine is CH3CH2CH2NH2 (3C), the amide must be CH3CH2CH2CONH2 (butanamide, C4H9ON). Check formula: C4H9ON = butanamide (CH3CH2CH2CONH2): C4H9NO - yes, this matches C4H9ON. Step 3 - The first reaction converts (X) C4H7OCl with NH3 to give butanamide C4H9ON. For NH3 to convert an acid chloride to an amide, (X) must be butanoyl chloride (CH3CH2CH2COCl). Check formula of butanoyl chloride: C4H7OCl - C=4, H=7, O=1, Cl=1. Yes, this matches. Step 4 - Butanoyl chloride (CH3CH2CH2COCl) reacts with NH3 to give butanamide (CH3CH2CH2CONH2, C4H9ON). Then Hofmann rearrangement with Br2/KOH gives CH3CH2CH2NH2 (propylamine). This pathway is fully consistent. Step 5 - Identify option (a): The structure shown is a ketone (C=O) with a Cl on the alpha carbon of a 4-carbon chain, which corresponds to 1-chlorobutan-2-one or similar - but looking more carefully, option (a) shows a straight chain with a ketone carbonyl and Cl directly on the carbonyl carbon in an acyl chloride arrangement. Actually, option (a) depicts what appears to be CH3CH2CH2COCl (butanoyl chloride) - an acyl chloride where the carbonyl carbon bears the Cl, matching C4H7OCl. This is butanoyl chloride. Step 6 - Why other options fail: - Option (b) is (CH3)2C-COCl (pivaloyl chloride, (CH3)2CHCOC... actually 2,2-dimethylpropanoyl chloride): Hofmann on its amide would give (CH3)2CHNH2 (isopropylamine), not n-propylamine. - Option (c) is a cyclopropane with Cl and OH, which would not cleanly give butanamide with NH3. - Option (d) Cl(CH2)4CHO is a chloroaldehyde, not an acid chloride; reaction with NH3 would not give a simple amide suitable for Hofmann rearrangement to n-propylamine. Therefore, the correct answer is A.