See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Identify the reaction. CH3MgBr (a Grignard reagent) reacts with CH3CH2C(=O)Cl (propanoyl chloride / ethanoyl chloride with an ethyl group — i.e., propanoyl chloride). Grignard reagents react with acid chlorides: the first equivalent of RMgBr adds to give a ketone intermediate, and since excess Grignard is typically present (or the reaction proceeds), a second equivalent adds to give a tertiary alcohol. However, with one equivalent, the product is a ketone. Step 2: Product identification. CH3MgBr adds to propanoyl chloride (CH3CH2COCl). The acid chloride first reacts with CH3MgBr to give the ketone: CH3CH2-C(=O)-CH3, which is methyl ethyl ketone (butан-2-one, CH3COCH2CH3). Step 3: Enumerate enols of butan-2-one (CH3-CO-CH2-CH3). Enols are formed by tautomerization: moving the carbonyl to an enol (C=C-OH). There are two sets of alpha carbons: - Alpha carbon on the methyl side (CH3): gives enol CH2=C(OH)-CH2CH3. This double bond is terminal; no geometric isomerism possible (one end has two H substituents). This gives 1 enol structure (no E/Z). - Alpha carbon on the ethyl side (CH2): gives enol CH3-C(OH)=CH-CH3. This double bond has different groups on each carbon: C1 has CH3 and OH; C2 has CH3 and H. This allows E/Z (cis/trans) geometric isomerism, giving 2 enol structures (E and Z). Step 4: Total enols = 1 (from methyl side) + 2 (E and Z from ethyl side) = 3. Step 5: Why other options fail: - (a) 2: Undercounts; misses the geometric isomers or one set. - (c) 4 and (d) 5: Overcount; no additional stereoisomers exist beyond the E/Z pair. Therefore, the correct answer is B.