See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

The compound given is 4-heptanone: CH3-CH2-CO-CH2-CH2-CH3. The carbonyl carbon is at position 4 (a ketone). To form enol tautomers, a proton is removed from an alpha carbon and the OH is formed on the carbonyl carbon, creating a C=C double bond. Step 1: Identify the alpha carbons. The carbonyl group (C=O) at C4 has two sets of alpha carbons: - Alpha carbon on the left side: C3 (the CH2 adjacent to C=O on the ethyl side, i.e., CH3-CH2- side) - Alpha carbon on the right side: C5 (the CH2 adjacent to C=O on the propyl side, i.e., -CH2-CH2-CH3 side) Step 2: Enol formation by removing H from left alpha carbon (C3): This gives: CH3-CH=C(OH)-CH2-CH2-CH3 The double bond is between C3 and C4. C3 was CH2 (two H's removed one at a time) — but when we form the enol, C3 becomes CH= and C4 becomes C(OH)=. C3 in the enol: CH3-CH=, so C3 has one H and is connected to CH3 and C4(OH)=. This double bond (C3=C4) — C3 is connected to CH3 and H; C4 is connected to OH and -CH2CH2CH3. This gives E and Z stereoisomers because both carbons of the double bond have two different substituents: - C3: H and CH3 - C4: OH and -CH2CH2CH3 So this gives 2 enols (E and Z). Step 3: Enol formation by removing H from right alpha carbon (C5): This gives: CH3-CH2-C(OH)=CH-CH2-CH3 The double bond is between C4 and C5. C5 becomes =CH- with one H. C4: connected to OH and -CH2CH3 C5: connected to H and -CH2CH3 Both carbons of the double bond have two different substituents: - C4: OH and -CH2CH3 - C5: H and -CH2CH3 This also gives E and Z stereoisomers — 2 more enols. Step 4: Total enol forms = 2 (from left alpha) + 2 (from right alpha) = 4. Why other options fail: - (a) 2: Counts only one direction of enolization without stereoisomers. - (b) 3: Undercounts; misses one stereoisomer pair. - (d) 5: Overcounts; there is no additional site of alpha deprotonation. Therefore, the correct answer is C.