The ketone shown can form enols by loss of an alpha hydrogen. Count all distinct enol forms (OH on a carbon that was the carbonyl carbon), counting E and Z isomers of the same constitution separately. Give the answer as an integer.

Numerical answer type, tolerance 0.5.

Answer: 4

The ketone is hexan-3-one. Enolisation toward the ethyl side gives CH3CH=C(OH)CH2CH2CH3 (E and Z), and toward the propyl side gives CH3CH2C(OH)=CHCH2CH3 (E and Z). Each double bond has different substituents on both carbons, so each is E/Z. Total = 2 + 2 = 4.

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The ketone shown can form enols by loss of an alpha hydrogen. Count all distinct enol forms (OH on a carbon that was the carbonyl carbon), counting E and Z isomers of the same constitution separately. Give the answer as an integer.

Answer: 4

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