Write the IUPAC name of the compound shown: the ketone is the principal group and gets the lowest locant, then the double bond gets the lowest locant, then the prefixes. Add the locants of the double bond, the ketone and the halogen substituent. Give the sum as an integer.

Numerical answer type, tolerance 0.5.

Answer: 8

The compound is 4-bromopent-1-en-3-one. The ketone is at C3 in both directions; the tie is broken by the double bond, which gets locant 1 (not 4). Bromine is at C4. Sum of locants = 1 + 3 + 4 = 8.

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Write the IUPAC name of the compound shown: the ketone is the principal group and gets the lowest locant, then the double bond gets the lowest locant, then the prefixes. Add the locants of the double bond, the ketone and the halogen substituent. Give the sum as an integer.

Answer: 8

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