Answer:
7-hydroxy-pent-3-en-2-one.
Explanation:
Hello,
In this case, for the given compound, we must notice some special sections:
- A carbonyl group at the right side.
- A hydroxyl at the left side.
- A double bond at the middle.
In such a way, since carbonyl group has the priority, we should consider the molecule as a seven-carbon main chain that has a double bond at the fourth carbon and a hydroxyl at the seventh carbon. Thereby, the beginning of the chain is at the left-sided carbon for which the IUPAC name must be:
7-hydroxy-pent-3-en-2-one.
Best regards.
