Question

The IUPAC name of the compound

• A. 3-keto-2-methylhex-4-enal
• B. 5-formylhex-2-en-3-one
• C. 5-methyl-4-oxohex-2-en-5-al
• D. 3-keto-2-methylhex-5-enal

Answer 1

The Correct Option is A

To determine the IUPAC name of the compound, we start by analyzing its structure:

1. Identify the longest carbon chain that contains the highest priority functional group. Here, the longest chain has six carbon atoms, thus the root name is "hex".
2. Identify the functional groups present. The structure contains:
   • A ketone group (C=O) at the third carbon.
   • An aldehyde group (CHO) at the end of the chain.
   • A double bond between the fourth and fifth carbon.
   • A methyl group attached to the second carbon.
3. Number the carbon chain to give the lowest possible numbers to the functional groups, while considering the aldehyde (-CHO) as the principal functional group since it is of a higher priority over ketones and double bonds. So, the aldehyde carbon is numbered as C1.
4. Name the substituents with their respective positions:
   • Aldehyde: Presence at the terminal, named as "al".
   • Ketone: On the third carbon, termed as "3-keto".
   • Methyl group: On the second carbon, named as "2-methyl".
   • Double bond: Between carbons four and five, named as "4-en".

Bringing all parts together, the IUPAC name becomes 3-keto-2-methylhex-4-enal.

Let's rule out the other options:

• 5-formylhex-2-en-3-one - Incorrect because it misplaces the ketone and aldehyde groups.
• 5-methyl-4-oxohex-2-en-5-al - Incorrect naming order and misplacing of functional groups.
• 3-keto-2-methylhex-5-enal - Incorrect position of the double bond.

Thus, the correct answer is 3-keto-2-methylhex-4-enal.