Question

Identify IUPAC name of the following compound.
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• A. 4-Methylpent-3-en-2-one
• B. 2-Methylpent-2-en-4-one
• C. 4-Ethyl-but-3-en-2-one
• D. 4-Methyl-but-2-en-3-one

Hint:

- Priority: C=O $>$ C=C - Always number to give lowest locant to functional group

Answer 1

The Correct Option is A

Concept:

• Longest chain containing functional group is selected
• Ketone ($>C=O$) gets priority in numbering
• Double bond and substituents are numbered accordingly

Step 1: Identify parent chain
The longest chain containing the carbonyl group has 5 carbons $\Rightarrow$ pentanone.

Step 2: Identify functional group
Ketone group ($C=O$) is present $\Rightarrow$ suffix = -one

Step 3: Number the chain
Numbering is done such that:
• Carbonyl carbon gets lowest number Thus: \[ \text{C=O at position 2} \]

Step 4: Identify double bond
Double bond is between C3 and C4: \[ \Rightarrow \text{pent-3-en-2-one} \]

Step 5: Identify substituent
Methyl group present at C4: \[ \Rightarrow \text{4-methyl} \]

Step 6: Write final name
\[ \text{4-Methylpent-3-en-2-one} \]