JEE Main · 2024 · 30 Jan 2024 · Shift 1 · General Organic Chemistry

To draw the structure of 4-Methylpent-2-enal, we need to follow IUPAC nomenclature:

1. "Pent" → 5 carbon atoms in the main chain
2. "2-en" → Double bond starting at carbon-2
3. "al" → Aldehyde group at carbon-1 (always at the end)
4. "4-Methyl" → Methyl substituent at carbon-4

Numbering from the aldehyde end (C-1):

• C-1: $-\mathrm{CHO}$ (aldehyde)
• C-2: $=\mathrm{CH}-$ (double bond)
• C-3: $-\mathrm{CH}-$
• C-4: $-\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)-$ (methyl branch)
• C-5: $-\mathrm{C}{\mathrm{H}}_3$

Wait, let me reconsider. With 5 carbons and a methyl at position 4:

• C-1: CHO
• C-2: =CH
• C-3: -CH-
• C-4: -CH(CH₃)- means the main chain has only 4 carbons

Actually, the correct structure is:
$(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{CH}-\mathrm{CH}=\mathrm{CH}-\mathrm{CHO}$

This matches option (4), which shows an isopropyl group attached to the C=C-CHO system.

Therefore, the correct answer is (4).