See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the compound. The structure shows an aldehyde group (OHC/CHO) at one end, two carbon-carbon double bonds, and two methyl substituents on the carbon bearing both methyls near the other end. Writing out the carbon skeleton from the aldehyde end: C1(=O)H – C2=C3 – C4H2 – C5H2 – C6=C7(CH3) – but the image shows two methyl groups on the last carbon making it C(CH3)2, giving an 8-carbon chain when the methyls are treated as part of the principal chain or as substituents. Step 2 – Select the principal chain. The longest carbon chain containing the CHO group and both double bonds must be identified. The aldehyde carbon is C1. Counting: C1(CHO)-C2=C3-C4-C5-C6=C7-C8 with two methyl groups. Wait — the terminal carbon has two CH3 groups. If the two methyls are substituents on C7 and C8 is not present, then the chain is 7 carbons (hept) with gem-dimethyl. But if one methyl is incorporated into the chain, the longest chain is 8 carbons (oct) with one methyl substituent at C7. Step 3 – Apply IUPAC rules. The principal chain must include the CHO group (suffix -al) and be the longest possible chain containing the maximum number of double bonds. Taking the chain as 8 carbons: C1(CHO)=O, C2=C3 (double bond at 2), C4, C5, C6=C7 (double bond at 6), C8, with a methyl branch at C7. This gives: octa-2,6-dien-1-al with a methyl at C3 and a methyl at C7 → 3,7-dimethylocta-2,6-dienal. Step 4 – Number correctly. Numbering from the CHO end (C1) gives double bonds at positions 2 and 6, and methyl groups at positions 3 and 7, yielding the name 3,7-dimethylocta-2,6-dienal. Step 5 – Eliminate wrong options. Option (b) uses incorrect suffix placement (-8) and wrong locants. Option (c) uses 'formyl' as a prefix, which is incorrect when –CHO is the principal characteristic group (it should be the suffix -al). Option (d) uses the non-IUPAC term 'aldo' as a prefix, which is not acceptable in IUPAC nomenclature. Therefore, the correct answer is A.