The correct IUPAC name for the following structure is

The correct answer is Option (4) → 4-Bromo-3-methyloctanal

Note: None of the given options was correct in the NTA question paper. To provide additional question practice to the students, the question has been slightly modified so that the question has one clear answer.

To name a compound by the IUPAC system, first identify the principal functional group, since it gives the suffix and decides the numbering. Here the chain ends in a –CHO group, so the compound is an aldehyde and the name ends in -al. The carbon of the –CHO group is always C1 of the chain, so no locant is written for it. Then the longest chain that contains the –CHO carbon is counted, and the substituents are named in alphabetical order.

Numbering from the –CHO carbon of the drawn structure:

• C1 is the –CHO carbon and C2 the next \( \ce{CH2} \).

• C3 carries the \( \ce{-CH3} \) group: 3-methyl.

• C4 carries the Br atom: 4-bromo.

• After C4 the chain continues through three \( \ce{CH2} \) groups and ends in a \( \ce{CH3} \) group: C5, C6, C7 and C8.

    \( \displaystyle \ce{CH3-CH2-CH2-CH2-CH(Br)-CH(CH3)-CH2-CHO} \)

The main chain therefore has eight carbon atoms, and the name is 4-bromo-3-methyloctanal.

Option (4) 4-Bromo-3-methyloctanal — Correct. The –CHO group gives the suffix -al and C1; the longest chain through it has eight carbon atoms (octanal); the substituents are methyl at C3 and bromo at C4, written in alphabetical order (bromo before methyl).