The IUPAC name of the compound is

The correct answer is Option (1) → 2-Bromo-3-methylpent-2-en-1-ol

IUPAC rules for an unsaturated alcohol: the parent chain is the longest chain that contains the carbon bearing –OH and the C=C bond; it is numbered so that the –OH carbon gets the lowest number; the suffix -ol is added to the -ene name, and substituents are written in alphabetical order as prefixes.

The compound is \( \ce{CH3-C(C2H5)=C(Br)-CH2OH} \), six carbons in all. The longest chain runs from \( \ce{CH2OH} \) through both doubly bonded carbons and into the ethyl group:

    \( \mathrm{\overset{1}{C}H_2OH-\overset{2}{C}(Br)=\overset{3}{C}(CH_3)-\overset{4}{C}H_2-\overset{5}{C}H_3} \)

• Parent chain: 5 carbons with C=C at C-2 and –OH at C-1: pent-2-en-1-ol.

• Substituents: Br on C-2 and the left-over \( \ce{CH3} \) on C-3: 2-bromo-3-methyl (b before m).

So the name is 2-Bromo-3-methylpent-2-en-1-ol.