Q.The IUPAC name of the following compound is
(a pent-2-ene drawn with the LEFT double-bond carbon carrying Cl pointing up and H3C pointing down, and the RIGHT double-bond carbon carrying CH2-CH3 pointing up and I pointing down)

Step 1. Read the drawn structure: a pent-2-ene backbone, CH3-C(Cl)=C(I)-CH2-CH3, with the LEFT double-bond carbon carrying Cl (up) and CH3 (down), and the RIGHT double-bond carbon carrying an ethyl group -CH2CH3 (up) and I (down).

Step 2. Number the five-carbon chain from the methyl end (either end gives the same locant set {2,3} for the substituents, since the double bond sits centrally): C1 = CH3, C2 = C(Cl)=, C3 = C(I)=, C4-C5 = the ethyl group's own two carbons.

Step 3. Name the substituents alphabetically: chloro (at C2) before iodo (at C3), giving the base name 2-chloro-3-iodopent-2-ene.

Step 4. Assign CIP priority at each double-bond carbon: at C2, Cl outranks CH3 (C1); at C3, I outranks the ethyl group. In the drawing, Cl sits UP and I sits DOWN -- the two higher-priority groups are on OPPOSITE sides, which is the E (and, since each carbon has only one non-hydrogen priority pair, colloquially trans) configuration; the two chain-continuing groups (the C1 methyl, down, and the ethyl group, up) are likewise on opposite sides, confirming an extended, trans-style zig-zag. …

Number the chain to give the double bond (and hence the substituents) the lowest locants, name the substituents alphabetically, then separately assign …

• Numbering the chain from the ethyl end instead of the methyl end, which would (incorrectly, since it does not give the lowest locant to the double bond) produce locants like 3,4 instead of 2,3. …