Question

The IUPAC name of the following compound is \[ CH_3-C\equiv C-CH=CH-CH=CH-CH_3 \]

• A. Octa-2-yn-4,6-diene
• B. Octa-2,4-dien-6-yne
• C. Octa-4,6-dien-2-yne
• D. Octa-6-yn-2,4-diene

Hint:

For compounds containing both \(C=C\) and \(C\equiv C\) bonds: Choose the numbering with the lowest set of locants If a tie occurs, \[ C=C \;\gt \; C\equiv C \] i.e., the double bond gets priority over the triple bond while assigning locants.

Answer 1

The Correct Option is B

Concept: For compounds containing both double and triple bonds:

• Choose the longest chain containing all multiple bonds.

• Number the chain to give the lowest set of locants.

• If a tie occurs, double bonds receive lower numbers than triple bonds.

• Write ``en'' before ``yne'' in the name.

Step 1: Determine the parent chain. The compound contains eight carbon atoms. \[\begin{aligned} \text{Parent chain} = \text{Octa} \end{aligned}\]

Step 2: Locate the multiple bonds. Numbering from the left gives \[\begin{aligned} \text{Triple bond at }2 \end{aligned}\] and \[\begin{aligned} \text{Double bonds at }4,6 \end{aligned}\] Numbering from the right gives \[\begin{aligned} \text{Double bonds at }2,4 \end{aligned}\] and \[\begin{aligned} \text{Triple bond at }6 \end{aligned}\]

Step 3: Apply IUPAC priority rule. Both numberings give the same set of locants \((2,4,6)\). In such a tie, double bonds receive lower numbers. Hence the preferred numbering is \[\begin{aligned} \text{Double bonds: }2,4 \end{aligned}\] \[\begin{aligned} \text{Triple bond: }6 \end{aligned}\]

Step 4: Write the IUPAC name. \[\begin{aligned} \boxed{\text{Octa-2,4-dien-6-yne}} \end{aligned}\] This corresponds to option (B).