Question

Choose the correct name for compound given below : 

• A. (4E)-5-Bromo-hex-4-en-2-yne
• B. (2E)-2-Bromo-hex-2-en-4-yne
• C. (2E)-2-Bromo-hex-4-yn-2-ene
• D. (4E)-5-Bromo-hex-2-en-4-yne

Hint:

IUPAC priority: "ene" before "yne" only when locants are identical. Remember to use "E/Z" for substituted alkenes.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
IUPAC nomenclature for unsaturated hydrocarbons (en-ynes) requires choosing the longest carbon chain containing both multiple bonds and numbering it to give the multiple bonds the lowest possible locants.
Step 2: Detailed Explanation:
1. Identify the Chain: The longest continuous carbon chain containing both the double and triple bond has 6 carbons, making the parent name "hex".
2. Numbering:
- If we number from left to right (starting from the triple bond end): Multiple bonds are at C2 (yne) and C4 (ene). Locants: (2, 4).
- If we number from right to left (starting from the double bond end): Multiple bonds are at C2 (ene) and C4 (yne). Locants: (2, 4).
- Tie-breaker Rule: If locants are the same from both ends, the double bond is given priority for the lower number. Therefore, we number from right to left.
3. Substituent and locants: The double bond is at C2, the triple bond is at C4. The bromine atom is attached to C2.
4. Stereochemistry: The double bond has high priority groups (Br and the rest of the chain) on opposite sides, which corresponds to the 'E' (Entgegen) configuration.
5. Combining names: 2-Bromo + hex + 2-en + 4-yne \(\to\) 2-Bromo-hex-2-en-4-yne.
Step 3: Final Answer:
The IUPAC name is (2E)-2-Bromo-hex-2-en-4-yne.