JEE Main 2021 · 31 August · morning · Single correct · +4 / −1
Choose the correct name for compound given below :
- (a) (4E)-5-Bromo-hex-4-en-2-yne
- (b) (2E)-2-Bromo-hex-4-yn-2-ene
- (c) (2E)-2-Bromo-hex-2-en-4-yne
- (d) (4E)-5-Bromo-hex-2-en-4-yne
Answer and solution
Answer (c) (2E)-2-Bromo-hex-2-en-4-yne
Solution
Chemistry ma’am · 5 steps
Both directions give the multiple bonds locants 2 and 4, so break the tie in favour of the double bond.
- Find the chain. The structure is CH₃-C≡C-CH=C(Br)-CH₃. The longest chain holding both multiple bonds has 6 carbons, so the parent is hex-en-yne with a bromo substituent.
- Number the chain. From the left end the triple bond starts at C2 and the double bond at C4. From the right end the double bond starts at C2 and the triple bond at C4. Both give the set {2, 4}. When there is a tie, the double bond gets the lower number, so number from the right.
- Write the name. Numbered from the right, Br sits on C2, the double bond is at C2 and the triple bond is at C4. The ending is always written "en-yne", which gives 2-bromohex-2-en-4-yne.
- Assign E or Z. On C2, Br outranks CH₃. On C3, the alkyne carbon C4 outranks H. In the drawing Br points up and to the right while the alkyne chain points to the left, so they are on opposite sides of the double bond. That makes it E.
- So. The name is (2E)-2-Bromo-hex-2-en-4-yne, which is option (c).
The idea to keep. In an en-yne, give the lowest locants to the multiple bonds as a set. If both directions tie, the double bond gets the lower number. Always write "en" before "yne" in the name.
Answer from the NTA final answer key. Solution by PrepWiser.