NEETChemistryGeneral Organic Chemistry
Match List-I with List-II: List-I (Molecule) List-II (IUPAC Name) (A) CH ₂= CH - C CH (I) Pent-1-en-4-yne (B) CH ₂= CH - CH ₂- C CH (II) 2-Methylbuta-1,3-diene (C) CH ₃- CH = CH - C CH (III) But-1-en-3-yne (D) CH ₂= C ( CH ₃)- CH = CH ₂ (IV) Pent-3-en-1-yne Choose the correct answer from the options given below:
Options
- A(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
- D(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
Correct answer
A. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Step-by-step solution
According to IUPAC nomenclature rules for compounds containing both double and triple bonds: 1. The parent chain is numbered to give the lowest locant set to the multiple bonds. 2. If there is a choice (tie), the double bond is given the lower number. For (A) CH ₂= CH - C CH : Numbering from left gives locants 1,3. Numbering from right gives 1,3. Due to the tie, the double bond gets priority. Name: But-1-en-3-yne. For (B) CH ₂= CH - CH ₂- C CH : Numbering from left gives 1,4. Numbering from right gives 1,4. Tie-bre