Q7 · CSIR-NET Chemistry, December 2013
Paper: CSIR-NET December 2013 · Subject: Organic Chemistry · Chapter: General Organic Chemistry · Topic: Nomenclature Polycyclics · Marks: 2 · Difficulty: Medium
The IUPAC name of the compound given below is 
(a)(2E, 4E)-3-chlorohexa-2, 4-diene-1, 6-diol
(b)(2Z, 4E)-3-chlorohexa-2, 4-diene-1, 6-diol
(c)(2Z, 4Z)-4-chlorohexa-2, 4-diene-1, 6-diol
(d)(2E, 4Z)-4-chlorohexa-2, 4-diene-1, 6-diol
Answer
Answer (as printed): B
Explanation
Number from the end that gives Cl the lowest locant: HOCH₂(1)–CH(2)=C(Cl)(3)–CH(4)=CH(5)–CH₂OH(6), so the name is 3-chlorohexa-2,4-diene-1,6-diol (this rules out options c and d).
C2=C3: on C-2, CH₂OH > H; on C-3, Cl > C-4. CH₂OH and Cl are drawn on the same (lower) side, so this bond is 2Z.
C4=C5: on C-4, C-3 > H; on C-5, CH₂OH > H. C-3 (lower) and CH₂OH (upper) are on opposite sides, so this bond is 4E.
Answer: (2Z,4E)-3-chlorohexa-2,4-diene-1,6-diol (b).
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