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The IUPAC name of the following compound is
A
trans-3-iodo-4-chloro-3-pentene
B
cis-3-chloro-3-iodo-2-pentene
C
trans-2-chloro-3-iodo-2-pentene
D
cis-3-iodo-4-chloro-3-pentene
Detailed Solution
The longest chain containing the double bond is $CH_3-C(Cl)=C(I)-CH_2-CH_3$, which has five carbons: pentene.
Numbering from the methyl end gives the double bond position 2 (pent-2-ene); numbering from the other end would give 3, so the first numbering is chosen.
With this numbering, chlorine is on C-2 and iodine is on C-3.
Substituents are written in alphabetical order: chloro before iodo, giving 2-chloro-3-iodo-2-pentene.
Geometry: on C-2 the higher-priority group is Cl (over $CH_3$) and on C-3 it is I (over $CH_2CH_3$). In the structure Cl and I lie on opposite sides of the double bond, so the isomer is trans.
IUPAC name: trans-2-chloro-3-iodo-2-pentene
Numbering from the methyl end gives the double bond position 2 (pent-2-ene); numbering from the other end would give 3, so the first numbering is chosen.
With this numbering, chlorine is on C-2 and iodine is on C-3.
Substituents are written in alphabetical order: chloro before iodo, giving 2-chloro-3-iodo-2-pentene.
Geometry: on C-2 the higher-priority group is Cl (over $CH_3$) and on C-3 it is I (over $CH_2CH_3$). In the structure Cl and I lie on opposite sides of the double bond, so the isomer is trans.
IUPAC name: trans-2-chloro-3-iodo-2-pentene