Question: 58. Which one of the following compounds can exist as cis-trans isomers?
(1) Pent-1-ene
(2) 2-Methylhex-2-ene
(3) 1,1-Dimethylcyclopropane
(4) 1.2-Dimethylcyclohexane
Answer: Option (4)
Explanation:
Cis-trans (geometrical) isomerism is possible when:
(a) There is restricted rotation (due to a double bond or a ring structure), and
(b) Each of the two atoms (or ring positions) involved is attached to two different groups.
Let us analyze each option:
(1) Pent-1-ene: Structure is CH_3–CH_2–CH_2–CH_2–CH_3 with a double bond at C-1, i.e., CH_2=CH–CH_2–CH_2–CH_3.
At the terminal carbon of the double bond (CH_2), both substituents are H atoms. Since one of the double-bonded carbons does not have two different groups, cis-trans isomerism is not possible.
(2) 2-Methylhex-2-ene: The structure around the double bond at C-2 is such that the carbon C-2 is attached to two identical groups (two CH_3 groups: one from the main chain and one as a methyl substituent). When a double-bonded carbon has two identical groups, cis-trans isomerism is not possible.
(3) 1,1-Dimethylcyclopropane: Both methyl groups are attached to the same carbon (position 1) of the ring. Since cis-trans isomerism in cycloalkanes depends on substituents being on different ring carbons (and their relative positions “above” or “below” the ring), 1,1-disubstitution does not show cis-trans isomerism.
(4) 1,2-Dimethylcyclohexane: This is a disubstituted cycloalkane with the two methyl groups on adjacent carbons (C-1 and C-2). In a ring, the substituents can be arranged so that both are on the same side of the ring (cis) or on opposite sides (trans). Thus, 1,2-dimethylcyclohexane can exist as cis and trans isomers.
Therefore, the compound that can exist as cis-trans isomers is 1,2-dimethylcyclohexane.