Which one of the following compounds can exist as cis-trans isomers?

The correct option is 1.2-Dimethylcyclohexane.

To determine which compound can exist as cis-trans (geometric) isomers, let us analyze each of the options by looking at their structures:

1. Pent-1-ene: The structure of pent-1-ene is CH₂=CH-CH₂-CH₂-CH₃. The first carbon atom (C-1) of the double bond is bonded to two identical hydrogen atoms. For a compound to exhibit cis-trans isomerism about a double bond, each carbon of the double bond must be bonded to two different groups. Since C-1 is bonded to two identical groups (-H and -H), pent-1-ene cannot exhibit cis-trans isomerism.

2. 2-Methylhex-2-ene: The structure is (CH₃)₂C=CH-CH₂-CH₂-CH₃. Here, the C-2 carbon of the double bond is bonded to two identical methyl groups (-CH₃). Because C-2 is bonded to two identical groups, this compound cannot exhibit cis-trans isomerism.

3. 1,1-Dimethylcyclopropane: In cyclic compounds, cis-trans isomerism arises when there are substituents on different carbon atoms of the ring. In 1,1-dimethylcyclopropane, both methyl groups are attached to the same carbon atom (C-1). Therefore, there is no spatial variation possible for the groups relative to each other, and it cannot exist as cis-trans isomers.

4. 1,2-Dimethylcyclohexane (written as 1.2-dimethylcyclohexane): In this compound, the two methyl groups are attached to adjacent, different carbon atoms (C-1 and C-2) of the cyclohexane ring. This allows for two different spatial arrangements:
- cis-isomer: Both methyl groups point to the same face of the ring (either both pointing up or both pointing down).
- trans-isomer: The two methyl groups point to opposite faces of the ring (one pointing up and the other pointing down).

Therefore, 1,2-dimethylcyclohexane can exist as cis-trans isomers.