Geometrical isomerism, also known as cis-trans isomerism, occurs in compounds where there is restricted rotation around a bond, typically a double bond or a ring structure, and at least two different groups are attached to the carbon atoms involved in the double bond or in a ring system. Let's analyze each of the given options: Option A: 2-methylbut-2-ene The structure of 2-methylbut-2-ene is CH3−C(CH3)=CH−CH3. It is evident from the structure that both carbons forming the double bond have the same group (CH3) on one side. To have geometrical isomerism, each carbon of the double bond needs different substituents. Therefore, 2-methylbut-2-ene cannot exhibit geometrical isomerism. Option B: 2-methylpropene The molecule 2-methylpropene is given by the formula CH3−C(CH3)=CH2. Like Option A, this molecule has the same substituent (CH3) on one side of the carbon-carbon double bond, with the other side bearing an H atom. Since identical groups are attached to one carbon/carbon double bond, geometrical isomerism is not possible. Option C: Cyclohexene Cyclohexene is a cyclic compound containing a double bond within a six-membered ring, represented as C6H10. Despite the presence of a ring, which may limit rotation, cyclohexene does not show geometrical isomerism because it lacks the necessary different substituent groups on the carbons bonded by the double bond; it only has hydrogen atoms attached to them. Option D: 1,2-dibromopropene 1,2-dibromopropene has the structure CH2=CH−CHBr2. The double bond is between carbon 1 and carbon 2. Carbon 1 has a hydrogen and carbon 2 has two bromine atoms. Since different groups are attached to both carbons of the double bond, it satisfies the conditions for geometrical isomerism. Thus, it can exhibit cis (H and Br on the same side) and trans ( H and Br on opposite sides) forms. Therefore, among the provided options, Option D: 1,2-dibromopropene is the correct answer since it can show geometrical isomerism.