Indicate whether each of the following molecules is capable of geometrical isomerism. For those that are, draw the structures: (a) 2 -methylpropene, (b) 1,3-diiodobenzene, (c) 1,2-dichloropent-1-ene, (d) 3-methyl-1-butyne.

2-methylpropene has the formula \(CH_3C(CH_3)=CH_2\). It has a double bond between the second and third carbon atoms. For geometrical isomerism, we need to consider if there are different groups attached to each end of the double bond. In this case, one side of the double bond has a methyl group and hydrogen atom attached, and the other side has a hydrogen atom and another carbon with two hydrogen atoms and a methyl group attached. Since both sides of the double bond have different groups attached, 2-methylpropene is capable of geometrical isomerism. The two possible geometrical isomers are: 1. E-2-methylpropene: The methyl groups are on opposite sides of the double bond. 2. Z-2-methylpropene: The methyl groups are on the same side of the double bond.

1,3-diiodobenzene has a benzene ring with two iodine atoms replacing hydrogen atoms in positions 1 and 3. In this case, the presence of aromatic ring \(C_6H_6\) does not allow for any restricted rotation of bonds. Thus, 1,3-diiodobenzene is not capable of geometrical isomerism.

1,2-dichloropent-1-ene has the formula \(CH_2=CH(Cl)CH_2CH_2Cl\). It has a double bond between the first and second carbon atoms. For geometrical isomerism, we need to pay attention to the attachments to the double bond. In this case, one side of the double bond has a hydrogen atom and a chlorinated carbon atom, while the other side has a hydrogen atom and a carbon atom with chlorine. Since both sides of the double bond have different groups attached, 1,2-dichloropent-1-ene is capable of geometrical isomerism. The two possible geometrical isomers are: 1. E-1,2-dichloropent-1-ene: The chlorine atoms are on the opposite sides of the double bond. 2. Z-1,2-dichloropent-1-ene: The chlorine atoms are on the same side of the double bond.

3-methyl-1-butyne has the formula \(CH_3C \equiv CCH_2CH_3\). It has a triple bond between the second and third carbon atoms. No geometrical isomerism is expected due to the presence of a triple bond and absence of a repeated functional group on one side of the triple bond. Thus, 3-methyl-1-butyne is not capable of geometrical isomerism.