Assigning dichloroethene isomers from dipole moments and boiling points
Three unlabelled samples of dichloroethene, C2H2Cl2, are labelled A, B and C. The only structural isomers are 1,1-dichloroethene, cis-1,2-dichloroethene and trans-1,2-dichloroethene. Measurements give: A has dipole moment 0 D and boiling point 47.5 °C; B has dipole moment 1.3 D and boiling point 31.7 °C; C has dipole moment 1.9 D and boiling point 60.3 °C. All three are planar around the C=C bond. A student predicts that the most polar isomer must have the highest boiling point, but the data do not fully support a simple correlation: B is more polar than A yet boils lower than A. The samples are to be used to demonstrate how molecular shape, not just formula, controls physical properties. You have access to the dipole-moment data and boiling points only; no spectra are available.
1. Assign each of A, B and C to one of the three isomers. Justify each assignment using the dipole moments and the shapes of the molecules.
Read the worked answer
Treat each C–Cl bond dipole as a vector pointing from C to Cl. The C–H bond dipoles are much smaller and do not change the qualitative conclusion.
A has μ = 0 D. The only isomer in which the two C–Cl bond dipoles can cancel exactly is trans-1,2-dichloroethene. The two C–Cl bonds point in opposite directions across the planar C=C bond, so their vector sum is zero. Therefore A is trans-1,2-dichloroethene.
C has the largest dipole moment, 1.9 D. In cis-1,2-dichloroethene, the two C–Cl bond dipoles are on the same side of the C=C bond and their components add rather than cancel. This gives the largest resultant dipole. Therefore C is cis-1,2-dichloroethene.
B has an intermediate dipole moment, 1.3 D. In 1,1-dichloroethene, both chlorine atoms are on the same carbon. The two C–Cl bond dipoles are at about 120° to each other, so they add to give a resultant that is nonzero but smaller than in the cis isomer. Therefore B is 1,1-dichloroethene.
2. Explain why the boiling points are in the order C > A > B, even though B has a larger dipole moment than A.
Read the worked answer
Boiling involves separating molecules from the liquid, so it depends on the total strength of intermolecular attractions, not on the dipole moment alone.
C (cis-1,2-dichloroethene) has the largest net dipole moment and is the least symmetrical isomer. Its dipole–dipole attractions are strongest, so it has the highest boiling point, 60.3 °C.
A (trans-1,2-dichloroethene) has zero net dipole moment, so dipole–dipole attractions are absent. However, its highly symmetrical shape allows the molecules to pack efficiently, increasing the number and strength of dispersion-force contacts. This packing advantage is enough to give A a higher boiling point (47.5 °C) than B.
B (1,1-dichloroethene) has a net dipole moment, but the charge separation is concentrated at one end of the molecule and the shape is less effective for close packing. Its dispersion forces and packing are poorer than those of A, so despite the dipole–dipole attraction its boiling point is the lowest, 31.7 °C. The student’s simple rule fails because it ignores packing efficiency and dispersion forces.
3. Predict which isomer is most soluble in water and justify your answer. Include an explanation of why all three isomers are only sparingly soluble.
Read the worked answer
C (cis-1,2-dichloroethene) is expected to be the most soluble in water. It has the largest net dipole moment, so it can interact most strongly with polar water molecules through dipole–dipole forces. The dipole of water can align favourably with the larger molecular dipole of C, giving a more exothermic solvation energy than for A or B.
B (1,1-dichloroethene) is intermediate: it is polar, but its dipole is smaller than that of C. A (trans-1,2-dichloroethene) has zero net dipole and should be the least soluble of the three in water.
All three are only sparingly soluble because none can form hydrogen bonds with water. They are organic molecules with no O–H or N–H group, and the C–Cl bonds are not good hydrogen-bond acceptors in the same way as O–H or N–H groups. The energy gained from dipole–dipole interactions with water is not enough to overcome the energy needed to break the strong hydrogen-bond network of liquid water, so the solubility remains low.