Compare the properties of propan-2-ol (I) and the hexafluoro analog (II).
(a) Compound II has almost triple the molecular weight of I, but II has a lower boiling point. Explain.
(b) Explain why the dipole moment of Compound II is much lower than the dipole moment of I, despite the presence of the six electronegative fluorine atoms.
(c) Why is II a stronger acid than I?
Molecular weight of Compound II is higher than that of compound I but the boiling point of compound II is lower, the reason behind this can be explained as follows:-
It is known that in the absence of other intermolecular force (van der waals force, dipole-dipole interaction or H-bond), the higher the molecular mass the greater the boiling point.So the boiling of compound II should be high. Here both the compound I and II have OH group and forms H-bond, but the formation of H -bond in compound II is more difficult due to the presence of highly electronegative F substituents with greater electron clouds. So that the difference between the dipole moment among this two compound is also very high. Hence, even the molecular weight of compound II is large due to the increase in vanderwaal’s forces the dipole moment of compound II is very small. It is known that dipole moment is directly proportional to the boiling point. Thus due to smaller dipole moment (0.32D), the boiling point of compound II is lower
(b) The dipole moment of compound II is...