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Solutions for Structure and Synthesis of Alcohols
52 continued—See the note at the bottom of the previous page regarding (d) and (e).
OH
Ph
(d)
CH3
2) H2O2 , HO−
1) BH3 • THF
OH
syn addition of H and OH with anti-Markovnikov orientation
(e)
H2O
Cl2
Cl
chlorohydrin formation with anti addition of Cl and OH
CH3
OH
MCPBA(f) O
PhMgBr
ether
New C—C bond requires Grignard; 
OH is not on same carbon where 
new C—C is formed suggesting the 
epoxide intermediate.
H3O+
53 OH
CH3H3C
mol. wt. 60 g/mole
boiling point 82 °C
dipole moment 1.66 D
pKa 16.5
OH
CF3F3C
mol. wt. 168 g/mole
boiling point 58 °C
dipole moment 0.32 D
pKa 9.3
(a) Boiling point is a rough indicator of the strength of intermolecular forces, of which three common ones 
apply to organic compounds: van der Waal's forces (or London forces), the weakest; dipole-dipole 
interactions for molecules with a permanent dipole moment; and hydrogen-bonding in compounds with OH 
or NH bonds, the strongest. Both I and II have the alcohol functional group and are likely to form hydrogen 
bonds with their neighbors, although an argument could be made that the slightly larger F substituents with 
the greater electron clouds would make hydrogen bonding more difficult in II. Rather than stretch a 
possibility, let's focus on something obvious: there is a huge difference in the dipole moments of these two 
compounds. Even though the molecular weight of II is much greater, thereby increasing van der Waal's 
forces, the dipole moment of II is very small. We must conclude that the significantly decreased dipole-
dipole interaction of II is more important in boiling point than the increased van der Waal's interaction, a 
weaker force.
I II
(b) I has a large dipole moment because of the bond polarizations due to the electronegative oxygen. In II, 
however, the bond polarization in the alcohol is counteracted by six F atoms; recall that F is the most 
electronegative element in the periodic table. The oxygen pulling in one direction is balanced against the six 
F atoms pulling partially away from the oxygen. (If you have studied vectors in physics, consider each 
polarized C—F bond as a vector with a "down" component and a "left" or "right" component; you will see 
that some portion of the left CF3 group cancels the same portion of the right CF3, leaving only a portion of the 
C—F polarization to cancel the C—O polarization. Make a model!)
(c) The strength of an acid is determined by the stability of its conjugate base. The anion of I has no 
particular stabilization. The anion of II, however, has six F atoms pulling electron density by the inductive 
effect; that is, the negative charge on the oxygen is partially shared by the six electronegative F atoms pulling 
electron density through sigma bonds. The anion of II is much more stable than the anion of I, making II the 
stronger acid.
plus the enantiomer
plus the enantiomer
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