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Solutions for Reactions of Alkenes
I II
The other two double bonds have to be in the ring, but where? The products do not have branched chains, 
so double bonds must appear at both C-1 and C-4. There are only two possibilities for this requirement.
or
Ozonolysis of I would give fragments containing one carbon, two carbons, and seven carbons. Ozonolysis 
of II would give fragments containing one carbon, four carbons, and five carbons. Aha! Our mystery 
structure must be II.
(Editorial comment: Science is more than a collection of facts. The application of observation and logic to 
solve problems by deduction and inference is a critical scientific skill, one that distinguishes humans from 
algae.)
63 continued
OH
C
OHO
HO
C
H
O
HO
64 In this type of problem, begin by determining which bonds are broken and which are formed. These 
will always give clues as to what is happening.
Protonated epoxide opens to give 
the most stable carbocation (3°).
H+ goes to most
electronegative 
atom.
formed
broken
3° carbocation looks 
for electrons, finds 
them at nearby alkene, 
forming a 6-membered 
ring (yes!)—leaves a 
3° carbocation.
H O
H
H
O
H
H
H
COOH
HOOC
H
OH
HO
H
COOH
OHHO
HOOC
H
H2O
CH3CO3H
OsO4
H2O2
COOH
HOOC
HOOC
COOH
trans + anti → meso
trans + syn → racemic
65 See the solution to Problem 35 for simplified examples of these reactions.
(b)
(a)
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