1 (174)

Prévia do material em texto

Solutions for Structure and Synthesis of Alkenes
46 continued
H
H
H2O
C
H
H
H
H
OSO3H
H2O
OH
H HH
OH
H
H
C
H
47
– H2O
∆
+ HSO4
−
E1 works well because only one carbocation and 
only one alkene are possible. Substitution is not a 
problem here. The only nucleophiles are water, 
which would simply form starting material by a 
reverse of the dehydration, and bisulfate anion. 
Bisulfate anion is an extremely weak base and poor 
nucleophile; if it did attack the carbocation, the 
unstable product would quickly re-ionize, with no 
net change, back to the carbocation.
(e)
The conformation must be considered because the leaving 
group must be in an axial position. The tert-butyl group is so 
large that it must be equatorial, locking the conformation into 
the chair shown. As a result, only the Cl at position 4 is axial, 
so that is the one that must leave, not the one at position 3. 
Two isomers are possible but it is difficult to say which would 
be formed in greater amount.
Cl
Cl
(CH3)3C
(CH3)3C
Cl
Cl
1
3 4
ax
eq
KOH
∆
Cl
(CH3)3C
Cl
(CH3)3C
H
H(f) Br
Br
KOH
∆
H
H
Br
H
H
Br
Only one of the two Br atoms is in 
the required axial position for E2 
elimination. There are axial H 
atoms at the two adjacent carbons, 
so two alkene products can be 
formed. Predicting major and minor 
would be difficult.
CH3 C
CH3
HO
C
OH
CH3
CH3 C CH3
CH3
OH2
C
HO
CH3
CH3
C CH3
CH3
CH3
C
O
CH3CH3 C
O
C
CH3
CH3
CH3
H2O
CH3 C
OH
C
CH3
CH3
CH3
CH3 C
CH3
HO
C
CH3
CH3
The driving force for this rearrangement is the great stability of the resonance-stabilized, protonated 
carbonyl group.
methyl shift
48
H
− H2O
H OSO3H
H OSO3H
169