O
1) NaCN
2) H2O
H
?
O
O
H
H
This chirality center
will be inverted
2˚
(Less hindered)
3˚
@
O
@
O
C
H
H
C
H
HO
O
N
C
Proton transfer
N
N
Nucleophilic attack
Q
For each of the following reactions, predict the product and draw a mechanism for
its formation:
O
(a)
Me
O
(c)
Me
1) PhMgBr
2) H2O
1) NaSH
2) H2O
O
(e)
Et
Me
?
?
1) NaSH
2) H2O
O
(b)
1) NaCN
2) H2O
Me
O
(d)
?
1) LAH
2) H2O
Me
O
(f)
Et
H
H
?
?
1) LAH
Me
2) H2O
?
Predict the product of the reaction below and draw the likely mechanism for its formation:
O
Et
Me
?
[H2SO4]
Et
EtOH
H
Nu attack
where stability of C+ is more
Back side attack
Proton transfer
H
H
O
Et
H
Me
14.20
!
SN 2
!
O
Et
Et
Et
Me
Et
H
SN1
having
character
Et
O
H
OH
Me
Et
Et
O
Et
!
H
H
For each reaction, predict the product and draw a mechanism for its formation:
O
HCl
(a)
?
O
Me
Et
(c)
O
(e)
H
O
Me
O
(b)
[H2SO4]
EtOH
[H2SO4]
MeOH
?
?
Me
O
Me
(d)
?
HBr
H2 O
Me
Et
?
[H2SO4]
O
HBr
(f)
(a) Under acidic conditions, the epoxide is protonated,
thereby generating a very powerful electrophile (a
protonated epoxide). Since the starting epoxide is
symmetrical, regiochemistry is not a concern in this case.
That is, the nucleophile can attack the epoxide at either
position, giving the same product either way.
Stereochemistry is also not an issue in this case, because
the product does not contain any chiral centers.
Et
Me
?
competition is between a secondary position and a
tertiary position, electronic factors dominate and the
tertiary position is attacked. Back-side attack causes
inversion of configuration at the chiral center being
attacked. Finally, a proton is removed (the most likely
base is the solvent, ethanol).
protonated epoxide). Methanol (CH3OH) is a weak
nucleophile, and we must decide which position will be
attacked. In this case, one position (left) is tertiary, and
the other position (right) is secondary. When the
competition is between a secondary position and a
tertiary position, electronic factors dominate and the
tertiary position is attacked. Back-side attack causes
inversion of configuration at the chiral center being
attacked. Finally, a proton is removed (the most likely
base is the solvent, methanol).
O
OH
[H2SO4]
H
Me
MeOH
H
O
H
H
Me
MeO
H
MeOH
Me
O
(b) Under acidic conditions, the epoxide is protonated,
thereby generating a very powerful electrophile (a
protonated epoxide).
To determine where the
nucleophile (bromide) attacks, we must decide whether
steric or electronic effects dominate. In this case, one
position (left) is primary and the other position (right) is
secondary. Under these conditions, steric effects will
dominate, and the attack is expected to occur at the less
substituted position. The position being attacked is not a
chiral center. There is an existing chiral center, but that
center is not attacked, so we do not expect the
configuration of that chiral center to change.
(d) Under acidic conditions, the epoxide is protonated,
thereby generating a very powerful electrophile (a
protonated epoxide).
Water (H2O) is a weak
nucleophile, and we must decide which position will be
attacked. In this case, one position (left) is tertiary, and
the other position (right) is secondary. When the
competition is between a secondary position and a
tertiary position, electronic factors dominate and the
tertiary position is attacked. Back-side attack causes
inversion of configuration at the chiral center being
attacked. Finally, a proton is removed (the most likely
base is the solvent, water).
OH
H
Me
MeOH
Me
H
(f) Under acidic conditions, the epoxide is protonated,
thereby generating a very powerful electrophile (a
protonated epoxide). Bromide is a nucleophile, and we
must decide which position will be attacked. In this
case, one position (left) is tertiary, and the other position
(right) is secondary. When the competition is between a
secondary position and a tertiary position, electronic
factors dominate and the tertiary position is attacked.
Back-side attack causes inversion of configuration at the
chiral center being attacked.
O
Et
H
Me
Br
Et
HBr
Br
H
H
Br
O
Et
(e) Under acidic conditions, the epoxide is protonated,
thereby generating a very powerful electrophile (a
(c) Under acidic conditions, the epoxide is protonated,
thereby generating a very powerful electrophile (a
protonated epoxide).
Ethanol (EtOH) is a weak
nucleophile, and we must decide which position will be
attacked. In this case, one position (left) is tertiary, and
the other position (right) is secondary. When the
H
Me
O
H
Me
OH
H
Me