Chem 331, PS7 answers
Draw the structure of the products
The syntheses in problems 2-4 all give racemic products
1
O
OH
LiAlH4 is a strong reducing
agent, and reduces both
esters and ketones
LiAlH4
O
OH
Et
O
2
O
OH
NaBH4
O
NaBH4 only reduces the
ketone
O
Et
O
O
MCPBA
1 equiv.
3
1 equiv.
Et
O
A
epoxidation is selective for
the more substituted alkene
H
–
H Al H
H
LiAlH4
OH
B
reduction at the more sterically
accessible center
4) Propose a multistep synthesis
O
Retrosynthesis: this is your scratch work! We find that there is more than one possibility
This type of arrow
Williamson
ether synthesis
a
refers to a "backwards" transformation
Br
OH
O
a
O
b
BrMg
PBr3
b
Br
Williamson
ether synthesis
OH
Points about the above retrosynthesis
• Path a is better than b because it is shorter, and because it uses an allylic bromide rather than a secondary bromide
(we expect a better yield with the more reactive allylic bromide in an SN2 reaction).
• We worked our way back to a 5 carbon intermediate epoxide, because we knew that this could be constructed from
our starting alkene.
The forward synthesis: ( this is your formal answer!)
O
O
Cl
OH
O
(MCPBA)
BrMg
OH
ether
racemic
1) NaH
2)
note: NaH is being used
as a base to deprotonate the
alcohol
OH
NaH
Br
O
O–
Na+
5) Draw the structure of A and provide detailed arrow pushing mechanisms for its formation.
O
–O
H+
MgBr
O
(1 equiv)
HO
(2 equiv)
either answer was accepted as A
BrMg
O–
O
O
O
O
BrMg
leaving group
BrMg
–O
BrMg
BrMg
the first equivalent
H+
HO
H+
–O
H2O
The key is that a ketone is generated while the
Grignard reagent is still present. Thus, a second
Equivalent can add
the second
equivalent
6) Draw the structure of B and provide detailed arrow pushing mechanisms for its formation. Explain why A [your
answer from question 1] is not formed instead.
–O
O
O–
O
H+
Li
OH
(1 equiv)
(excess
>2 equiv)
either answer was accepted as B
O
Li
O
O
H
O–
O–
O
Li
O–
X
O2-
Li
Li
Li
the first equivalent
the second
equivalent
H3O+
:
O
H
+
O
+ OH
****
:
:
H2O
this does NOT occur
O2- cannot be a
leaving group
this species is
stable until we add
H+ in a separate
step
: OH
2
:
:
OH
H+
ketone
****
at this point, any excess aryllithium will be react with H3O+
H+
Li
H
Thus, because the ketone is only generated under acidic conditions, the organometallic reagent
gets protonated before it has the chance to add a second time!
OH
7) Propose a multistep synthesis of D using C as a starting material.
O
O
H
H
D
C
OH
(racemic)
Retrosynthesis:
X
O
Problem: we need to selectively add to
the epoxide and not to the aldehyde
O
H
H
D
Solution: use a acetal protecting group
O
OH
O
H
C
Forward Synthesis
O
H
HO
OH
H+
O
O
MCPBA
H
O
H
O
O
MgBr
O
O
H
OH
stable to Grignard
reagents
H3O+
O
H
D
OH
8) Deduce the structures of E–I
MCPBA
Me
Me
O
NMe2
NaOAc
E
CH3MgBr (1 equiv)
Et2O
1) NaOH
F
2) H+
G
C4H10O2
H+
H
G (the compound you
just made above)
(1 equiv)
1H
I
C7H14O2
NMR spectrum of I
3.38 (m, 2H)
1.28 (s, 6H)
1.15 (d, 6H, J=5.9 Hz)
ester hydrolysis
MCPBA
Me
Me
O
H
E
O
NMe2
(1 equiv)
CH3 MgBr (1 equiv)
Et2 O
HO
H
F
–OAc
CH3
H
1) NaOH
O
2) H+
racemic
O
H 3C
H
NaOAc
+ HO
OH
O
HO
OH
G
C4H 10O2
H+
O
G
H
H 3C
O
CH3