Chem 331, PS #8
due Mon, Nov 11 , 2002
use extra paper (use a staple or paper clip)
Key
Name
1. Match each compound with its IR spectrum.
OH
For A, steric hinderance prevents
H-bonding. Thus, the OH stretch
is sharp.
OH
A
B
B
3300 cm–1
A
3620 cm–1
SHARP!
The syntheses in problems 2-4 all give racemic products
2)
O
OH
LiAlH4 is a strong reducing
agent, and reduces both
esters and ketones
LiAlH4
O
OH
Et
O
3)
HO
O
NaBH 4
O
NaBH 4 only reduces the
ketone
O
Et
MCPBA
1 equiv.
4)
1 equiv.
Et
O
O
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
5) 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
BrMg
OH
ether
racemic
1) NaH
(MCPBA)
2)
note: NaH is being used
as a base to deprotonate the
alcohol
OH
NaH
Br
O
O–
Na+