Chem 335 S01 MT#2: Page 1
UNIVERSITY OF VICTORIA
CHEMISTRY 335, S01
SYNTHETIC METHODS IN ORGANIC CHEMISTRY
MIDTERM EXAM #2 — MARCH 12, 2010
NAME:
STUDENT ID:______________
INSTRUCTOR: FRASER HOF
TOTAL MARKS = 35
DURATION: 50 MINUTES
THIS EXAMINATION PAPER HAS 6 PAGES, INCLUDING THIS COVER PAGE. COUNT THE
NUMBER OF PAGES IN THIS EXAMINATION PAPER BEFORE YOU START TO WRITE, AND
REPORT ANY DISCREPANCY IMMEDIATELY TO THE INVIGILATOR.
QUESTIONS ARE TO BE ANSWERED IN THE SPACE PROVIDED ON THE EXAM FORM.
QUESTIONS ANSWERED IN PENCIL WILL NOT BE ELIGIBLE FOR RE-GRADING.
Chem 335 S01 MT#2: Page 2
1. (13 marks) Fill in the boxes with the missing reagents or products (1 mark each). Show only the
MAJOR stereoisomer produced by the given reaction conditions. For all products that you fill in,
indicate whether the compound is achiral, racemic, or a single enantiomer (0.5 marks each).
a)
1) LDA, –78°C
2) BnBr
N
Me
Bn
N
Me
O
O
sing. ent.
b)
1) LDA, –78 °C
2)
O
H
O
HO
O
(±)
c)
O
1) Br2, H2O
2) NaOH
sing. ent.
d)
Et3N,
Cl B
O
Me
O
Me
Cy
B
O
Cy
Me
H
O
OH
Me
achiral
Me
(intermediate, fill in)
Me
(±)
Chem 335 S01 MT#2: Page 3
1. (continued)
e)
O
mCPBA
EtO2C
EtO2C
(±)
(±)
f)
OH
O
O
O
O
I2, NaHCO3
O
AIBN, Bu3SnH
I
sing. ent.
OH
OH sing. ent.
OH
g)
EtO2C
O
N
1) NaOEt
2) MeI
EtO2C
Me
O
N
sing. ent.
2. (4 marks) Give the product of the following reaction. In the space below the reaction, include
stereochemical 3D diagrams of starting materials and products, and explain the source of the observed
stereoselectivity in words.
O
HO Et
EtMgBr
attack via least hindered path toward π*
O Me
O
H
OHMe
Ph H
Me
Ph
Ph
H O
Me Ph
Et
H
HO
H
Et
Me
Ph
Chem 335 S01 MT#2: Page 4
3a. (2 marks) Draw a 3-dimensional representation of the molecule shown below and indicate which of
the two faces of the carbonyl is least hindered.
more hindered
O
O
O
H
H
H
O
O
O
H
less hindered
3b. (5 marks) Plan the synthesis shown below. Whether you go in the forward or reverse direction,
show all intermediates, and include reagents and conditions for all reactions.
H
O
O
H
??
CO2Me
HO
H
O
CO2Me
H
TBAF
NaBH4
H
CO2Me
TBDMSO
H
HO
H
H
MeO2C
O
O
Et3N, TBDMSCl
CO2Me
CO2Me
NaOMe, heat
H
O
TBDMSO
H
O
HCl, H2O
H
TBDMSO
O
H
Chem 335 S01 MT#2: Page 5
4. The chiral α-chloroketone shown below is represented in the conformation in which it is most likely
to be attacked by a nucleophile.
O
Nu O–
Nu
Cl
Cl
H
H
4a. (4 marks) Draw the bonding/antibonding orbitals requested separately on each of the four side-view
representations below.
Cl
Cl
Cl
or
O
Et
O
H
Et
C=O π orbital
C-Cl σ orbital
Cl
H
O
Et
C-Cl σ orbital
Cl
O
Et
H
O
Et
C=O π∗ orbital
H
H
C-Cl σ∗ orbital
Key points: bonding orbitals have overlap between atoms… otherwise they don’t make bonds. The
Burgi-Dunitz angle does matter. Phases matter too.
4b. (2 marks) Explain why this is the conformation that will be attacked by the incoming nucleophile.
In this conformation, the σ * orbital is aligned with the π* orbital. This overlap stabilizes the π* orbital
and makes it easier to attack by incoming nucleophiles.
Chem 335 S01 MT#2: Page 6
5a. (3 marks) The reaction below, used in the synthesis of the natural product chlorofusin, is a close
relative of iodolactonization called a bromoetherification. Give the detailed mechanism for this reaction,
including all intermediates, proton transfers, and byproducts.
Br
O
Br2, K2CO3
N
mechanism?
OH
KHCO3
KBr
N
(±)
Br Br
Br
N
OH
Br
N
O H
N
O–
Br–
2 K+
CO32–
KBr
KHCO3
Note that in the first step, the arrow departing the alkene and making a new bond to bromine should be
leaving from the actual pi electrons… a chemdraw glitch just won’t let it come through that way on this
printout.
5b. (2 marks) There are four alkene carbons in the starting material above. Explain why the oxygen
atom attacks exclusively at the one indicated.
This is the only carbon atom that leads to the formation of a 5-membered ring. (Others would give 6, 7,
and 8-membered rings.) 5-membered rings form more quickly than the others. (Baldwin’s rules). It’s
because of the increased entropic penalty for forming any of the larger rings.
END