Chem 331: Problem Set #4 (Hour Test practice questions) O (2S, 4Z)-4-bromo-2-methyl-4-hexenal Br

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Chem 331: Problem Set #4 (Hour Test practice questions)
1) Draw the structure:
(2S, 4Z)-4-bromo-2-methyl-4-hexenal
O
Br
2) Give the IUPAC name for each compound
Br
O
O
CHO
(2S)-2-bromopropyl (3S)-3-methylpentanoate
(1S,3R)-3-methylcyclohexane
carboxaldehyde
OH
CH3
Cl
H
H
Br
HS
Br
H
CH3
H
(1R,4S,5R)-4-bromo-5-mercapto-2-cycloheptenol
(2R,4S)-2-bromo-4-chloropentane
3) For each pair, indicate which is more stable. Use a clear picture to explain why
Both cyclohexanes are 'locked' with
equatorial tBu's and axial methyl groups
so the answer lies in the double bond
Br
Br
t-Bu
Br
Br
Brax
H
H
Breq
Breq
Breq
H
Brax
H
H
H
CH3
H
Brax
H
Brax
H
CH3
H
tBu
t-Bu
most stable form has no
axial Br's
Breq
t-Bu
H
H
sp 2 center
There is no
axial group here,
and therefore there
is only one
1,3-diaxial interaction
make a model to
convince yourself!
two 1,3-diaxial
interactions
Chem 331: Problem Set #4 (Hour Test practice questions)
4) 1,2-cycloheptadiene is not a stable molecule. Use a clear picture and less than 20 words
to explain why.
Begin by drawing the MO for the allene (C=C=C)
fragment
C
C
C
the central carbon is sp hybridized,
and the ideal geometry for this fragment
is linear
C
C
C
C
that puts these two carbons far away
from one another, and you only have 2
carbons left to form the 7-membered ring.
It is impossible to do it without severly
bending the allene away from its ideal
geometry of 180 ° (approximately a 30 °
distortion is required)
BUILD A MODEL!
5) Provide a mechanism
Br
Bu3SnH
O
O
H
O
AIBN
H
NC
N N
•
H
CN
Br
SnBu3
multiple
bond
addition
Br-atom
extraction
•
O
O
O
H-atom
extraction
O
Bu3Sn
• SnBu3
H
CN
AIBN
• SnBu3
:N N: +
O
H
H
HH
O
H
HH
H-abstraction
H
•
H
O
H
+
O
•
O
O
multiple
bond
addition
Having trouble getting started? Map it out first. Start with the acetal as your marker. You can see
that it is one carbon away from a 5 membered ring
Br
O
1
3
2
O
3
O
1
2
O
Now, we can see that the central 5-membered ring of the product is the ring from the starting
material
6
O
7
1
2
3
5
6
Br
7
O
O
3
5
4
1
O
2
4
finish labeling. We can see that a bond must be formed between the atoms indicated below
Br
e
c
6
b
a
d O
7
3
5
4
1
2
O
e
6 7
c
b
3
5
a
4
2
d
O
O
O
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