Organic Chemistry I
CHEM 2323
Packet #9
1.
A solution of 1-methylcyclohexene (0.1 M) dissolved in acetone solvent was reacted with
1 equivalent of bromine. Draw the preferred product.
Br
Br2
Br
If 0.1 M of lithium iodide was also dissolved in the acetone solvent, how would the preferred
product change? Discuss what factors need to be considered.
!+
Br
Br
!+
or
I
I
Br
If 0.1 M lithium iodide is added, nucleophilic iodide is present in addition to the bromide present
after the bromine reacts with the 1-methylcyclohexene to generate the bromonium ion. The
bromonium ion will react preferentially with the more nucleophilic iodide to generate the
product shown. Must show anti addition with the iodide reacting at the more substituted carbon.
What changes would occur if 0.1 M of lithium chloride was dissolved in acetone instead?
Would have the same consideration as with lithium iodide, but now have a less nucleophilic
chloride present relative to bromide. The bromide would react preferentially to generate the
product as shown in the first example. Would obtain some product with a lower yield, however,
with the chloride reacting in an anti addition at the more substituted carbon.
2.
An unknown compound (1) with a molecular formula of C8H15Br1 was found to be
saturated. When 1 was reacted with sodium methoxide, two enantiomers (2 and 3) with
molecular formula C8H14 were obtained. When either 2 or 3 were hydrogenated they both
yielded cis-1,3-dimethylcyclohexane. Also when either 2 or 3 were reacted with ozone followed
by dimethyl sulfide, a dialdehyde (C8H14O2) was obtained. Propose structures for compounds 1,
2 and 3.
1
Br
NaOCH3
1) O3
2) CH3SCH3
OHC
CHO
CHO
H2
Pt
CHO
2
3
Compound 1 can be either of two diastereomers. Draw the structure of the two diastereomers.
One of the diastereomers reacts faster with sodium methoxide to yield compounds 2 and 3.
Which diastereomer reacts faster? Why?
Br
Br
The diastereomer with the bromine in the axial position as drawn will react faster as the bromine
is positioned for an anticoplanar elimination with a greater percentage.
3.
An unknown compound (1, C5H8) was deprotonated with NaNH2 quantitatively and then
reacted with 1-bromopropane to generate compound 2. When 2 was reacted with potassium
permanganate only butanoic acid (a linear four carbon carboxylic acid, C4H8O2) was obtained.
Compound 2 was reduced under two different conditions to generate alkenes 3 and 4. When
compound 3 was reacted with Br2 a meso compound was obtained, while when compound 4 was
reacted with Br2 a chiral compound was obtained. What are the structures for compounds 1-4
and what reducing conditions were used to generate compounds 3 and 4 respectively?
Br2
Na, NH3, CH3OH
Br
1) NaNH2
2) CH3CH2CH2Br
reduction to alkene
1
2
KMnO4
3
Pd/BaSO4
quinoline
Br
meso
Br2
Br
Br
4
O
chiral
OH