Tutorial 9 Solutions
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CHEMISTRY 1AA3-TUTORIAL 9 SOLUTIONS-WEEK O
WEEK OF MARCH 19, 2001
1. a) One of the isomers of C6H14 gives two (and only two!) different
monochlorinated products when treated with Cl2 in the presence of ultraviolet
light. Write a reaction equation, showing the structures of the starting materials
and products clearly.
Cl
Cl2
Cl
+
hυ
2,3-dimethylbutane
1-chloro-2,3-dimethylbutane
2-chloro-2,3-dimethylbutane
(Names not required.)
b) Two different isomers of C6H14 give three monochlorinated products when
treated with Cl2 in the presence of ultraviolet light. Write reaction equations for
both of these isomers, showing the structures of the starting materials and
products clearly.
Cl
Cl2
hυ
hexane
+
Cl
+
3-chlorohexane
Cl
2-chlorohexane
1-chlorohexane
AND
Cl
Cl2
hυ
2,2-dimethylbutane
Cl
Cl
+
+
3-chloro-2,2-dimethylbutane
1-chloro-2,2-dimethylbutane
4-chloro-2,2-dimethylbutane
2. Suppose you mixed equimolar amounts of methane, ethane, and bromine, and
then irradiated the mixture with ultraviolet light. Give all possible
monobrominated products, and explain which would be formed in the greatest
amount.
CH4 +
CH3CH3
Br2
hυ
CH3Br
minor
+
CH3CH2Br
major
Note first that if there are equimolar amounts, then we have twice as many moles of
alkane (in total) as we have of bromine. That is, there isn’t enough bromine to go
around! The methane and ethane will have to compete for bromine, and the major
product will reflect the “winner”.
Tutorial 9 Solutions
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Statistically speaking, if all H’s were equally reactive, we would expect a
bromomethane: bromoethane ratio of 4H:6H, i.e. 40%:60%. However, the
.
intermediate in the free-radical halogenation of methane is a methyl radical ( CH3),
which is less stable than the 1o radical formed in the free-radical halogenation of
.
ethane. Because the CH3 radical is less stable, we expect even less than 40%
bromomethane.
3. Look at the reaction coordinate diagram (or “energy profile”) shown below and
answer the following questions:
d
PE
b
c
e
a
Rxn coordinate
a) Is this reaction exothermic or endothermic? How do you know?
Because the energy of the products (“e”) is higher than the energy of the starting
materials (“a”), this is an endothermic reaction.
b) How many steps are involved in the reaction mechanism? How do you
know?
There are two steps. Each has its own activation barrier, so there are two “humps”
in the diagram.
c) Which step is fastest? How do you know?
The second of the two steps is faster, because it has a lower Ea (activation energy).
That is, the height from “c” to “d” is less than the height from “a” to “b”.
Tutorial 9 Solutions
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4. Which alkyl halide would you expect to react more rapidly by an SN2
mechanism? Explain briefly.
a)
CH3CH2CH2CH2Br or
CH3CH2CHCH3
Br
The primary alkyl halide would react more rapidly than the secondary alkyl halide,
because it is less crowded (backside attack is less hindered).
CH3CH2CH2Cl
b)
or
CH3CH2CH2Br
-
The bromide would react more rapidly, because Br is a better leaving group than
-
Cl .
CH3
c)
CH3CH2CH2Cl
or
CH3CCH2Cl
CH3
Although both of these alkyl halides are primary, backside attack on 1chloropropane would be less hindered. (1-Chloro-2,2-dimethylpropane has a
bulky tert-butyl group on the carbon bearing the leaving group.)
5. Which SN1 reaction would you expect to take place more rapidly? Explain
briefly.
a)
(CH3)3CBr
or (CH3)3CCl
CH3OH
CH3OH
-
The bromide would react more rapidly, because Br is a better leaving group than
-
Cl .
b)
(CH3)3CBr
AgNO3
CH3CH2OH
or (CH3)3CBr
AgNO3
CH3CH2OH/H2O
The SN1 mechanism has a carbocation intermediate, which is more stable in more
polar solvent mixtures (note that water is more polar than ethanol). Because the
intermediate is more stable, the reaction is faster.
c)
(CH3)3CCl
or (CH3)2CHCl
H2O
H2O
The 3o carbocation formed as the intermediate in the reaction of the 3o alkyl halide
would be more stable, and therefore the reaction would be faster.
Tutorial 9 Solutions
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6. Give the products of the following substitution reactions, and state the
mechanism (SN1 or SN2) by which the product is formed.
a)
OH
Cl
H2O
SN1; 3o alkyl halide
O
b)
Br
c)
I
NaOCCH3
SN2; 1o alkyl halide
O
OCCH3
NaSCH2CH3
SN1; 3o alkyl halide
SCH2CH3
d)
CH2Cl
KOCH(CH3)2
SN2; 1o alkyl halide
CH2OCH(CH3)2
e)
I
KOH
(SN2 only)
OH
(Note the inversion
of configuration; the "OH"
is on a dashed bond)