ChemActivity 20
ChemActivity
Radical Halogenation of Alkanes
1
20
Part A: Radical Halogenation Reactions
(How can we add a functional group to an unfunctionalized alkane?)
Model 1: Polar and Non-polar Bond Breakage
= "double barbed" arrow (or full arrow)
= "single barbed" arrow (or half arrow)
Polar Bond Breakage
Non-polar Bond Breakage
Z
Z
Z
Z
Z
Z
Z
Z
Y
Z
Y
Z
Y
Z
Y
Z
for example:
for example:
Br
Br
Br
Br
Br
Br
Br
Br
H
Br
H
Br
H
Br
H
Br
Critical Thinking Questions
1. Add formal charges, if necessary, to the example species in Model 1 (everything
below the words “for example.” )
2. A full arrow depicts the movement of a pair of electrons. What does a half arrow
depict? A half (single barbed) arrow depicts the movement of one electron.
3. Circle any radical species in Model 1. (A radical species is a species that has an
unpaired electron.)
Model 2: Showing Radical Bond Formation
new σ bond made from two radical electrons
Rxn I
Br
Rxn II
Y
Br
Z
Br
Br
Y
Br
Z
Br
new σ bond made from one radical electron and one electron from a breaking σ bond
Critical Thinking Questions
4. Rxn II in Model 2 is one of two sigma bond forming reactions that could have taken
place with these two reactants. Use curved half-arrows to show the mechanism of a
reaction that yields Y— Br and Z radical from these same reactants.
Y
Z
Br
Y
Z
Br
ChemActivity 20
Radical Halogenation of Alkanes
2
Model 3: Reactions of Bromine Radical with Various Alkanes
For each reaction below, the most likely products are shown.
H
Rxn III
Rxn IV
H
H
C
C
H
H
primary
Br
H
H
Br
C
C
H
H
H
H
C
C
C
H
H
H
H
Br
H
H
H
H
H
C
C
C
Br
CH3
H
H
secondary
H
CH3
H3C
C
H
H
H
CH3
H2C
Br
Br
H
H
H3C
Rxn V
H
H2C
C
CH3
tertiary
H
Critical Thinking Questions
5. Label the carbon radicals in Model 3 as methyl, primary, secondary, tertiary, allyl or
benzyl.
a) Which carbon radical in Model 3 is closest to having an octet? Explain your
reasoning. (Recall that alkyl groups are electron donating.)
The tertiary radical is closest to having an octet. This can be explained by noting that, as with
carbocations, alkyl groups are slightly electron donating.
b) The following is a list of radicals from highest to lowest potential energy. Is
this list consistent with your conclusion in part a)? Explain.
highest V.E.
H
CH3
1o
2o
3o
or
allyl
benzyl
X (halogen)
lowest V.E.
Yes. The 3o radical is lower in P.E. than the 2o or 1o radicals.
c) Which reaction in Model 3 is most likely? Explain
Rxn V is most likely because it is least up-hill in P.E.. (It is easiest to make the 3o radical.)
6. Add curved half arrows to Rxn III in Model 3 so as to illustrate the mechanism of
product formation.
a) Draw the other sigma bond forming reaction that could have taken place in
Rxn III, but did not. (Note: all 6 H’s of ethane are equivalent.)
Br
H
H
H
C
C
H
H
H
H
Br
H
H
C
C
H
H
H
b) Construct an explanation for why Rxn III in Model 3 happens instead of the
reaction you drew in part a).
Rxn III is more favorable than the reaction shown above because the H radical is very high in P.E.
and very difficult to make relative to even the 1o C radical.
ChemActivity 20
Radical Halogenation of Alkanes
3
Model 4: Light Induced “Homolytic” Bond Cleavage
When a solution of Br2 is exposed to a strong light source a very small percentage of
the Br2 molecules break apart to form two bromine radicals.
Br
Strong Light
Source
Br
Br
(hν)
(ratio of reactants to products)
billions and billions
“Strong Light Source” is usually abbreviated with the letters hν.
Br
very few
Information
No evidence for methyl or 1o carbocations has been found to date. However, the
same is not true for methyl and 1o radicals: both are known. In fact there is evidence for
all of the following radicals (listed below from highest to lowest potential energy).
highest energy
H
CH3
2o
3o
allyl
benzyl
X (halogen)
lowest energy
Critical Thinking Questions
7. When a solution of Br2 and methane is exposed to a strong light source one of the
major products is CH3Br.
H
H
C
H
(excess)
Br
Br
H
H
Br
C
hν
H
+ other
products
H
Use curved arrows to show a reasonable multi-step mechanism for this reaction.
Br
hν
Br
Br
(excess)
this radical also reacts with a molecule of CH4
Br
Br
H
H
C
H
Br
since Br2 is very
abundant in the reaction
mixture, the radical is
most likely to react with a
molecule of Br2
H
H
C
H
H
H
the following "chain termination step"
can occur with low levels of Br2
Br
H
C
H
H
Br
Br
Br
Br
H
C
H
H
H
C
H
H
this radical reacts with a
molecule of CH4
ChemActivity 20
Radical Halogenation of Alkanes
4
Model 5: Collision Statistics
A reaction between two radicals may seem very likely from an energy standpoint
because in such a reaction two high P.E. species combine to form one low P.E. species.
for example:
rxn is very very down hill
Br
H3 C
H3 C
Br
but statistically unlikely!!
However, radicals are so high in energy that they react with anything they “bump
into.” This means a radical has a short life-span and usually does not have time to “find”
another radical.
Critical Thinking Questions
8. Picture a crowded stadium with 100,000 people. Five of these people are violently
reactive and will start a fight with the first person they bump into-getting themselves
thrown out of the stadium. What are the chances that two of the five "violently
reactive people" get in a fight with one another?
It is very unlikely that the two "fighters" would react with one another in a crowded stadium since
each of them will fight with the first person they bump into, and then be thrown out.
9. The first two steps of the mechanism from CTQ 7 are drawn for you below.
reaction mixture after step 2
hν
Br
Br
Br
huge excess
of Br2
Br
Br
Br
Br
step 1
H
H
H
C
step 2
Br
Br
H
Br
step 3
H
H
C
C
H
H
H
H
a) Based on the information in Model 5, circle any step in your mechanism on
the previous page that has a low probability of occurring.
Any step in which two radicals react with one another is very unlikely at the beginning of the reaction
when there is still lots of Br2 available.
b) ·CH3 is like the "fighter" in the stadium analogy. It is a very reactive radical,
and it will react with the first thing it bumps into. What is the most prevalent
species in the reaction mixture ("stadium")? Add this species to the box
labeled "reaction mixture after step 2."
Br2 is in large excess and is the most prevalent molecule in the reaction mixture.
c) Show the mechanism of the statistically most likely reaction involving ·CH3
and draw the products of step 3. Notes: This reaction leads to CH3Br. There
are radical species left over at the end of step 3.
H
ChemActivity 20
Radical Halogenation of Alkanes
5
Part B: Radical Chain Reactions
(What is the mechanism of a radical halogenation chain reaction?)
Model 6: The Three Parts of a Chain Reaction
A radical chain reaction is a reaction with
the following parts:
1. Initiation = net generation of
radicals
2. Propagation = 1 radical consumed;
1 radical produced
3. Termination = net consumption of
radicals
Initiation of
Chain Reaction
Propogation of
Chain Reaction
Termination of
Chain Reaction
Gap causes termination
(another initiation required)
Critical Thinking Questions
10. The reaction in CTQ 7 (shown again below) is a radical chain reaction.
H
H
C
H
(excess)
Br
Br
H
H
+ other
products
Br
C
hν
H
H
a) Use curved arrows to show the initiation step.
(excess)
Br
hν
Br
Br
initiation step
Br
b) Use curved arrows to show the most likely propagation steps leading to the
product, H3C-Br.
Br
this radical also reacts with a molecule of CH4
Br
Br
H
H
C
Br
Br
H
since Br2 is very
abundant in the reaction
mixture, the radical is
most likely to react with a
molecule of Br2
C
H
Br
H
H
C
H
H
H
H
Br
H
this radical
reacts with a
molecule of
CH4
c) A collision between two radicals is unlikely, but it does happen every once in
a while. Eventually such reactions will stop the chain reaction. List at least
two possible chain termination steps (they don’t have to lead to H3C-Br).
Br
H
C
H
Br
H
C
H
the following "chain termination step"
can occur with low levels of Br2
H
H
d) A variety of less favorable propagation reactions lead to side products. List
one possible propagation step that leads to a product other than H3C-Br.
H
H
C
H
H
H
C
H
H
H
H
Br
H
C
H
Br
H
Br
H
C
H
H
H
C
C
H
H
H
H
C
H
H
H
H
C
H
H
H
C
H
H
H
H
H
Br
H
Br
Br
Br
Br
Br
C
Br
H
ChemActivity 20
Radical Halogenation of Alkanes
6
11. Which of the following diagrams best tracks the products formed from a single
initiation event (breaking apart of one Br2 molecule) in a radical chain reaction.
The middle of the three accurately depicts the course of a radical chain reaction.
Br2
Br2
Br2
Br
Br
Br
Br
Br
Br + P
Br + Products
Br + Products
Br + P
Br + Products
Br + P
etc.
etc.
etc.
C.T.
Chain Termination
Chain Termination
Br + P
etc.
etc. etc.
C.T.
C.T. C.T.
etc.
Br + P
etc.
Br + P
etc. etc. etc.
C.T. C.T. C.T. C.T. C.T.
Model 7: Which H Will Be Replaced by X?
• In a radical halogenation reaction a halogen (F, Cl, Br or I) replaces an H.
• In most cases there is more than one type of H to choose from.
For example the hydrocarbon below has two different types of H's:
CH3
X2
CH
H3C
CH3
CH3
hν
C
X = F, Cl, Br or I
H3C
CH3
X
or
CH3
CH
H3C
X
C
H2
Three factors determine which H on a hydrocarbon will be replaced:
i.
The type of H. (benzyl, allyl, 3o, 2o, 1o or methyl)
ii.
The number of H's of a given type.
iii.
The identity of X. (We will discuss this on the next page.)
First Propagation Step
(H Abstraction by ·X)
H
H
H
C
H
H
H
C
C
X
C
H
H
H
H
Abstraction of a 3o H
gives a lower V.E.
radical intermediate.
o
But there is only ONE 3 H to choose from.
First Propagation Step
(H Abstraction by ·X)
H
H
H
C
Abstraction of
a
1o H leads
C
H
X
to a higher
C
C
H
H
V.E. radical
H
H
intermediate.
o
But there are NINE 1 H's to choose from!
H
H
Critical Thinking Questions
12. Explain why, for the example above, you would expect a 1:9 ratio of 3o to 1o H
substitution products if a 3o radical had about the same potential energy as a 1o radical.
You would expect nine times more primary substitutions if 1o and 3o radicals had the same likelihood
of forming because there are nine primary H's to choose from and only one tertiary H.
13. But a 3o radical is LOWER in potential energy than a 1o radical. When X = Cl, a 3o
radical intermediate is 5 times more favorable than a 1o radical intermediate. Given
this, explain why, for the example above, a 5:9 ratio of 3o to 1o H substitution
products is expected. (Consider both the type of H and the number of identical H's.)
There are 9 primary H's and 1 tertiary H, but the intermediate of tertiary substitution is five times
more likely to form. The probability of replacing a primary H is therefore 1 x 9 = 9, while the
probability of replacing the tertieary H is 5 x 1 = 5.
ChemActivity 20
Radical Halogenation of Alkanes
7
Model 8: Selectivity of the Photo-Halogenation Reaction
• Radical halogenation with F2 is unselective, violent and dangerous limiting the
usefulness of this reaction in organic synthesis.
• Radical halogenation with I2 is so slow as to be useless in organic synthesis.
• Radical halogenation with Br2 is just right! It is very useful in organic synthesis.
The rate is manageable and reactions with Br2 are very selective.
(A selective reaction gives close to 100% of a single product. An unselective reaction
gives a mixture of products.)
Table 8: Relative Preference of Halogens for 1o, 2 o, 3 o, Allyl & Benzyl Sites
X sub. for a H
Halogen (X)
X sub. for X sub. for X sub. for a X sub. for a
o
o
o
in benzyl
H in allyl
H in a 1
a H in a 2
H in a 3
Position
Position
Position
Position
Position
Fluorine (F)
1
1.2
1.4
1.5
1.6
Chlorine (Cl)
1
4
5
7
10
Bromine (Br)
1
100
2000
10,000
100,000
For Example:
CH3
Rxn VI
CH3
F2
CH
H3C
hν
CH3
C
H3C
Rxn VII
Cl 2
hν
CH3
C
H3C
CH3
Br2
CH
H3C
CH3
C
H3C
Cl
CH
H3C
9
product ratio
Cl
C
H2
CH3
Br
CH3
2000
F
C
H2
CH3
CH3
hν
9
product ratio
CH3
5
Rxn VIII
H3C
CH3
CH
H3C
CH
CH3
1.4
CH3
CH3
F
CH
H3C
product ratio
9
Br
C
H2
Critical Thinking Questions
14. Fluorination and Chlorination are less selective than Bromination. Do the product
ratios for Rxn VI-VII support this statement? Explain.
Yes. The bromination reaction gives a much high % yield of a single product (>99% of the tertiary
substitution product and less than 1% of the primary substitution product). In contrast, the fluorination
gives 10% tertiary and 90% primary and the chlorination gives 36% tertiary and 63% primary. The latter
two are mixtures of two products in significant amounts.
15. Explain how the ratio 1.4 to 9 was arrived at in Rxn VI in Model 8.
There are 9 primary H's and 1 tertiary H.
The probability of each type of product =
(relative probability of forming the intermediate radial) x (# of H's of that type)
Probability of making a 1o substitution product = 1 x 9 = 9
Probability of making a 3o substitution product = 1.4 x 1 = 1.4
16. Is your explanation consistent with the 2000:9 ratio in Rxn VIII?
Yes. 1 x 9 = probability of making a primary substitution product
2000 x 1 = probability of making a tertiary substitution product
ChemActivity 20
Radical Halogenation of Alkanes
8
17. Consider the following two step synthesis. Note that the identity of the halogen (X2)
is not specified.
X2
CN
X
NC
hν
a) A student chooses Cl2 as the halogen. In the first step, he gets a mixture of three
mono-chlorinated products. Use the information in Model 8 to calculate the ratio
of these three products.
Cl
Cl
Cl
only this one gives the desired
product upon SN2 reaction with NC
5 x 2 = 10
4 x 4 = 16
1 x 12 = 12
so the ratio of 3o:2o:1o = 10:16:12
b) If the mixture above is treated with cyanide ion (NC–) a mixture of three different
cyano-products is observed in a ratio of 5:8:6. Is this ratio consistent with your
answer to part a)? Yes. 5:8:6 is the same as 10:16:12
c) Explain why replacing Cl2 with a different halogen (specify which one) would
give a much better yield of the desired product.
Use of Br2 would give a much higher yield of the desired product relative to the side products since
radical bromination is much more selective.
d) Calculate the ratio of desired mono-CN product to other mono-CN products if this
other halogen were used.
2000 x 2 = 4000
100 x 4 = 400
1 x 12 = 12
CN
NC
4000
= 91%
4000+400+12
400
= 9%
4000+400+12
NC
12
= <1 %
4000+400+12
Exercises for Part A
1. Along with the methyl radical, there is a Br radical in the box in CTQ 9. What
happens if Br radical reacts with the most abundant species in the reaction mixture
(Br-Br)? Draw this reaction and explain why it not very interesting.
2. Construct an explanation for why the solvents on the left are suitable for radical
halogenation reactions, but the solvents on the right are not. (Note that benzene is
resonance stabilized, making it very unreactive.)
Cl
Cl
Cl
benzene
H
C
carbon
tetrachloride
suitable solvents
Cl
Cl
Cl
cyclohexane
hexane
C
H
O
dichloromethane diethyl ether
NOT suitable solvents
ChemActivity 20
Radical Halogenation of Alkanes
9
3. Consider the partial resonance representation of a benzyl radical species.
a) Use curved half-arrows to generate the other three resonance structures.
b) On the incomplete composite structure of this benzyl radical (shown below),
mark each carbon that has partial radical character with a delta radical.
δ = "delta radical"
c) Explain why a benzyl radical is much lower in potential energy than the
radical drawn below. (Hint:are there important resonance structures for the
species below that show the spreading out of the radical over several atoms?)
4. Based on the analogy to carbocations, circle the species below that is lower in
potential energy.
a) Construct an explanation for your choice. (Hint: use curved arrows to draw
any other resonance structures for each of the radical species above.)
b) Which one of these radicals is an allyl radical? Explain why.
c) Draw an allyl carbocation and explain the similarity between an allyl radical
and an allyl carbocation.
5. Give an example of each of the following and label each example with one of the
following words: primary radical, secondary radical, tertiary radical, methyl radical,
benzyl radical, allyl radical.
6. Read the assigned pages in the text and do the assigned problems.
ChemActivity 20
Radical Halogenation of Alkanes
10
Exercises for Part B
7. Consider the following reaction.
X2
hν
3
Note: only H's attached to sp C's can be substituted for by X
a) There are two different H's in this molecule that can be replaced by X. Add
them to the drawing above and label them Ha and Hb.
b) There are two different H's on this molecule that will not react with X radical.
Label them Hc and Hd.
c) Specify whether Ha and Hb are primary, secondary, tertiary, allyl or benzyl.
d) Calculate the relative amounts of each of the two different products when the
reaction is run with Cl2.
e) Calculate the relative amounts of each of the two different products when the
reaction is run with Br2.
f) Which reaction is more selective? Rxn in part d) or Rxn in part e) [circle
one].
8. Consider the following reaction:
H
C
H
H
H
C
C
H
H
H
H
Br2
Br
hν
H2
C
CH
H 3C
CH3
H 3C
Product I
Br
C
H2
Product II
product ratio
a) Use curved arrows to show the mechanism for formation of Product I (include
the initiation step and at least one termination step).
b) Use curved arrows to show the mechanism for formation of Product II
(include the initiation step and at least one termination step).
c) Using the data in Table 8, predict the ratio of Product I:Product II and put the
appropriate numbers in the box labeled “product ratio.”
9. A selective reaction gives mostly a single product. An un-selective reaction gives a
mixture of various products. In general, which reaction below is most selective?
a)
radical fluorination of an alkane
b)
radical chlorination of an alkane
c)
radical bromination of an alkane
ChemActivity 20
Radical Halogenation of Alkanes
11
10. Use curved half arrows to show the most likely radical reaction including the most
likely mono-halogenation product using…
a) Cl2 and light
b) Br2 and light.
11. S-3-methylhexane undergoes radical bromination to form a racemic mixture of two
products. Draw both products and explain why a racemic mixture is formed rather
than pure S or pure R.
12. Design a synthesis of each of the following target molecules starting from
cyclohexane. You may use any reagents in your proposed syntheses (carbon or noncarbon containing).
OH
H3C
C
C
OH
Br
OH
+ enantiomer
+ enantiomer
13. Read the assigned pages in the text and do the assigned problems.
ChemActivity 20
Radical Halogenation of Alkanes
12