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Chapter 11
Free Radical Substitution and Addition Reactions
from
Organic Chemistry
by
Robert C. Neuman, Jr.
Professor of Chemistry, emeritus
University of California, Riverside
orgchembyneuman@yahoo.com
http://web.chem.ucsb.edu/~neuman/orgchembyneuman/
Chapter Outline of the Book
I. Foundations
- Organic Molecules and Chemical Bonding
- Alkanes and Cycloalkanes
- Haloalkanes, Alcohols, Ethers, and Amines
- Stereochemistry
- Organic Spectrometry
II. Reactions, Mechanisms, Multiple Bonds
- Organic Reactions *( Not yet Posted )
- Reactions of Haloalkanes, Alcohols, and Amines. Nucleophilic Substitution
- Alkenes and Alkynes
- Formation of Alkenes and Alkynes. Elimination Reactions
- Alkenes and Alkynes. Addition Reactions
- Free Radical Addition and Substitution Reactions
III. Conjugation, Electronic Effects, Carbonyl Groups
- Conjugated and Aromatic Molecules
- Carbonyl Compounds. Ketones, Aldehydes, and Carboxylic Acids
- Substituent Effects
- Carbonyl Compounds. Esters, Amides, and Related Molecules
IV. Carbonyl and Pericyclic Reactions and Mechanisms
- Carbonyl Compounds. Addition and Substitution Reactions
- Oxidation and Reduction Reactions
- Reactions of Enolate Ions and Enols
- Cyclization and Pericyclic Reactions *( Not yet Posted )
V. Bioorganic Compounds
- Carbohydrates
- Lipids
- Peptides, Proteins, and α−Amino Acids
- Nucleic Acids
Note: Chapters marked with an () are not yet posted.
11: Free Radical Substitution and Addition Reactions
11.1 Free Radicals and Free Radical Reactions 11 - 3
Free Radicals (11.1A) 11 - 3
Halogen Atoms
Alkoxy Radicals
Carbon Radicals
11.2 Halogenation of Alkanes with Br
Bromination of Ethane (11.2A) 11 - 7
Mechanism
Initiation Step
Propagation Steps
The CH 3
- CH
Radical
Radical Chain Reactions (11.2B) 11 - 10
Propagation Steps Repeat
Many Chains Occur Simultaneously
Termination Reactions (11.2C) 11 - 11
Combination Reactions
Disproportionation Reactions
Polybromination (11.2D) 11 - 12
11.3 Alternate Bromination Sites 11 - 13
General Mechanism for Propane Bromination (11.3A) 11 - 14
Origins of 1-Bromopropane and 2-Bromopropane (11.3B) 11 - 14
Propagation Reactions
Termination Reactions
Polybromination
Relative Yields of 1-Bromopropane and 2-Bromopropane (11.3C) 11 - 17
11.4 Relative Reactivity of C-H Hydrogens 11 - 18
C-H Bond Strengths (11.4A) 11 - 18
Bond Strengths
C-H Bond Strength and Alkane Structure
Relative Reactivities of C-H's
Radical Stability (11.4B) 11 - 21
Relative Stabilities of Alkyl Radicals
Origin of Radical Stability Order
11.5 Alkane Halogenation with Cl
, F
, or I
Chlorination (11.5A) 11 - 23
Relative Product Yields in Chlorination and Bromination
Cl
is More Reactive and Less Selective than Br
Correlation Between Reactivity and Selectivity
Fluorination and Iodination of Alkanes (11.5B) (continued) 11 - 25
11: Free Radical Substitution and Addition Reactions
•Free Radicals and Radical Reactions
•Halogenation of Alkanes with Br 2
•Alternate Bromination Sites
•Relative Reactivity of C-H Hydrogens
•Halogenation with Cl 2
, F
2
, or I 2
•Radical Additions to Alkenes
•Halogenation with Other Reagents (Appendix A)
•Halogen Atom Reactivity and Selectivity (Appendix B)
11.1 Free Radicals and Free Radical Reactions
Many reactions in earlier chapters have ionic reagents and ionic intermediates. The reactions in
this chapter involve electrically neutral free radicals. These reactions include free radical
halogenations of alkanes and free radical additions to alkenes.
Alkane Halogenation
R
C-H + X
→ R
C-X + H-X
Alkene Addition
R
C=CR
+ X-Y → R
CX-CYR
Some aspects of these reactions cause them to be more complex than ionic reactions. In order to
address these details adequately without overwhelming this general presentation, we include
some topics in " Asides " (in small font) in the chapter text, while some are in Appendices at the
end of the chapter.
Free Radicals (11.1A)
Important free radicals that we see in this chapter include halogen atoms (X
. ), alkoxy radicals
(RO
. ), and carbon free radicals (R 3
C
. ).
Halogen Atoms. The atoms in column 7A (or 17) of a periodic table are the halogen atoms.
Of these, chlorine (Cl) and bromine (Br) atoms are particularly important in the free radical
reactions that we describe here. To clearly contrast them with halide ions (X:
often write halogen atoms as X
. where the (
. ) is an unshared electron.
As with all atoms, each halogen atom has the same number of electrons as it has protons and that
is why it is electrically neutral. In contrast, halide ions (X:
- ) are negatively charged because each
has one more electron than it has protons (Table 11.1).
Table 11.1. Comparison of Halogen Atoms (X
. ) and Halide Ions (X:
- ).
X
. protons electrons X:
- protons electrons
F
. 9 9 F:
Cl
. 17 17 Cl:
Br
. 35 35 Br:
I
. 53 53 I:
We represent halide ions as X:
- that shows the reactive unshared electron pair (e.g. see Chapter
7). We obtain the symbol X
. for the neutral halogen atom by simply removing one electron with
a - 1 charge (an e
We can also visualize the meaning of X
. by picturing its formation from its parent molecular
halogen X 2
X
or X-X or X:X → X
X
The covalent bond between the two halogen atoms (X-X) is an electron pair (X:X). When that
bond breaks homolytically (undergoes homolysis ), each halogen atom retains one of the two
electrons in that bond.
Alternatively we can visualize the formation of the molecular halogens X 2
from individual
halogen atoms.
X
X → X:X or X-X or X 2
Halogen atoms atoms are highly reactive. They do not exist alone, but in molecules such as X 2
H-X, or CH 3
- X where they are bonded to other atoms. We will see at the end of this section
why organic chemists also refer to halogen atoms as free radicals.
Alkoxy Radicals. Another free radical in this chapter is the alkoxy (or alkoxyl ) radical (RO
. ).
We saw alkoxide ions (RO:
- ) in earlier chapters where they were nucleophiles and also strong
bases. Alkoxy radicals (RO
. ) are also highly reactive, but they are electrically neutral.
You can see that they are electrically neutral by imagining their formation from alkoxide ions by
removal of one e
In contrast, if we irradiate (CH 3
3
C-I with light (when it is in a non-polar solvent) we make
iodine atoms (I
. ) and (CH 3
3
C
. radicals.
(CH
C-I light
or → (CH 3
C
I
(CH
C:I
The I
. and (CH 3
3
C
. each retain one electron of the two originally in the C-I bond. Each of those
species is electrically neutral because each has an equal number of protons and electrons.
Counting Protons and Electrons. You can count up protons and electrons in (CH 3
) 3
C
. and in I
. in
order to verify that each species is electrically neutral. But it is easier to simply recognize that since I
. is
electrically neutral (see Table 11.1), (CH 3
) 3
C
. must also be electrically neutral since they both come from
the electrically neutral molecule (CH 3
) 3
C-I.
Why Call Them "Radicals"? We can explain why R 3
C
. species are called radicals (or free radicals ) by
understanding that the symbol "R" that we have used so often is derived from the word "Radical". Early
chemists referred to the organic parts of molecules as "Radicals" and wrote general examples of these
molecules such as CH 3
) 3
C-I, as R-OH and R-I, respectively.
They called the CH 3
group in CH 3
- OH the "methyl radical", and the (CH 3
) 3
C group in (CH 3
) 3
C-I
the "t-butyl radical". Using the general formula R-I for (CH 3
) 3
C-I, we can symbolize how light causes it
to react to form I
. as we show here.
light
R-I → R
.. I
When the R-I bond breaks, R
. becomes a "free" radical (R
. ). Now days, organic chemists reserve the terms
"radical" or "free radical" to refer to neutral species such as (CH 3
) 3
C
. and have extended those terms to
include neutral species such as RO
. and X
. .
11.2 Halogenation of Alkanes with Br
Free radical halogenation reactions of alkanes and cycloalkanes are substitution reactions in which
a C-H is converted to a C-X.
R
C-H + X
→ R
C-X + H-X
While any of the molecular halogens F 2
, Cl 2
, Br 2
, and I 2
will halogenate alkanes and cycloalkanes,
Br 2
or Cl 2
are used most often. We will use bromination (X = Br) to illustrate alkane
halogenation. We discuss chorination with Cl 2
, and possible halogenation using the other
molecular halogens in later sections.
Bromination of Ethane (11.2A)
We describe the general mechanism of alkane halogenation using bromination of ethane
(CH
3
CH
3
) to give bromoethane (CH 3
CH
2
Br).
Figure 11.
h ν
CH
- CH
→ CH
- CH
The Symbol h ν. This reaction occurs when we irradiate a mixture of ethane and Br 2 , either as gases or
in a solvent, with ultraviolet (UV) or visible light. The symbol h ν represents UV or visible light energy
since the energy ( E ) of light is proportional to its frequency ( ν) ( E = h ν )(Chapter 5). We call this
reaction a photochemical reaction because it is initiated by light, but we will also see many radical
reactions that are not photochemical reactions.
Mechanism. The overall reaction for photochemical bromination of ethane includes several
separate steps. We will group the first three of these steps (Figure 11.02 [next page]) into two
categories called initiation and propagation. In order to emphasize that these species are
electrically neutral, we have omitted the traditional "+" signs used in chemical reactions.
Figure 11.02. Initiation and Propagation Steps for Bromination of Ethane.
Initiation
h ν
Br-Br → Br
Br (Step1)
Propagation
CH
- CH
Br → CH 3
- CH
H-Br (Step 2)
CH
- CH
Br-Br → CH 3
- CH
Br (Step 3)
It is important for you to note that the two products of ethane bromination, ( CH 3
- CH
2
- Br and
H-Br ) (Figure 11.01), are formed in different reaction steps. H-Br is formed in Step 2 of this
three step sequence, while CH 3
- CH
2
- Br is formed in Step 3 of the same sequence. We will see
that this a characteristic of all chain reactions is that reaction products are formed in different
steps.
Curved arrows with full arrowheads show the movement of a pair of electrons (two electrons) in a chemical
process as we showed in earlier chapters
The CH 3
- CH
2
. Radical. The ethyl radical (CH 3
- CH
2
. ) shown above is a neutral (uncharged)
chemical species that forms when Br
. removes (abstracts) a neutral H atom from a neutral ethane
molecule. Its geometry might be either planar or pyramidal ( tetrahedral ) (Figure 11.03b).
Figure 11.03b
If it is pyramidal , the C
.
atom is sp
3
hybridized (Chapter 1) and the single unpaired electron is in
an sp
3
orbital. If it is planar , the C
.
atom is sp
2
hybridized (Chapter 1) and the single unpaired
electron is in a 2p atomic orbital perpendicular to the plane containing the C-H and C-C bonds.
Experimental results and calculations indicate that alkyl radicals generally prefer to be planar.
The ethyl radical. like most alkyl radicals, is very reactive because of its unshared electron.
Alkyl radicals rapidly react with other molecules or other radicals that provide another electron
to form a chemical bond. During ethane bromination, the ethyl radical reacts primarily with
molecular bromine (Br 2
) by abstracting a Br to form a C-Br bond (Step 3, Figure 11.02). We
provide more details for that reaction in Figure 11.04 using the curved "half-arrowhead" arrows
that are sometimes used to show the movement of electrons in radical reactions.
Figure 11.
A Comment About Radicals. If you have already studied Chapter 5 (Organic Spectrometry), then you
should note that this ethyl radical, formed as an intermediate in bromination of ethane, is identical to ethyl
radicals formed in the mass spectral fragmentation reactions of alkanes such as butane, pentane, hexane, etc.
that we described in that chapter.
Radical Chain Reactions (11.2B)
Free radical alkane halogenation reactions are chain reactions.
Propagation Steps Repeat. The Br
. that forms in Step 3 (see below) reacts with another
ethane molecule in Step 2 (see below) and the resulting ethyl radical reacts with a new Br 2
in
Step 3 to once again form Br
. .
Propagation
CH
- CH
Br → CH 3
- CH
H-Br (Step 2)
CH
- CH
Br-Br → CH 3
- CH
Br (Step 3)
This cycle of "Step 2 followed by Step 3" repeats many times giving high yields of the product
CH
3
CH
2
Br from Step 3, and the product HBr from Step 2 (Figure 11.05).
Figure 11.
Since free radicals are intermediates in each of Steps 2 and 3, and because these two steps seem
to form a continuous reaction "chain", we call the bromination of ethane a radical chain
reaction.
We call Steps 2 and 3 propagation reactions because in each of them, one radical species
generates another radical keeping the "chain" alive. For example, Br
. reacts with ethane to give
CH
3
- CH
2
. in Step 2, while CH 3
- CH
2
. reacts with Br 2
to regenerate Br
. in Step 3.
Many Chains Occur Simultaneously. We call homolytic decomposition of Br 2
into Br
.
atoms (Step 1, Figure 11.02) an initiation reaction because it gives the radicals that start the
"chains". When we irradiate the reaction mixture, many Br 2
molecules simultaneously undergo
homolytic scission into Br
. atoms (although this represents only a small fraction of the Br 2
an ethyl radical (CH 3
CH
2
. ) as we show in Figure 11.08 using curved arrows.
Figure 11.
In Step 5a, the Br
. that abstracts H becomes H-Br while the CH 3
- CH
2
. that loses the H becomes
ethene (CH 2
=CH
2
). In Step 6a, an ethyl radical (CH 3
- CH
2
. ) abstracts an H from another CH 3
CH
2
. giving ethane and ethene.
Termination Reaction Products. The combination reactions involving Br
. (Step 4 and Step 5) are
chemically "invisible" because Step 4 forms Br 2
that is one of the starting materials, while Step 5 leads to
bromoethane (CH 3
CH 2
Br) that is one of the reaction products. But this is not the case for some of the
other termination products.
For example, we did not show the termination reaction products from Steps 5a and 6a ( ethene, ethane,
and butane ) in the overall reaction for bromination of ethane (Figure 11.01). Ethane is the starting
material, so its formation in a termination reaction is invisible. However, butane and ethene are undesired
side products. We ignore them because their yields are very low compared to that of the desired organic
product bromoethane (CH 3
CH 2
Br). The propagation steps (Steps 2 and 3 in Figure 11.02) leading to
bromoethane occur many times before Step 5a and/or 6a occurs.
Because two radicals that react with each other can no longer propagate chains (Steps 2 and 3), all
three of these combination reactions are called termination reactions. Each termination reaction stops
two chains.
Polybromination (11.2D)
A disadvantage of radical halogenation reactions, such as bromination of ethane, is that the
desired product (in this case CH 3
CH
2
Br) may be further brominated. The 5 H's that remain on
bromoethane (CH 3
CH
2
Br) are reactive toward Br
. like those on ethane. As a result, it is difficult
to prevent the bromination of bromoethane to give dibromo ethanes (Figure 11.09)[next page].
Figure 11.
These dibromo ethanes in turn may undergo further bromination to tribromo ethanes. Ultimately
the products can also include tetrabromo ethanes, pentabromo ethane, or hexabromo ethane.
Minimizing Polybromination. While we cannot prevent the formation of these polybromoalkanes , we
can minimize them by using a large excess of the alkane reactant compared to Br 2
. We can also minimize
polybromination by permitting the monobromination reaction to proceed only partly to completion.
Each of these strategies causes the unreacted alkane to have a higher concentration than that of the
bromoalkane product. As a result, bromine atoms in the reaction mixture encounter and react with
unbrominated ethane molecules much more frequently than with bromoethane molecules that are present in
much lower concentration.
11.3 Alternate Bromination Sites
All 6 H's on ethane (CH 3
CH
3
) are chemically equivalent so abstraction of any of them gives only
bromoethane. However, this is not true for most other alkanes. For example, propane has two
different types of H's and its bromination simultaneously gives both 1-bromo-propane and 2-
bromopropane as reaction products.
CH
- CH
- CH
Br + H-Br
1 - bromopropane
hν
CH
- CH
- CH
→ and
CH
2 - bromopropane
Propagation Reactions. Abstraction of H from either CH 3
group of propane gives a 1 -
propyl radical, while abstraction of H from the CH 2
group gives a 1 - methylethyl radical
(commonly called the 2 - propyl or isopropyl radical) (Figure 11.11).
Figure 11.
When the 1 - propyl radical reacts with Br 2
, the product is 1 - bromopropane , while reaction of the
1 - methylethyl radical with Br 2
leads to 2 - bromopropane (Figure 11.12).
Figure 11.
The Br
. resulting from either of these reactions abstracts an H from propane to form either a 1-
propyl radical, or a 1-methylethyl radical, that subsequently react with Br 2
to repeat this
sequence.
Termination Reactions. Since - propyl radicals and 1 - methylethyl radicals are present together
in this reaction mixture, there are many different combination and disproportionation reactions
that are possible. In each case, two radicals react to give non-radical products that terminate their
radical chains.
The Various Termination Reactions. The termination combination reactions (Step 5 in Figure 11.10)
give both 1 - bromopropane as and 2 - bromopropane.
Figure 11.
In the disproportionation reactions (Step 5 in Figure 11.10) Br
.
abstracts an H from either of the two
different alkyl radicals to give propene and H-Br. These reactions are analogous to the disproportionation
reaction between Br
.
and ethyl radical that gives ethene (Step 5A, Figure 11.07).
Termination reactions between propyl radicals (Step 6 in Figure 11.10) include several combination
and disproportionation reactions analogous to those for ethyl radicals in Figures 11.06 and 11.07.
Figure 11.
11.4 Relative Reactivity of C-H Hydrogens
The relative product yields in free radical alkane halogenation reactions depend not only on the
numbers of H's available for abstraction, but also on the relative strengths of their C-H bonds.
C-H Bond Strengths (11.4A)
The H's on the CH 2
group of propane are more reactive than those on the CH 3
groups because of
differences in the relative bond strengths of these C-H bonds.
Bond Strengths. C-H bond strengths are the amount of energy required to break the
indicated C-H bond to give the corresponding carbon radical and H
. .
CH
- CH
- CH
- CH
- CH
and H
⎪ 1 - propyl
H
CH
- CH-CH
- CH-CH
and H
⎪ 1 - methylethyl
H
While these specific reactions do not occur during bromination of propane, the lower bond
strength of the CH 2
bonds compared to the CH 3
bonds, tells us that it is easier for Br
. to abstract
an H from CH 2
than from CH 3
. We see this direct correlation between C-H bond strength and
C-H reactivity in bromination of alkanes when we examine product distributions in alkane
bromination.
C-H Bond Strength and Alkane Structure. Alkane C-H bond strengths (Table 11.2)
generally decrease in the order CH 3
- H > RCH
2
- H > R
2
CH-H > R
3
C-H (Table 11.2 [next page]).
While there is only one example for R 3
C-H, the two examples for R 2
CH-H, and three examples
for RCH 2
- H, show that the same trends apply to different alkyl groups (R).
Table 11.2. Approximate Bond Strengths for Various C-H Bonds in Alkanes
C-H Bond Type Compound Bond Strength (kJ/mol)
CH 3
RCH 2
CH 2
CH 3
CH 2
CH 2
(CH 3
) 3
CCH 2
R 2
CH-H (2°) (CH 3
) 2
CH-H 413
(CH 3
CH 2
)(CH 3
)CH-H 411
R 3
C-H (3°) (CH 3
) 3
C-H 404
When one alkyl group (R) replaces an H on CH 4
, we call the 3 remaining H's on RCH 3
primary
( 1 °) H's. Similarly, the 2 H's on R 2
CH
2
are secondary ( 2 °) C-H's, while the single H on R 3
C-H
is a tertiary ( 3 °) C-H. You can see that 1 °, 2 °, and 3 ° are "numbers" that directly reflect the
number of R groups substituted for H's on CH 4
and that the trend in C-H bond strength
is methyl > 1 ° > 2 ° > 3 °.
Table 11.2 includes the two different types of C-H bonds in propane that we mentioned earlier
(Figure 11.15). The more reactive C-H's on propane's CH 2
group (shown as (CH 3
2
CH-H in
Table 11.2) are secondary (2°) C-H's, while the less reactive C-H's on propane's CH 3
groups
(shown as CH 3
CH
2
CH
2
- H) are primary (1°) C-H's.
The C-H reactivity trend is 2° > 1° opposite to the trend in C-H bond strengths that is 1° > 2°.
We can extend this correlation for propane bromination to other alkanes as we describe in the
next section.
Example
Q: According to the data in Table 11.2, the bond strength for the indicated C-H bond in
(CH 3
CH 2
)(CH 3
)CH-H is 411 kJ/mol. Predict the approximate bond strengths for all of the other
types of C-H bonds in this compound.
A: The structure of the compound showing all of the different types of H's is
CH 3
(a) (b) (b) (a)
The compound has only two different types of H's, those labelled (a) and (b). Each of the H's labelled
(b) corresponds to the indicated C-H and has the bond strength of 411 kJ/mol. The bond strengths of
the H's labelled (a) can be approximated by those on the CH 3
group on propane (CH 3
CH 2
CH 2
which are given as 420 kJ/mol in Table 11.2.
Relative Reactivities of C-H's. The relative reactivity order of alkane C-H bonds in all radical
bromination reactions is R 3
C-H > R
2
CH-H > RCH
2
- H > CH
4
, and this is exactly opposite the
C-H bond strength order shown in Table 11.2. We show relative product yields for bromination
of three different alkanes in Figure 11.16 [next page] that confirm this reactivity order. These
include bromination of propane , of butane , and of 2 - methylpropane under comparable reaction
conditions.
