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Transcript
CH. 4
Chapter 4
Alcohols and Alkyl Halides
Alcohols and alkyl halides are very important functional groups. A functional group is an
atom or group of atoms that undergoes certain reactions that are typical of that functional
group. It is important to recognize functional groups since it makes the organization and
learning of organic chemistry much easier. There are several million organic compounds
that are known and more are discovered or synthesized everyday but there is a limited
number of functional groups.
Functional groups we have already seen:
Alkanes – essentially have no functional group. They contain only C and H single bonds.
Alkenes – have the carbon-carbon double bond as the functional group.
Alkynes – have the carbon-carbon triple bond as the functional group.
Arenes – have the benzene ring as a functional group.
New functional groups in this chapter:
Alcohols – contain the hydroxyl group, -OH.
Alkyl halides – contain a halide. We use the symbol X to stand for any halogen (I, Br, Cl,
F).
“R” is used to refer to any alkyl group regardless of size. It can be primary, secondary or
tertiary, so a generalized alcohol is written as ROH and a generalized alkyl halide is
written as RX.
Other functional groups that will be studied throughout the rest of the course include:
Amines – contain a nitrogen, N. The amine may be primary (RNH2), secondary (R2NH),
or tertiary (R3N) where the R groups can be the same or different.
CH3CH2
N H
H
ethylamine
(primary amine)
CH3CH2
N CH3
CH3CH2
H
ethyl methylamine
N CH2CH3
CH2CH3
(secondary amine)
triethylamine
tertiary amine)
Ethers – contain a C-O-C linkage, R-O-R’, where R and R’ can be the same or different.
CH3CH2
O CH2CH3
diethyl ether
Epoxides – these are special 3-membered ring ethers.
R
R
O
R
R
1
CH. 4
Nitriles – these contain the CN functional group, RCN. This functional group is also
called a cyano-group.
CH3CH2CH2
C
N
Nitroalkanes – these contain the nitro group, -NO2.
O
CH3CH2
N
O
Thiols – these contain an –SH group RSH.
CH3CH2CH2
SH
Sulfide – these are thio ethers, R-S-R’.
CH3CH2CH2
S CH2CH2CH3
Carbonyl Derivatives – all contain the carbonyl group, C=O
Aldehydes – have the carbonyl group at the end of the chain, -CHO.
O
CH3CH2CH2
C
H
Ketones – have the carbonyl group attached to two other carbons, one on each side,
RCOR’.
O
CH3CH2CH2
C CH3
Carboxylic Acid – contain the functional group –CO2H.
O
CH3CH2CH2
C OH
Carboxylic Acid Esters (or Esters) – contain the functional group –CO2R.
O
CH3CH2CH2
C O CH2CH3
Amides – contain the functional group –CONR’2
CH3CH2CH2
C
O
O
O
NH2
CH3CH2CH2
C
NH
CH3
CH3CH2CH2
Carboxylic Acid Anhydrides – contain the functional group RCO2COR’
2
C
N CH3
CH2CH3
CH. 4
O
CH3CH2
O
C O C CH2CH3
Nomenclature
Alkyl halides: There are two ways to name alkyl halides.
1. Name the alkyl group first, then as a separate word name the halide.
CH3F
CH3CH2CH2CH2Br
methyl fluoride
butyl bromide
2. The systematic IUPAC name is to treat the halide as a substituent on the longest alkyl
chair that contains the halide. Alphabetize it along with any other substituents. Number
the chain so that the substituents get the lowest number at the first point of difference
between the two possible directions for numbering.
CH3 Cl
CH3CH2CH2CH2CH2F
CH3
1
1-fluoropentane
CH
2
CH
3
CH2 CH3
4
5
3-chloro-2-methylpentane
Alcohols:
For the systematic IUPAC name, identify the parent alkane as the longest chain that
contains the hydroxyl group. Drop the “e” from the alkane and add the suffix “-ol”. The
alcohol group takes precedence over alkyl and halide substituents so number the parent
chain in the direction that gives the alcohol the lowest number, giving the terminal carbon
the number one. List the other substituents alphabetically, specifying their locations by
number.
In cyclic alcohol, the hydroxyl group always gets number one, though the number is not
written. Number the other substituents so as to obtain the lowest number at the first pint of
difference between the two possible numbering directions.
OH
CH3 CH2 CH2 OH
2
1
3
1-propanol
CH3 CH
2
3
CH3
1
2-propanol
CH3
6
CH3
Br
CH
5
CH CH2 CH
2
3
4
OH
OH
4-bromo-5-methyl-2-hexanol
1
CH3
1
Cl
3
2
CH2CH3
3-chloro-3-ethylcyclohexanol
Classes of Alcohols and Alkyl Halides
A primary alcohol has a hydroxyl group attached to a primary carbon and a primary alkyl
halide has a halogen attached to a primary carbon.
3
CH. 4
CH3
CH2 CH2 OH
CH2 CH2 Br
CH3
primary alcohol
primary alkyl haldie
A second alcohol has a hydroxyl group attached to a secondary carbon and a secondary
alkyl halide has an alkyl group attached to a secondary carbon.
OH
CH3
4
CH2 CH
2
3
Cl
CH3
1
CH3
4
secondary alcohol
CH2 CH
2
3
CH3
1
secondary alkyl halide
A tertiary alcohol has a hydroxyl group attached to a tertiary carbon and a tertiary alkyl
halide has a halide attached to a tertiary carbon.
OH
Cl
CH3
4
CH2 C
CH3
2
1
3
CH3
tertiary alcohol
CH3
4
CH2 C
CH3
2
1
3
CH3
tertiary alkyl halide
Both alcohols and alkyl halides have fairly large dipole moments.
δ− Cl
R
δ−O
C
H
R
R
δ+
R
C
δ+ R
R
For alkyl halides, note the trend in bond lengths. The C-F bond is the shortest and the C-I
bond is the longest as one would expect based on the much larger size of iodine versus
fluorine.
R
R C
Carbon-halogen
bond length
R
F
<
R
1.4 A°
R
R
R C Cl
<
R C
R
1.79 A°
Br
R
1.97 A°
Physical Properties:
Compare boiling points:
CH3CH2CH3
CH3CH2F
CH3CH2OH
b.p. -42°C
b.p. -32°C
b.p. 78°C
4
<
R C
I
R
2.16 A°
CH. 4
Fluoroethane is more polar than propane and has a slightly higher boiling point due to
increases in dipole-induced dipole interactions but the alcohol has a much higher boiling
point due to intermolecular hydrogen bonding. In order for hydrogen bonding to occur a
hydrogen must be attached to an electronegative atom like oxygen or nitrogen or fluorine
that has a lone pair. Each hydrogen bond is worth about 20 KJ/mol or ~ 5 Kcal/mol in
energy. Carbon-hydrogen bonds are not polarized enough to engage in hydrogen bonding.
CH3CH2
O
H
H
CH3CH2
O
O H
O
CH2CH3
hydrogen bond
H
O H
O
N H
N
N H
O
N
For alkyl halides, the boiling point increases with the increasing size of the halogen.
boiling point
°C
CH3 Cl
-24
CH3 F
-78
CH3 Br
CH3 I
3
42
This is because the larger halogens are more polarizable. In the very large iodine, the
electrons are much farther from the nucleus and more easily distorted. Therefore, there are
stronger induced dipole-induced dipole interactions.
The boiling point also increases with the increasing number of halogens.
CH3Cl
boiling point
°C
CH2Cl2
CHCl3
CCl4
61
77
40
-24
Again, this is due to increased induced dipole-induced dipole interactions.
But fluorine is an exception due to the fact that fluorine is not very polarizable, though
CF3CF3 still has a higher boiling point than ethane, CH3CH3, since it is more polar.
CH3CH2F
boling point, °C
-32
CH3CHF2
CH3CF3
-47
-25
CF3CF3
-78
CH3CH3
-89
Water solubility:
All alkyl halides are insoluble in water, like alkanes. Low molecular weight alcohols
(methanol, ethanol, 1-propanol, 2-propanol) are miscible with water. Miscible means they
are soluble in all proportions and form one layer with water. This is due to intermolecular
hydrogen bonding. As the hydrocarbon chain gets longer, the solubility decreases.
5
CH. 4
Only 0.5 ml dissolves in one liter of water.
CH3CH2CH2CH2CH2CH2CH2CH2 OH
polar
non-polar
Preparation of Alkyl Halides from Alcohols and HX
General reaction:
+
R OH
R X
H X
+
H O
H
The reactivity order for the hydrogen halides parallels their acidity:
H
I
H Br >
>
H Cl
>>
strongest acid
H F
weakest acid
The reactivity order for the alcohols is:
R
H
R C OH
>
R C OH
R
>
>
R C OH
R
secondary
tertiary
H
H
H C OH
H
H
methyl
primary
Ex:
CH3
CH3
C OH
+
room temp.
H Cl
(rt)
CH3
t-butanol
CH3
CH3
C Cl
+
H2O
CH3
t-butyl chloride 78-88%
Sometimes we write the reagents over the arrow in order to save space.
CH3CH2CH2CH2 OH
NaBr, H2SO4
heat
CH3CH2CH2CH2
Br + H2O + Na+ -OSO3H
Mechanism of the Reaction of t-Butanol and HCl
When t-butanol reacts with hydrochloric acid to form t-butyl chloride, this is a substitution
reaction. We substitute the hydroxyl group with the chlorine. The mechanism is a
Nucleophilic Substitution Unimolecular or SN1 mechanism. The mechanism of a chemical
reaction is a detailed, step-by-step description of how the reaction occurs. As chemists, we
can never actually observe a reaction occurring and so we have to infer the actual
6
CH. 4
mechanism based on what intermediates are formed and based on our understanding of
chemical principles. Generally most chemists will agree on a mechanism and it is critical
in the study of organic chemistry to learn the proper mechanism and to keep track of how
the electrons are moving in a given reaction by means of the curved arrow formalism.
Step One: The SN1 mechanism under consideration here involves three separate steps.
Step one is proton transfer from the hydrochloric acid to the oxygen lone pair of the
alcohol. We protonate the hydroxyl group to make it into a good leaving group in step 2.
CH3
CH3
CH3
C OH
CH3
+
CH3
H Cl
H
C O
+
Cl
CH3 H
acid
base
This is a simple acid-base reaction. It is a very fast reaction and therefore has a low
activation energy (EACT). The rate of the reaction is determined by the height of energy
barrier. In general, proton transfer reactions are the fastest reactions in chemistry. The
reaction is exothermic, which means heat is given off and it is favorable. The H-Cl bond is
weak and easily broken. The hydrogen forms a stronger bond with oxygen.
Potential Energy
If we make a graph of what happens to the potential energy as the reaction proceeds we see
that initially the potential energy increases slightly. This is the activation energy or energy
barrier to the reaction. The potential energy then reaches a maximum. We define this
maximum as the transition state or TS. This literally is the point of transition between
starting materials and the product.
Transition State (TS)
EACT
CH3
CH3 C OH H Cl
CH3
Exothermic Reaction,
early transition state
CH3 H
CH3 C O
CH3H
Reaction Progress
We cannot actually isolate the transition state to study its structure. It has a very short
lifetime and is an unstable structure that is literally making the transition from starting
material to product. It is very important to understanding the full mechanism to have some
sense of the structure of the transition state because it is the height of the maximum point
of the transition state, the activation energy, that determines the rate and the more stable
the transition state the lower it is in energy.
7
CH. 4
Hammond’s Postulate: this is a theory or postulate that predicts the structure of the
transition state based on whether a reaction is exothermic or endothermic. It states that if
two species are similar in energy, they are similar in structure. In general:
For an exothermic reaction, the transition state is similar in energy to the starting materials
so it resembles the starting materials in structure. Bond breaking and bond formation have
not proceeded very far. We call this an early transition state and show it using dotted lines
for the bonds that are breaking and forming.
CH3 H
CH3
C O
δ+
CH3
H
δ+
symbol for transition state
Cl
δ−
The oxygen-hydrogen bond is still much longer than the H-Cl bond;
bond formation has not proceeded very far.
For an endothermic reaction, we have a late transition state. Bond breaking and bond
formation is nearly complete and the transition state resembles the product.
We will see an example of an endothermic reaction and a late transition state in step two of
our SN1 mechanism.
Step Two: Step two is the slow, rate-determining step. It is a unimolecular reaction in
which the protonated alcohol from step one leaves as the neutral water to form a
carbocation.
CH3
CH3
H
C O
CH3 H
CH3
CH3
C
CH3
H
+
O
H
This step is endothermic since a bond is being broken but no bond is being formed. It
requires energy to break a bond. The carbocation product of this reaction is a high energy
intermediate. The potential energy versus reaction progress diagram shows that the
transition state is similar in energy to the carbocation product. So there is lots of positive
charge on the carbocation intermediate.
8
CH. 4
Potential Energy
TS - has lots of carbcation character
because bond cleaving is almost complete.
CH3
CH3 C
CH3
EACT
CH3
CH3
CH3 H
CH3 C O
CH3 H
C
δ+
CH3
H
O δ+
H
Bond breaking is almost complete so there
is lots of positive charge on the carbon.
Reaction Progress
Carbocations are planar, trigonal structures with an sp2 carbon that has an empty p-orbital
perpendicular to the plane of the carbon atom and its three substituents. The carbocation
has only six electrons around it so it is a Lewis Acid. Another term for a Lewis acid is
electrophile. An electrophile is a species that “loves” (from the Greek philos) electrons.
CH3
sp2 carbon
CH3
The empty p orbital is perpendicular to the plane of the
three atoms attached to the sp2 carbon.
CH3
empty p orbital
Step 3: The chlorine attacks the empty p orbital on the carbocation to form a new Cl-C
bond. This step is very fast and thermodynamically very favorable since a bond is being
formed and no bond is being broken. The chlorine is a nucleophile. Nucleophiles are
Lewis bases that have a lone pair and/or a minus charge. They donate electrons to the
electrophile. We always show the arrow for the electrons as moving from the nucleophile
lone pair to the electrophile.
CH3
CH3
C
CH3
+
Cl
fast
CH3
CH3
C Cl
CH3
This step is fast and has a low activation energy. It is exothermic.
9
CH. 4
Potential Energy
EACT3
CH3
CH3 C
CH3
CH3
CH3 C Cl
CH3
Reaction Progress
The overall potential energy diagram for all three steps is as follows:
EACT3
CH3
CH3 C
CH3
Potential Energy
EACT2
EACT1
CH3 H
CH3 C O
CH3H
CH3
CH3
C Cl
CH3
Reaction Progress
The second step, the formation of the carbocation has the highest activation energy and
therefore it is the slowest step and determines the overall rate of the reaction.
Carbocation Stability
Alkyl groups stabilize a carbocation by donating electrons and helping to spread out the
positive charge.
10
CH. 4
CH3
C CH3
>
CH3
CH3
tertiary carbocation
most stable
C
H
>
CH3
secondary carbocation
CH3
C
H
>
H
primary carbocation
H C
H
H
methyl carbocation
least stable
There are two ways that the alkyl groups release electrons.
(1) The Inductive Effect: This is due to the polarization of sigma bonds. The electrons in
a C-C bond are more polarizable than the electrons in a C-H bond. So replacing H’s with
alkyl groups has the effect of increasing the polarizability of the bonds to the carbocation
and allows for better electron donation through the sigma bonds. This has the result of
making the carbocation less electron deficient and therefore more stable and lower in
energy.
CH3
C
CH3
CH3
(2) Hyperconjugation: This is a resonance effect due to overlap of the six C-H bonds with
the empty p orbital of the carbocation. For this overlap to be effective the C-H bonds must
be on the carbon directly connected to the carbocation.
The two electrons from the C-H bond on the alkyl group attached to
the carbocation can donate into the empty p orbital of the
carbocation. The two C-H electrons are partially delocalized onto the
p orbital. The more alkyl groups that are attached to the carbocation,
the more electron donation there is.
CH3
H
CH3
H
Another representation of this effect that is sometimes shown is as follows:
H
H
H
C CH3
H
CH3
H
C CH3
H
CH3
Now we can understand why tertiary alcohols react faster than secondary alcohols and why
secondary alcohols react faster than primary ones. It is due to the fact that the tertiary
alcohols, upon protonation and loss of water, form the more stable tertiary carbocations.
Tertiary carbocations are lower in energy than secondary and primary and methyl
carbocations and form faster.
11
CH. 4
A graphical comparison of the relative energies of the various carbocations shows that the
tertiary is much lower in energy and furthermore has a lower activation energy (EACT3°) so
that it forms much faster.
potential energy
H C H
H
R C H
H
R C R
R
EACT methyl
EACT1°
EACT2°
R C R
R
EACT3°
H
3°
H H
O
R C R
R
H
O
R C H
R
2°
H
H
O
R C H
H
1°
methyl
H H
O
H C H
H
Generally speaking, the primary and methyl carbocations are too high in energy to be
formed and are never observed. So with primary and methyl alcohols there is a different
mechanism.
SN2 Mechanism: Nucleophilic Substitution Bimolecular
With methyl alcohol and primary alcohol we have a different mechanism called the SN2
mechanism, where the “2” refers to the fact that the rate determining step has two
molecules coming together (i.e. bi-molecular).
The first step is the same as in the tertiary alcohol case: the alcohol oxygen is protonated so
as to make it into a good leaving group.
fast
CH3CH2CH2 O
+
H
H Cl
H
CH3CH2CH2 O
+
Cl
H
In the second step, the chlorine anion produced in the first step directly attacks the carbon
bearing the protonated hydroxyl group to replace it. The carbon bearing the oxygen has a
partial positive charge since the oxygen is more electronegative than carbon.
12
CH. 4
H
CH3CH2CH2 O
slow
CH3CH2CH2
H
CH3CH2CH2 O
H
δ+
Cl + H2O
H
The negatively charged chlorine is attracted to the
partial (+) charge on the carbon bearing the oxygen.
Cl
As will be discussed in much more detail in Chapter 8, the attack by the nucleophilic
chlorine occurs from the backside. The transition state is pictured below.
CH2CH2CH3
Cl
δ−
C
H H
H
O
δ+ H
In the transition state breaking of the C-OH2 bond is more or
less sequenced with formation of the C-Cl bond. Think of the
chlorine electrons as pushing out the electrons of the C-O
bond. The OH2 group leaves with these two electrons.
The potential energy versus reaction progress diagram shows two transition states
corresponding to the two steps. The second step has the higher activation energy and is
therefore the slow step and overall rate-determining step.
Cl−
Potential Energy
δ
CH2CH2CH3
H
C
O
+
δ H
HH
EACT2
EACT1
CH3CH2CH2OH
HCl
H
CH3CH2CH2 O
H
CH3CH2CH2
Cl
Reaction Progress
Other ways to convert alcohols to alky halides
(1) Thionyl Chloride: We can treat alcohols with boiling thionyl chloride (SOCl2) to
convert the alcohol to an alkyl chloride. This method is considerably milder than using
concentrated hydrochloric acid and is useful when the molecule contains sensitive
functional groups that would react with the strong acid. This is usually used with primary
and secondary alcohols since, as we have seen, tertiary alcohols react easily at low
temperatures with hydrochloric acid. Often a weak base such as potassium carbonate
(K2CO3) or pyridine (C5H5N) is used to neutralize the HCl that is produced.
13
CH. 4
O
ROH
+
Cl
heat
RCl
S Cl
SO2
+
HCl
+
For example:
O
OH
CH3CH2
CH
CH2CH3
+
Cl
Cl
heat
S Cl
K2CO3
CH3CH2
CH
CH2CH3
(2) Phosphorus Tribromide:
A mild way to convert alcohols to alkyl bromides involves the treatment of the alcohol
with phosphorus tribromide, PBr3. One equivalent of phosphorus tribromide will convert
three molecules of the alcohol.
3 R OH
+
PBr3
3 R Br
+ H3PO3
phosphorus acid
Halogenation of Alkanes
A hydrogen atom on a simple alkane can be replaced with a halogen atom.
R H
+
R X
X2
+
H X
X = Br, Cl
The order of reactivity for the halogens is: F2 > Cl2 > Br2 >I2.
Fluorine is too reactive. It is a strong oxidizing reagent and the reaction is strongly
exothermic and difficult to control. Iodine is unreactive. The reaction is endothermic and
not useful. Therefore, the only useful halogenations occur with bromine and chlorine.
This is not a nucleophilic substitution reaction like the reactions we just studied with
alcohols and halogen acids. The C-H bond is not polarized and H:- is not a leaving group.
This is an example of a free radical reaction.
Chlorination of methane:
The reaction can give up to four products due to over reaction. On a large industrial
scale the products can be separated by distillation.
H C
H
H +
Cl
Cl
heat
uV light
H C Cl
+
H C Cl
Cl
H
14
Cl
Cl
H
H
H
+
H C Cl
Cl
+
Cl
C Cl
Cl
CH. 4
This is an example of a free radical reaction. A radical is a species that has a free unpaired
electron. There are several examples of stable radicals, the most common of which is
molecular oxygen. Nitrogen dioxide, NO2, is also a stable radical and the important cellcell signaling molecule, nitrogen monoxide, NO, is a radical as well.
N O
O O
O N O
molecular oxygen
nitrogen dioxide
nitrogen monoxide
Carbon free radicals are usually much less stable. They are missing one electron and are
generally very reactive intermediates that exist only for a short time. Carbon radical
stability parallels that of carbocation stability. Since carbon radicals are missing one
electron and so are electron deficient species, they are stabilized by electron donating
substituents such as alkyl groups. Therefore, tertiary radicals are the most stable and
methyl radicals are the least stable. Note also that the carbon radical has no charge.
R C
R
R
>
R C
H
R C
>
H
>
H
R
tertiary radical
most stable
H C
H
H
methyl radical
least stable
The methyl radical is planar, trigonal with the hybridization very nearly sp2. The t-butyl
radical is slightly pyramidal but still flattened and closer to sp2 than sp3 hybridization.
H
H
C
H
Alkyl groups stabilize the radical just like they stabilize carbocations. Evidence for this
stabilization can be seen in terms of the Bond Dissociation Enthalpy (BDE). For
homolytic cleavage each atom gets one of the two electrons in the bond between the two
atoms. The heat change, ΔH, in this reaction is the bond dissociation energy. Note that by
convention in radical reactions, we use a single-headed arrow and that the product(s) of
hemolytic cleavage are uncharged species.
X
Y
X
+
Y
ΔH = BDE
In homolytic cleavage, each
atom gets one electron.
Note that no charge develops and that the arrows are single
headed arrows, showing that only one electron is moving.
Recall that in heterolytic cleavage one atom gets both electrons.
15
CH. 4
heterolytic cleavage
X
+
X
Y
One atom gets both electrons and charges develop
in the product(s).
Y
It is generally the case that hemolytic cleavage requires less energy than heterolytic
cleavage.
For carbon atoms, the bond dissociation enthalpies are known and can be found in tables.
Compare the bond dissociation energy of a primary versus a secondary carbon.
CH3CH2CH2
CH3
CH
H
CH3
CH3CH2CH2 +
CH3
H
C
CH3
ΔH = +410 KJ/mol
H
+
ΔH = +397 KJ/mol
H
H
We see the cleavage of a secondary carbon is 13 KJ/mol more favorable, implying that the
secondary carbon radical is more stable than the primary by 13 KJ/mol.
We can use the bond dissociation enthalpies to compare the stabilities of a primary versus
a tertiary carbon radical.
CH3
CH
CH3
CH2 H
CH
H
C
CH3
H
ΔH = +410 KJ/mol
CH3
CH3
CH3
+
CH2
CH3
CH3
C
CH3
+
H
ΔH = +380 KJ/mol
CH3
We see that the tertiary radical is more stable than the primary radical by 30 KJ/mol and
more stable than the secondary radical by 17 KJ/mol.
Halogenation of alkanes occurs by means of a radical chain reaction. This is shown below
for the chlorination of methane.
16
CH. 4
Inititation:
Cl
Cl
heat
2 Cl
or UV light
H
(1)
Propagation:
+
Cl
H C
H
H
H
H
(2)
H C
+
+
H
Cl
Cl
H C Cl
Cl
+
Cl
H
H
The first step is the initiation step. This is the hemolytic cleavage of the chlorine molecule
into two chlorine radicals. This bond cleavage occurs due to heating or to ultra-violet
(UV) light. Generally, this step occurs only once.
The next two steps occur over and over again. These are the propagation steps. In each
propagation step a new radical is formed which then goes on to react to generate a new
radical, which continues the chain reaction.
We can also get dichloromethane, tricholormethane and even tetrachloromethane. This
occurs when the initial chloromethane continues to react and is more likely to occur toward
the end of the reaction as the concentration of starting methane begins to decrease and the
concentration of chloromethane increases.
H
step 1
H C
H
H +
Cl
Cl
H C
Cl
H Cl
Cl
H
step 2
H C
H
+ Cl
Cl
H C Cl
+
Cl
Cl
The chain reaction can be broken by termination steps. Termination steps are those in
which two radicals come together to react with each other. This does not generate a new
radical and so the chain is broken. Some termination steps for the chlorination of methane
are:
17
CH. 4
H
H C
H H
H
+
C
H
H C C
H
H H
H
H
H
H C
H C Cl
Cl
+
H
H
Cl
H
Cl
Cl
+
Cl
We can use bond dissociation enthalpies to calculate the overall in enthalpy, ΔH, for the
reaction. In general, the overall change in enthalpy, ΔH, is equal to the sum of the bond
dissociation enthalpies of the bonds broken plus the sum of the bond dissociation
enthalpies of the bonds formed. We will use the convention that a bond broken is a
positive number, since energy is required to break a bond, and a bond formed is a negative
number, since heat is given off when a bond is formed.
ΔH = ΣBDEbonds broken + ΣBDEbonds broken
Where bonds broken = (+)
bonds formed = (-)
Typically, we ignore the initiation step since it happens only once, where as the
propagation steps can happen thousands of times.
H
step 1
Cl
+
H C
H
H
H C
H
+
H
step 2
H C
H
ΔH = 435 - 431 = +4 KJ/mol
form H-Cl bond
break methyl C-H bond
+435 KJ/mol
H
H Cl
-431 KJ/mol
H
+
Cl
Cl
break Cl-Cl bond
+242 KJ/mol
H C Cl
+
Cl
H
form C-Cl bond
ΔH = -107 KJ/mol
ΔHrxn = -107 + 4 = -103 KJ/mol
-349 KJ/mol
To find the overall enthalpy of reaction, add step 1 and step 2 to get (-107 KJ/mol + 4
KJ/mol) -103 KJ/mol. The reaction is exothermic and therefore favorable.
If the same calculation is done for fluorination, we find that ΔH is -426 KJ/mol. This is a
very large amount of energy that is released for each step. As more molecules react, more
energy is released, the reaction mixture heats up and reacts even faster. It quickly becomes
uncontrollable. Free radical fluorinations are dangerously explosive.
18
CH. 4
If we do the same calculation for bromination, we find that the ΔH = -30 KJ/mol,
considerably less than that for chlorination and much less than that for fluorination. As we
will see, free radical bromination reactions are much more selective than free radical
chlorinations.
For iodination, ΔH = +54 KJ/mol. In the case of iodination, the reaction is unfavorable and
free radical iodination reactions are not feasible.
Halogenation of Higher Alkanes
If the molecule is symmetrical, we get only one monochloro- product.
Cl
Cl2, UV light
But if the molecule is not symmetrical, a mixture of products can form. We see that the
major product is the one that results from the formation of the secondary radical.
Cl
CH3
CH2 CH2 CH3 Cl2, UV light
CH3
CH2 CH2 CH2 Cl
+
CH3
CH2 CH
CH3
72%
28%
Since there are six primary hydrogens and only four secondary hydrogens, we would
expect a mixture of 60% (6 of 10) 1-chlorobutane and 40% (4 of 10) of the 2-chlorobutane.
But we see 72% of 2-chlorobutane. This tells us that the secondary radical is formed faster
than the primary radical by a factor of 72/28 x 6/4 = 3.9/1.
Bromination is even more selective. Tertiary hydrogens are abstracted selectively. For
example, compare the product rations of free radical chlorination and free radical
bromination of 2-methylpropane.
H
CH3
C CH3
CH3
H
Cl
Cl2
UV
CH3
CH3
C CH3
CH3
37%
C CH2 Cl
CH3
63%
Here we see that there are nine primary hydrogens and only one tertiary hydrogen, so if
primary hydrogens and tertiary hydrogens reacted at the same rate we would expect a
product ratio of 10% 2-chloro-2-methyl propane and 90% 1-chloro-2-methylpropane. So
clearly, the tertiary position reacts faster by a ratio of 37/63 x 9/1 = 5.2.
19
CH. 4
For bromination the results are very striking: bromination is much more selective for
removal of the tertiary hydrogen than chlorination. Doing the same calculation that we did
before, we see that in the bromination reaction, the tertiary hydrogen reacts 99/1 x 9/1 =
891 times faster.
CH3
H
Br
H
Br2
C CH3
UV
CH3
CH3
CH3
C CH3
CH3
C CH2 Br
CH3
>99%
<1%
The reason for this is that the first step in the chlorination reaction is quite exothermic
(early transition state) for both primary and tertiary products while the first step in the
bromination reaction is endothermic (late transition state) for both products.
H
CH3
C CH3
H
+
Cl
CH3
CH3
break 1° C-H +422
C CH2
+
CH3
H Cl
ΔH = -9 KJ/mol
form H-Cl -431
H
CH3
C CH3
+
Cl
CH3
CH3
break 3° C-H +400
C CH3
+
CH3
H
CH3
C CH3
H Cl
ΔH = -31 KJ/mol
form H-Cl -431
H
+
Br
CH3
CH3
break 1° C-H +422
C CH2
+
CH3
H Br
ΔH = +56 KJ/mol
form HBr = -366
H
CH3
C CH3
+
CH3
break 3° C-H +400
Br
CH3
C CH3
CH3
+
H Br
ΔH = +34 KJ/mol
form HBr = -366
Hammond’s postulate tells us that we have a late transition state in an endothermic
reaction so that there is a large difference in energy between the transition states leading to
the primary radical and the tertiary radical.
20
CH. 4
H
CH3 C CH2
CH3
Potential Energy
CH3 C CH3
CH3
Br
+ H(CH3)3
Endothermic reaction with a late transition state so that
there is a lot of radical character in the transitions state
and the difference in stability of a primary versus tertiary
radical is important.
Reaction Progress
Potential Energy
The chlorination reaction is exothermic. Therefore there is an early transition state with
little radical character and the difference in energy between the transition states is small.
In the exothermic reation there is a relatively small difference
in energy in the transition states since bond breaking is not
very far advanced and there is little radical character.
Cl
+ H(CH3)3
H
CH3 C CH2
CH3
CH3 C CH3
CH3
Reaction Progress
21