Download 4.14 Halogenation of Alkanes RH + X2 → RX + HX RH + X2 → RX +

Energetics
RH + X2 → RX + HX
4.14
Halogenation of Alkanes
RH + X2 → RX + HX
explosive for F2
exothermic for Cl2 and Br2
endothermic for I2
Chlorination of Methane
carried out at high temperature (400 °C)
4.15
Chlorination of Methane
CH4 + Cl2 → CH3Cl + HCl
CH3Cl + Cl2 → CH2Cl2 + HCl
CH2Cl2 + Cl2 → CHCl3 + HCl
CHCl 3 + Cl2 → CCl4 + HCl
Free Radicals
4.16
Structure and Stability of Free
Radicals
contain unpaired electrons
..
..
: O O:
Examples: O2
.
.
..
.
: N O:
NO
..
: Cl .
Cl
..
Figure 4.17 Structure of Methyl Radical
Alkyl Radicals
R
.
C
R
R
Most free radicals in which carbon bears
the unpaired electron are too unstable to be
isolated.
Methyl radical is planar, which suggests that
carbon is sp2 hybridized and that the unpaired
electron is in a p orbital.
Alkyl radicals are classified as primary,
secondary, or tertiary in the same way that
carbocations are.
Alkyl Radicals
Alkyl Radicals
H
R
.
C
.
C
R
R
H 3C
H
less stable
than
H
Methyl radical
The order of stability of free radicals is the
same as for carbocations.
H 3C
less stable
than
.
C
CH3
.
C
H
Ethyl radical
(primary)
H 3C
A chemical bond can be broken in two
different ways—
ways—heterolytically or homolytically.
tert-Butyl
tert-Butyl radical
(tertiary)
In a homolytic bond cleavage, the two electrons in
the bond are divided equally between the two atoms.
One electron goes with one atom, the second with
the other atom.
In a heterolytic cleavage, one atom retains both
electrons.
Heterolytic
CH3
CH3
Homolytic
Alkyl Radicals
By "bond strength" we mean the energy
required to break a covalent bond.
.
C
less stable
than
H
Isopropyl radical
(secondary)
The order of stability of free radicals can be
determined by measuring bond strengths.
H
–
+
Homolytic
The species formed by a homolytic bond cleavage
of a neutral molecule are free radicals. Therefore,
measure enthalpy cost of homolytic bond cleavage to
gain information about stability of free radicals.
The more stable the free-radical products, the weaker
the bond, and the lower the bond-dissociation energy.
Measures of Free Radical Stability
Bond-dissociation enthalpy measurements tell
us that isopropyl radical is 13 kJ/mol more
stable than propyl.
.
CH3CH2CH2
.
+
CH3CHCH 3
.
H
+
410
397
H.
CH3CH2CH3
Measures of Free Radical Stability
Bond-dissociation enthalpy measurements tell
us that tert-butyl
tert-butyl radical is 30 kJ/mol more
stable than isobutyl.
.
(CH3)2CHCH 2
+
(CH3)3C .
H.
+
410
380
H.
4.17
Mechanism of Chlorination of
Methane
(CH3)3CH
Mechanism of Chlorination of Methane
Free-radical chain mechanism.
Initiation step: (Light or Heat is Necessary)
.. ..
..
..
: Cl: Cl
:
: Cl . + . Cl
.. :
.. ..
..
The initiation step "gets the reaction going"
by producing free radicals—
radicals—chlorine atoms
from chlorine molecules in this case.
Initiation step is followed by propagation
steps. Each propagation step consumes one
free radical but generates another one.
Mechanism of Chlorination of Methane
First propagation step:
H 3C : H
+
..
. Cl:
Cl:
..
Second propagation step:
.. ..
:
H3C . + : Cl: Cl
.. ..
H 3C .
..
Cl
+ H : Cl:
.. :
..
..
Cl:
H3C: Cl:
Cl: + . Cl:
..
..
Mechanism of Chlorination of Methane
First propagation step:
H 3C : H
+
..
. Cl:
Cl:
..
Second propagation step:
.. ..
:
H3C . + : Cl: Cl
.. ..
.. ..
: Cl :
H3C : H + : Cl
.. ..
Almost all of the product is formed by repetitive
cycles of the two propagation steps.
First propagation step:
H 3C .
..
Cl
+ H : Cl:
.. :
..
..
Cl:
H3C: Cl:
Cl: + . Cl:
..
..
..
..
H3C: Cl : + H : Cl:
Cl
..
.. :
Termination Steps
stop chain reaction by consuming free radicals
..
H3C . + . Cl:
Cl:
..
..
H3C: Cl:
Cl:
..
hardly any product is formed by termination step
because concentration of free radicals at any
instant is extremely low
H 3C : H
+
..
. Cl:
Cl:
..
H 3C .
Second propagation step:
.. ..
:
H3C . + : Cl: Cl
.. ..
..
Cl
+ H : Cl:
.. :
..
..
Cl:
H3C: Cl:
Cl: + . Cl:
..
..
..
..
H3C: Cl : + H : Cl:
Cl
..
.. :
.. ..
: Cl :
H3C : H + : Cl
.. ..
Question 11
The step shown below is a _____________
step of the free-radical chlorination of
chloromethane.
A) initiation
B) propagation
C) chain-terminating
D) bond cleavage
Question 16
For the free-radical reaction below, light is
involved in which of the following reaction
steps?
4.18
Halogenation of Higher Alkanes
A) Initiation only
B) Propagation only
C) Termination only
D) Initiation and propagation
Chlorination of Alkanes
can be used to prepare alkyl chlorides
from alkanes in which all of the hydrogens
are equivalent to one another
420°
420°C
CH3CH3 + Cl2
CH3CH2Cl + HCl
(78%)
hν
+ Cl2
Cl + HCl
Chlorination of Alkanes
Major limitation:
Chlorination gives every possible
monochloride derived from original carbon
skeleton.
Not much difference in reactivity of
different hydrogens in molecule.
(73%)
Example
Chlorination of butane gives a mixture of
1-chlorobutane and 2-chlorobutane.
Percentage of Product that Results from
Substitution of Indicated Hydrogen if
Every Collision with Chlorine Atoms is
Productive
(28%)
10%
CH3CH2CH2CH2Cl
CH3CH2CH2CH3
Cl2
10%
10%
10%
hν
CH3CHCH 2CH3
(72%)
10%
Cl
18%
4.6%
4.6%
18%
4.6%
4.6%
4.6%
18%
18%
4.6
=1
4.6
18
4.6
4.6%
10%
Relative Rates of Hydrogen Atom Abstraction
divide by 4.6
18%
10%
10%
10%
Percentage of Product that Actually Results
from Replacement of Indicated Hydrogen
4.6%
10%
= 3.9
A secondary hydrogen is abstracted 3.9 times
faster than a primary hydrogen by a chlorine
atom.
Similarly, chlorination of 2-methylbutane
gives a mixture of isobutyl chloride and
tert-butyl
tert-butyl chloride
(63%)
Question 10
CH3
CH3CCH2Cl
CH3
CH3CCH3
H
Cl2
A) two
B) three
hν
H
How many monochlorination products do you
expect to obtain from the chlorination of
2-methylbutane?
CH3
(37%)
C) four
D) five
CH3CCH3
Cl
Percentage of Product that Results from
Replacement of Indicated Hydrogen
Answer 10
How many monochlorination products do you
expect to obtain from the chlorination of
2-methylbutane?
A) two
7.0%
B) three
C) four
37%
D) five
Relative Rates of Hydrogen Atom Abstraction
Selectivity of Free-radical Halogenation
R3CH
divide by 7
7.0
7
=1
37
7
chlorination:
= 5.3
bromination:
5
1640
> R2CH2 >
RCH3
4
1
82
1
Chlorination of an alkane gives a mixture of
every possible isomer having the same skeleton
as the starting alkane. Useful for synthesis only
when all hydrogens in a molecule are equivalent.
A tertiary hydrogen is abstracted 5.3 times faster
than a primary hydrogen by a chlorine atom.
Bromination is highly regioselective for
substitution of tertiary hydrogens. Major synthetic
application is in synthesis of tertiary alkyl
bromides.
Synthetic Application of Chlorination of an Alkane
Cl
Cl2
hν
(64%)
Chlorination is useful for synthesis only when
all of the hydrogens in a molecule are equivalent.
Question 20
An alkane with a molecular formula of C8H18
reacts with Cl2 in the presence of light and
heat to give a single monochloride C8H17Cl.
What is the most reasonable structure for the
starting alkane?
alkane?
A) CH3CH2CH2CH2CH2CH2CH2CH3
B) (CH3CH2)2CHCH 2CH2CH3
C) (CH3)2CHCH 2CH2CH(CH 3)2
D) (CH3)3CC(CH3)3
Synthetic Application of Bromination of an Alkane
H
CH3CCH2CH2CH3
CH3
Br2
hν
Br
CH3CCH2CH2CH3
CH3
(76%)
(A) Barbamide, a cyanobacterial peptide
containing a trichloromethyl group and (B)
dysidenin, a barbamide-related compound
isolated from a sponge-cyanobacterial
association
Question 19
Which of the following best describes a mechanistic feature of
the free-radical bromination (Br2, light) of 2methylpropane?
A) The initiation step involves cleavage of a C-H bond.
B) The free-radical (CH3)3C· is produced in one
propagation step and reacts with Br 2 in another.
C) The reaction is characterized by the homolytic
cleavage of the C-Br bond.
D) The reaction is concerted; i.e., it occurs in a single step.
Bromination is highly selective for substitution
of tertiary hydrogens.
Major synthetic application is in synthesis of
tertiary alkyl bromides.