MORISSON   Organic Chemistry

MORISSON Organic Chemistry


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As with methane, iodination does not take place at all.
Depending upon which nydrogen atom is replaced, any of a number of iso-
meric products can be formed from a single alkane. Ethane can yield only one halo-
ethane; propane, /z-butane, and isobutane can yield two isomers each; -pentane
can yield three isomers, and isopentane, four isomers. Experiment has shown
that on halogenation an alkane yields a mixture of all possible isomeric products,
indicating that all hydrogen atoms are susceptible to replacement. For example,
for chlorination :
CH3CH3 -- .
"250*'
CH3CH2 Cl
Ethane
'
b.p. 13
Chloroethane
Ethyl chloride
CH3CH2CH3 l{^\y > CH 3CH2CH2-C1 and CH3CHCH3
Propane
'
b.p. 47 L
1-Chloropropane ^
W
.Propylchloride
2^hloropropane
Isopropyl chloride
55%
CH3CH2CH2CH3
ligh^2So> CH3CH2CH2CH2 Cl and CH3CH2CHCH3
n-Butane
'
b.p. 78.5 L
1-Chlorobutane e
-Butylchloride **!
870
wc-Bulyl chloride
96 ALKANES CHAP. 3
CH 3
CH 3CHCH 3
Isobutane
CH 3
CH 3CHCH 2
b.p. 69
l-Chloro-2-
methylpropane
Isobutyl chloride
,
'
CI and
CH 3
CH 3CCH 3
I
C
*P* 5I
2-Chloro-2-
methylpropane
/?/7-Butyl chloride
36%
Bromination gives the corresponding bromides but in different proportions:
CH3CH 3
.JTur
' CH,CH2Br
Ethane
CH 3CH2CH 3 j 27,> CH 3CH 2CH 2Br and CH 3CHCH 3
Propane
'
3/ 1
97%
CH 3CH 2CH2CH 3
Il8hf ,370 > CH 3CH 2CH 2CH 2Br and CH 3CH 2CHCH3
/r-Butane
'
2% 1DT
98%
CH 3 CH 3 CH,
CH3CHCH 3 hgh
*r2
127o
> CH 3CHCH 2Br and CH 3CCH 3
Isobutane
'
trace I
ur
over 99%
Problem 3.12 Draw the structures of: (a) the three monochioro derivatives
of ff-pentune; (h) the four monochioro derivatives of isopentane.
Although both chlorination and bromination yield mixtures of isomers, the
results given above show that the relative amounts of the various isomers differ
markedly depending upon the halogen used. Chlorination gives mixtures in which
no isomer greatly predominates; in bromination, by contrast, one isomer may
predominate to such an extent as to be almost the only product, making up 97-
99% of the total mixture. In bromination, there is a high degree of selectivity as
to which hydrogen atoms are to be replaced. (As we shall see in Sec. 3.28, this
characteristic of bromination is due to the relatively low reactivity of bromine
atoms, ancl is an example of a general relationship between reactivity and selec-
tivity.)
Chlorination of an alkane is not usually suitable for the laboratory prepara-
tion of an alkyl chloride; any one product is necessarily formed in low yield, and
is- difficult to separate from its isomers, whose boiling points are seldom far from
its own. Bromination, on the other hand, often gives a nearly pure alkyl bromide
in high yield. As we shall see, it is possible to predict just which isomer will pre-
dominate; if this product is the one desired, direct bromination could be a feasible
synthetic route.
On an industrial scale, chlorination of alkanes is important. For many
SEC. 3.20 MECHANISM OF HALOGENATION 97
purposes, for example, use as a solvent, a mixture of isomers is just as suitable as,
and much cheaper than, a pure compound. It may be even worthwhile^ when
necessary, to separate a mixture of isomers if each isomer can then be marketed.
Problem 3.13 How do you account for the fact that not only bromination but
also chlorination is a feasible laboratory route to a neopentyl halidc,
3.20 Mechanism of halogenation
Halogenation of alkanes proceeds by the same mechanism as halogenaiion
of methane:
Chain-initiating step
Chain-propagating steps
then (2), (3), (2), (3), etc., until finally a chain is terminated (Sec. 2.13
N
,
A halogen atom abstracts hydrogen from the alkane (RH) to form an alkyl radical
(R-). The radical in turn abstracts a halogen atom from a halogen molecule to
yield the alkyl halide (RX).
Which alkyl halide is obtained depends upon which alkyl radical is formed.
CH4 -
Methane
CH 3CH 3 -
Ethane
abstraction
CH3CH2CH3
Propane
CH 3 *
Methyl
radical
CH 3CH 2 -
Ethyl
radical
CH 3CH 2CH 2 -
//-Propyl
radical
CH 3X
Methyl
halide
CH 3CH2X
Ethyl
halide
abstraction
CH 3CHCH 3
Isopropyl
radical
CH 3CH 2CH 2X
/?-Prop>l
halidc
CH 3CHCH 3
X
Isopropyl
haltde
This in turn depends upon the alkane and which hydrogen atom is abstracted from
it. For example, w-propyl halide is obtained from a w-propyl radical, formed
from propane by abstraction of a primary hydrogen; isopropyl halide is obtained
from an isopropyl radical, formed by abstraction of a secondary hydrogen.
How fast an alkyl halide is formed depends upon how fast the alkyl radical is
formed. Here also, as was the case with methane (Sec. 2.20), of the two chain-
propagating steps, step (2) is more difficult than step (3), and hence controls the
rate of overall reaction. Formation of the alkyl radical is difficult, but once formed
the radical is readily converted into the alkyl halide (see Fig. 3.5).
98 ALKANES CHAP. 3
Difficult step
RCI +C1-
Progress of reaction >
Figure 3.5. Potential energy changes during progress of reaction:
chlorination of an alkane. Formation of radical is rate-controlling step.
3.21 Orientation of halogenation
With this background let us turn to the problem of orientation; that is, let us
examine the factors that determine where in a molecule reaction is most likely to
occur. It is a problem that we shall encounter again and again, whenever we study
a compound that offers more than one reactive site to attack by a reagent. It is an
important problem, because orientation determines what product we obtain.
As an example let us take chlorination of propane. The relative amounts of
/?-propyl chloride and isopropyl chloride obtained depend upon the relative rates
at which w-propyl radicals and isopropyl radicals are formed. If, say, isopropyl
radicals are formed faster, then isopropyl chloride will be formed faster, and will
make up a larger fraction of the product. As we can see, w-propyl radicals are
formed by abstraction of primary hydrogens, and isopropyl radicals by abstraction
of secondary hydrogens.
H H H H H H
abstraction 111 IIIQf'H
> H C C~C Cli> H C-C C CI
H H H
C C C H
I I
H H
Propane
H H H
/j-Propyl
radical
H H H
H H H
/i-Propyl
chloride
H H H
H-i-i-i-H _a H-i-i-i-H
'-
'
-'-H H
Isopropyl Isopronyl
SEC. 3.21 ORIENTATION OF HALOGENATION 99
Thus orientation is determined by the relative rates of competing reactions. In
this case we are comparing the rate of abstraction of primary hydrogens with the
rate of abstraction of secondary hydrogens. What are the factors that determine
the rates of these two reactions, and in which of these factors may the two reac-
tions differ?
First of all, there is the collision frequency. This must be the same for the two
reactions, since both involve collisions of the same particles : a propane molecule
and a chlorine atom.
Next, there is the probability factor. If a primary hydrogen is to be abstracted,
the propane molecule must be so oriented at the time of collision that the chlorine
atom strikes a primary hydrogen ; if a secondary hydrogen is to be abstracted, the
propane must be so oriented that the chlorine collides with a secondary hydrogen.
Since there are six primary hydrogens and only two secondary hydrogens in each
molecule, we might estimate that the probability factor favors abstraction of
primary hydrogens by the ratio of 6:2, or 3: 1.
Considering only collision frequency and our guess about probability factors,
we predict that chlorination of propane would yield w-propyl chloride and iso-
propyl chloride in the ratio of 3 : 1 . As shown on page 95, however, the two
chlorides are formed in roughly equal amounts, that is, in the ratio