Wurtz Reaction
Wurtz Reaction
An ethereal
solution of an alkyl halide (preferably the bromide or iodide) is treated with
sodium, e.g.,
R1X + R2X + 2Na → R1 – R2 + 2NaX
when we do not wish to specify a particular alkyl group, we use the symbol R. When we deal with two unspecified alkyl groups which may, or may not, be the same, we can indicate this by R1 and R2 : also, when dealing with compounds containing a halogen atom, and we do not wish to specify the halogen, we can indicate the presence of the unspecified halogen atom by means of X.
when we do not wish to specify a particular alkyl group, we use the symbol R. When we deal with two unspecified alkyl groups which may, or may not, be the same, we can indicate this by R1 and R2 : also, when dealing with compounds containing a halogen atom, and we do not wish to specify the halogen, we can indicate the presence of the unspecified halogen atom by means of X.
Consideration
of the equation given above shows that in addition to the desired alkane R1
– R2, there will also be present the alkanes R1 – R2
and R2 – R2. Unsaturated hydrocarbons are also obtained. Obviously,
then, the best yield of an alkane will be obtained when R1 and R2
are the same, i.e., when the alkane contains an even number of carbon atoms and
is symmetrical. It has been found that the Wurts reaction gives good yields
only for ‘even carbon’ alkanes of high molecular weight, and that the reaction
generally fails with tertiary alkyl halides.
Sodium is used in the Wurts reaction. Other metals, however, in a finely divided state, may also be used, e.g., Ag, Cu.
Two
mechanisms have been suggested for the Wurts reaction.
- The intermediate formation of an organo-metallic compound, e.g., the formation of n-butane from ethyl bromide.
C2H5 –Br +2Na. → C2H5-
Na+ + NaBr
C2H5-
Na+ + C2H5Br → C2H5
– C2H5 + NaBr
- The international formation of free redicals, e.g.,
C2H5 –Br +2Na.
→ C2H5 + NaBr
C2H5 + C2H5 → C2H5 – C2H5
One of the
properties of free radicals is disproportionation, i.e., intermolecular hydrogenation, one
molecule acquiring hydrogen at the expense of the other, e.g.,
C2H5 + C2H5 → C2H6 – C2H4
This would
account for the presence of ethane and ethylene in the products. According to
Morton, ethane and ethylene may be produced as follows :
This is a
bimolecular elimination mechanism. On the other hand, the mechanism of alkane
formation is generally accepted as usually being a bimolecular nucleophilic
substitution:
The
free-radical mechanism, however, has been shown to operate in some cases
(Bryce-Smith, 1963).
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