Markovnikov's Rule:
In additions of HX to unsymmetrical alkenes, the H+ of HX goes to the
double-bonded carbon that already has the greatest number of hydrogens
Animated Tutorial
Examples
Symmetrical Alkenes
Ethene (ethylene) is a symmetrical molecule.
H |
H |
C
=
C
| H
| H
Each carbon atom is covalently bonded to two hydrogen atoms.
Ethene reacts with hydrogen bromide to produce bromoethane.
ethene
+
hydrogen bromide
→
bromoethane
C2H4
+
HBr
→
C2H5Br
H |
H |
C
=
C
| H
| H
+
H-Br
→
H |
Br |
H-
C
-
C
-H
| H
| H
Bromoethane is the only product possible for this reaction.
Unsymmetrical Alkenes
Propene is an unsymmetrical molecule.
H |
H |
C(1)
=
C(2)
-
C
-H
| H
| H
| H
C(1) is covalently bonded to two hydrogen atoms.
C(2) is covalently bonded to one hydrogen atom.
When propene reacts with a halogen halide such as hydrogen bromide, two structural isomers are possible products:
If the bromine atom bonds with C(1), the product is 1-bromopropane.
If the bromine atom bonds with C(2), the product is 2-bromopropane.
propene
+
hydrogen bromide
→
1-bromopropane
+
2-bromopropane
H |
H |
C
=
C
-
C
-H
| H
| H
| H
+
H-Br
→
Br |
H |
H |
H-
C
-
C
-
C
-H
| H
| H
| H
+
H |
Br |
H |
H-
C
-
C
-
C
-H
| H
| H
| H
Markovnikov's Rule states that the hydrogen adds to the carbon atom of the double bond which is already bonded to more hydrogen atoms, that is, the hydrogen atom adds to C(1) and the bromine atom adds to C(2) so the major product of the reaction will be 2-bromopropane.
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