"In the addition of an unsymmetrical reagent to an alkene, the less electronegative part of the reagent adds to the carbon atom in the double bond with the most hydrogen atoms already attached"
When is Markovnikov's rule used?
During the addition of an unsymmetrical reagent to an alkene translates as adding a compound with more than one type of atom in it to a compound with a pair of carbon atoms joined by a double bond. Typically the molecule being added will be a Hydrogen Halide: that is H-X where X is a halogen (group 7 element such as Chlorine), and it is for this hydrohalogenation that markovnikov first observed his rule.
An additional complication is that for the application of Markovnikov's rule to actually have any effect, the alkene must have 3 or more carbon atoms present (i.e. propene and above undergo markovnikov addition). An Alkene is not possible with only a single carbon atom; with just two then both theoretical products are identical so the rule makes no difference.
The rule is used when you wish to determine which of the possible products arises during the addition across the double bond. If this bond is broken, both of the Carbon atoms are looking for another atom to bond with. The attacking species supplies two different ones, and crucially the positioning of them within the product will alter that product's chemistry.
Consider for example the addition of Hydrogen Bromide (HBr) to Propene (CH3CHCH2)
One possible product is the primary haloalkane:
H H H H H
H C - C = C + HBR --> H C - C - C - Br
H H H H H H
NB: C-H bonds represented as CH for ease of ASCII.
The alternative product is the secondary haloalkane
H H H Br H
H C - C = C + HBR --> H C - C - C H
H H H H H H
Without Markovnikov's rule it would be impossible to state which would actually arise. A reasonable guess would be that each product is produced in equal volumes. In fact, in this example the secondary halogenoalkane
is overwhelmingly favoured.
How is Markovnikov's rule used?
Having identified that an addition reaction across an alkene double bond is occuring, it is necessary to identify the less electronegative part of the molecule.
Some common Electronegativities
Note that Electronegativity has arbitrary
units and is a combination of many factors: it simply allows for comparisons of the relative
standings of the elements. Note also that the higher the electronegativity value, the more negative the element will be in a compound due to its greater ability to attract bond electrons
It can also be seen from this table that typically Hydrogen will be the least electronegative part of the attacking species: certainly in all A-level
work I've done this has been the case.
Having identified the least electronegative part, count the number of hydrogens bonded to each of the double-bonded carbons in the organic
reagent. Whichever has the most gets the least electronegative half of the attacking compound, and hence the other carbon atom gets the more electronegative part. A quick shortcut for A-level is therefore that in general, the carbon atom with the most hydrogen atoms gets another one
- whilst it's not chemically accurate to say that like is seeking like or that the hydrogens try to group together, it's an easier way to remember what happens than applying the rule in full.
Applying the rule to our example above, carbon atom 2 (the middle one) only has a single hydrogen whilst carbon atom 3 (the far-right one) has 2. Therefore the Bromine adds to the middle carbon and the hydrogen adds to the end one. Note that although the first carbon atom has 3 hydrogens, it is not part of a double bond so the hydrogen cannot be added there.
Why does Markovnikov's rule apply?
For a reaction to proceed effectively the intermediate stages must be as stable as possible. The addition to an alkene is not a single step but rather a mechanism with two steps. Hence there is an intermediate product. The stability of this product depends on how the first step proceeds: in other words, to which carbon atom the hydrogen adds to. If one intermediate is significantly more stable than the other, then the reaction will tend to proceed via that intermediate. The major product is then the one which corresponds to that intermediate.
In our example reaction, the Hydrogen atom is less electronegative. This will give it a δ+ charge. As a result, it will be attracted to the region of negative charge created by the π bond present in the Carbon-Carbon double bond. By attaching to one of these carbons, the π bond is broken leaving a single bond between the carbons and thus creating a positive charge on the carbon that didn't bond to hydrogen, as it is now effectively missing a bond.
It is this carbon with a positive charge, known as a carbocation (cations are species that are attracted to cathodes; as cathodes are negative it follows that cations are positive), which is the intermediate. It's stability is greatly influenced by its ability to balance this charge or disperse it through the structure.
Consider the two carbocations possible from our earlier example. If hydrogen adds to the second carbon, we get this:
H H H
H C - C - C +
H H H
Whereas addition to the third carbon (as Markovnikov's rule suggests is best) gives this:
H H H
H C - C - C H
H + H
Why then is the second carbocation more stable? An important property of the methyl
- is that it is electron donating. Thus the second carbocation has two groups directly adjacent to the positive carbon which can supply electrons- in other words, reduce the positive nature of the atom. This makes it more stable than the first carbocation shown, as although this has a methyl group it is not directly connected to the positive carbon atom, and furthermore there is only one.
What Markovnikov's Rule is in effect saying is that when there is a choice between two products, the one formed from the more stable intermediate is favoured.
In practical terms, this product is the one formed when the less electronegative part (usually hydrogen) adds to the carbon which already has the most hydrogens.
When doesn't Markovnikov's rule apply?
The rule works for hydrohalogenation as shown, and works for acid-catalysed hydration and hypohalous adition too.
However, an exception exists: in the hydrobromination of an alkene in the presence of a peroxide the predominant product is the opposite to that predicted by Markovnikov addition, as in this case there is a different reaction path involving free radicals rather than carbocations.
It should also be noted that Markovnikov's rule is a practical rule of thumb rather than an absolute law. Some of the unfavoured product will form, as despite the carbocation being less stable it can still arise given sufficiently energetic particles and/or supporting conditions. Thus a pure product is not usually obtained, like most organic reactions.
This is largely a re-write as my original node was rather difficult to comprehend. It is based on my own study of Chemistry up to A-level standard and so if further exceptions or conditions exist then they are only absent here because they were not part of my course. Do message me with further details if you have them- but bear in mind that I haven't studied chemistry for over a year now!