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Additions to Alkenes: Regiochemistry

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This page covers the regiochemistry of HX additions across C=C double bonds. The reasons for the regioselectivity seen in these reactions will be discussed in terms of the reaction mechanism. The purpose of this page is to familiarize the student with the application of Markovnikov's Rule, and to help the student distinguish between Markovnikov and non-Markovnikov regioisomers as the products of addition reactions occurring under different conditions. (For a discussion of the stereochemical features of these reactions, click on "Additions to alkenes: stereochemistry," below)

Strongly Related Topics

Somewhat Related Topics

Glossary Terms
carbocations electrophile hydration
intermediate Markovnikov's rule mechanism
regiochemistry regioisomers regiospecific

Regiochemistry and Markovnikov's rule

One of the most fundamental reactions in organic chemistry is the addition of HX to an alkene. This reaction is illusturated below, using the addition of HCl to ethylene as an example:

Notice that ethylene is a symmetrical alkene: it has the same substituents (two hydrogen atoms) on either end of the C=C double bond. As such, it doesn't matter which carbon the "H" ends up attached to and which carbon the "Cl" ends attached to in this addition reaction -- the product is still the same, ethyl chloride:

But what about cases where the alkene is not symmetrical? In those cases, two different "regioisomers" could be formed. For example, see the addition of HCl to propene, shown below:

In fact, the only product that is formed in the reaction of propene with HCl is isopropyl chloride. None of the regioisomer (n-propyl chloride) is formed. As such, we call this a "regiospecific" reaction, since only one of several possible regioisomers was formed.

In these sorts of cases, when the degree of substitution on either end of the double is not identical, we can use "Markovnikov's rule" to predict which regioisomer will form predominantly, if not exclusively, in the addition of HX to an alkene. In simplest terms, Markovnikov's rule states that:

"In the addition of HX to an alkene, the acid hydrogen (H) becomes attached to the carbon with fewer alkyl substituents, and the halide (X) group becomes attached to the carbon with more alkyl substituents."

Another way of saying this is that "the hydrogen-rich atom becomes hydrogen-richer," i.e., the hydrogen of HX gets attached to the carbon that had more hydrogens in the first place.

Because there are only three possible degrees of alkyl substitution on either end of a C=C double bond -- no alkyl groups (and two hydrogens), one alkyl group (and one hydrogen), or two alkyl groups (and no hydrogens) -- it should be a simple matter to apply Markovnikov's rule. Some examples are shown below:

Self-test question #1

Although the addition of HBr to 1-pentene gives a good yield of 2-bromopentane, the addition of HBr to 2-pentene gives an approximately equal mixture of 2-bromopentane and 3-bromopentane. Can you explain why?


Self-test question #2

Hydrogen atoms usually aren't shown explicitly in line structures, and so students sometimes lose track of "where the 'H' went" in HX additions to alkenes. One way you'd be "forced" to keep track is by using "heavy hydrogen" (deuterium="D") instead of ordinary hydrogen in an addition reaction. Unlike regular hydrogen, deuterium atoms are shown explicity in line structures. For example, 1-methylcyclohexene and methylenecyclohexane each give the same product in their reactions with HBr, but they give different products in their reactions with DBr:


Self-test question #3

In the web page on "Nomenclature," you learned about(or reviewed) the trival names for twelve alkyl groups having five or fewer carbons. Consider only the bromides derived from each of those alkyl groups.

Of those twelve, only methyl bromide and neopentyl bromide could not possibly be derived by the addition of HBr--Markovnikov or non-Markovnikov--to an alkene of the appropriate structure. For each of the other ten alkyl bromides, determine:

  1. which ones could be formed by the Markovnikov addition of HBr to an alkene
  2. which ones could be formed by the non-Markovnikov addition of HBr to an alkene
  3. which ones could be formed by either method, and
  4. which is the only 1 alkyl bromide that can be formed by Markovnikov addition of HBr
If you can, try also to draw the structure(s) fo the alkene(s) that would be required for each of the transformations implied in part(1)-(3) of the question.


Self-test question #4

Replace this writing with the real question.


Self-test question #5

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Related reading in textbook (McMurry, Organic Chemistry, 4th ed.)

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