Electrophilic addition: Markovnikov's rule and the peroxide effect

Why alkenes add

An alkane is hard to attack, but an alkene is not. The pi bond sits above and below the plane of the molecule, and its electrons are held loosely. A reagent that is short of electrons can reach them easily. Such a reagent is an electrophile.

In an addition reaction the pi bond breaks, and two new sigma bonds form in its place. The pi bond is the weaker part of the double bond. A C=C bond has about $610\ \text{kJ/mol}$ and a C-C bond has about $350\ \text{kJ/mol}$. So the pi bond holds about $260\ \text{kJ/mol}$. Two new sigma bonds hold much more, so addition gives out heat.

The main additions

• Hydrogen, with nickel, platinum or palladium: an alkene becomes an alkane.
• Halogens. Bromine dissolved in an inert solvent such as tetrachloromethane adds to give a vicinal dibromide. The red-brown colour disappears. This is the standard test for a double bond.

• Hydrogen halides. The order of reactivity is $\text{HI} > \text{HBr} > \text{HCl}$.
• Water. Dilute sulphuric acid adds water. Propene gives propan-2-ol. Ethanol is made in industry by passing ethene and steam over a phosphoric acid catalyst at about $573\ \text{K}$ and $60$ to $70\ \text{atm}$.

The mechanism of electrophilic addition

Take hydrogen bromide and ethene. It goes in two steps.

1. The pi electrons attack the hydrogen of $\text{HBr}$. The H-Br bond breaks, and bromide leaves. A hydrogen atom is now joined to one carbon, and the other carbon has a positive charge: a carbocation. This step is slow.
2. The bromide ion, with a lone pair, attacks the positive carbon. The new C-Br bond forms.

Markovnikov's rule

What if the alkene is not symmetrical, as in propene, and the reagent is not symmetrical either, like $\text{HBr}$? The hydrogen can go to either carbon. Two products are possible:

• $\text{CH}_{3}\text{CHBrCH}_{3}$, 2-bromopropane.
• $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{Br}$, 1-bromopropane.

Vladimir Markovnikov noticed in 1870 that the first one is the main product. His rule, in its older form, says: the negative part of the reagent goes to the carbon that has fewer hydrogen atoms. That is the same as saying that the hydrogen goes to the carbon that has more hydrogen atoms.

The reason is in the mechanism. In step 1 the hydrogen ion can add to carbon 1 and leave a positive charge on carbon 2, or the other way round. Add it to the $\text{CH}_{2}$ end of propene, and the positive charge lands on the middle carbon. That is a secondary carbocation, which is more stable than the primary cation that forms the other way. A more stable intermediate forms faster, so it decides the main product.

So the modern form of the rule is: the more stable carbocation forms, and the halide adds to it.

Two cases to watch.

• Equal ends. In but-2-ene each carbon has one alkyl group, so both cations are secondary. There is one product, 2-bromobutane. In pent-2-ene the two carbon atoms also carry one alkyl group each, but the groups differ in size. A mixture of 2-bromopentane and 3-bromopentane forms.
• A cation that rearranges. 3-Methylbut-1-ene with hydrogen chloride gives mostly 2-chloro-2-methylbutane. The hydrogen first adds to carbon 1, which leaves a secondary cation on carbon 2. A hydrogen atom, with its pair of electrons, then moves from carbon 3 to carbon 2. This hydride shift puts the positive charge on carbon 3, which is tertiary and more stable. Chloride attacks there. A smaller amount of 2-chloro-3-methylbutane also forms, from the cation before it shifts.

Practise with new numbers

The peroxide effect

In 1933 Kharasch and Mayo found that hydrogen bromide adds to propene the other way round when an organic peroxide, such as benzoyl peroxide, is present. The product is 1-bromopropane. This is the peroxide effect, or the Kharasch effect. The rule is also called anti-Markovnikov addition.

The peroxide starts a free-radical chain.

1. Initiation. The weak O-O bond of the peroxide breaks into two radicals. A radical takes a hydrogen atom from $\text{HBr}$ and leaves a bromine atom, $\text{Br}^{\bullet}$.
2. Propagation. The bromine atom adds to the $\text{CH}_{2}$ end of propene. This gives the secondary radical $\text{CH}_{3}\text{CH}^{\bullet}\text{CH}_{2}\text{Br}$, which is more stable than the primary radical formed the other way. The radical takes a hydrogen atom from another $\text{HBr}$. That makes 1-bromopropane and a new bromine atom, which repeats the cycle.
3. Termination. Two radicals join.

Notice that the same rule still holds. In both pathways, the more stable intermediate, a cation or a radical, forms on the more substituted carbon. What changes is the first species to add. In the ionic path it is $\text{H}^{+}$, and in the radical path it is $\text{Br}^{\bullet}$.

The effect is seen only with HBr.

• $\text{HCl}$ has a strong H-Cl bond, $431\ \text{kJ/mol}$. The alkyl radical cannot take a hydrogen atom from it, and the chain breaks down.
• $\text{HI}$ has a weak H-I bond, $299\ \text{kJ/mol}$, but the iodine atom adds to the alkene so weakly that the step is not favourable. Iodine atoms mostly join to form $\text{I}_{2}$.

The $\text{HBr}$ bond, $366\ \text{kJ/mol}$, is just right. NCERT prints $430.5$, $363.7$ and $296.8\ \text{kJ/mol}$ for the three bonds. The values above are the ones in standard tables, rounded.

Try it in Python

The program applies the rule to four alkenes, with and without a peroxide.

Common mistakes

• Adding the peroxide effect to HCl or HI. It works only with HBr.
• Applying the rule to the wrong carbon. The hydrogen goes to the carbon that already has more hydrogen atoms.
• Forgetting rearrangement. A secondary cation can shift to a tertiary one, so the halide may not sit where the double bond was.
• Saying that the peroxide changes the mechanism to ionic. It changes it to a radical chain.
• Writing both products as equal. The main product comes from the more stable intermediate.

Remember these

• Alkenes add electrophiles. The pi bond is used up, and two sigma bonds form.
• $\text{Br}_{2}$ in an inert solvent loses its colour with an alkene. It is the test for unsaturation.
• Markovnikov: the hydrogen goes to the carbon with more hydrogen atoms. The halide goes to the carbon with more alkyl groups.
• The more stable carbocation forms. Secondary beats primary, and tertiary beats secondary. Cations can rearrange.
• Peroxide effect: only $\text{HBr}$, by a radical chain, giving the anti-Markovnikov product.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What happens to the pi bond of an alkene in an addition reaction?

2. Write the main product of the addition of hydrogen bromide to 2-methylpropene without a peroxide.

3. Propene reacts with hydrogen bromide in the presence of benzoyl peroxide. Name the main product and the first species that adds.

4. Why does the peroxide effect not occur with hydrogen chloride?

5. Which gives the more stable carbocation on addition of $\text{H}^{+}$ to but-1-ene: $\text{CH}_{3}\text{CH}_{2}\text{CH}^{+}\text{CH}_{3}$ or $\text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{CH}_{2}^{+}$?

References

• OpenStax Organic Chemistry (McMurry), chapter 7, the sections "Electrophilic Addition Reactions of Alkenes", "Markovnikov's Rule" and "Carbocation Rearrangements"; chapter 8, "Radical Additions and Chain-Growth Polymers".
• NCERT Chemistry, Class 11, Part II, the chapter "Hydrocarbons" (addition reactions of alkenes, the peroxide effect).
• Bond enthalpies of H-Cl, H-Br and H-I: OpenStax Chemistry 2e, chapter 7, table of average bond energies.