Electrophilic Addition to Alkenes: Regiochemistry and Stereochemistry

Why the alkene attacks first

The pi bond of an alkene is a region of high electron density sitting above and below the plane of the two carbons. It is loosely held compared with a sigma bond, so it behaves as a nucleophile. Every reaction in this chapter therefore begins the same way: the pi electrons reach out and attack an electrophile, and the pi bond is destroyed in the process.

That first step is what creates the intermediate. Almost every question you will be asked about alkene addition is really a question about which intermediate forms, so it is worth writing that step out explicitly rather than jumping to the product.

Markovnikov orientation is a statement about carbocation stability

When HBr adds to propene, the proton can add to either carbon of the double bond. Adding it to the terminal carbon leaves the positive charge on the middle carbon, giving a secondary carbocation. Adding it to the middle carbon leaves a primary carbocation. Secondary is more stable, so that pathway has the lower activation energy and dominates.

The traditional wording — that hydrogen goes to the carbon already carrying more hydrogens — is a description of the outcome, not the reason. Stating the reason is what earns marks, because it also tells you when the rule will not apply.

Peroxide effect: same reagent, opposite orientation

Add HBr to propene in the presence of organic peroxides and the product is 1-bromopropane — the anti-Markovnikov product. Nothing about Markovnikov has been violated, because the mechanism is no longer ionic. Peroxides generate bromine radicals, the bromine radical adds first, and it adds to the terminal carbon because that leaves the more stable secondary carbon radical.

The orientation rule is unchanged in spirit: the more stable intermediate still wins. What changed is which atom adds first.

Halogen addition and the bridged intermediate

Bromine adds to an alkene through a cyclic bromonium ion rather than an open carbocation. The bromine atom bridges both carbons, which blocks one face of the molecule completely. Bromide then has to attack from the opposite face.

The consequence is stereochemical and it is testable: addition of bromine to cyclohexene gives exclusively trans-1,2-dibromocyclohexane. No cis product forms at all. If a question gives bromine and asks about stereochemistry, anti addition is the expected answer, and the bridged ion is the reason.

Halohydrin formation: two rules acting at once

Running the bromine addition in water gives a bromohydrin. The bromonium ion still forms, so the two new groups are still anti to each other. But water now chooses which carbon to attack, and it attacks the more substituted one.

That looks backwards for a nucleophile until you notice that the bridged ion is unsymmetrical: the more substituted carbon carries more of the positive charge, so the C–Br bond on that side is longer and weaker. The nucleophile goes where the charge is. Both the regiochemistry and the stereochemistry come out of the same intermediate.

A working order for any addition question

1. Identify the electrophile and write the step where the pi bond attacks it.
2. Name the intermediate — open cation, bridged ion, or radical. This one decision settles most of the answer.
3. If it is an open cation, check for rearrangement. This is the single most common source of a wrong major product.
4. Apply the stereochemical consequence — bridged means anti, cyclic ester means syn, planar cation means a mixture.
5. State the reason, not just the rule. An answer that says “secondary carbocation is more stable than primary” scores where “by Markovnikov’s rule” alone often does not.

Frequently asked questions

Is Markovnikov’s rule ever actually broken?

The orientation it predicts is reversed under radical conditions, but the underlying principle — the more stable intermediate forms faster — holds in both cases. It is better to think of the rule as a special case of intermediate stability than as an independent law that sometimes fails.

Why does bromine give only the trans product with cyclohexene?

Because the bromonium ion physically covers one face of the ring. Bromide can only reach the carbon from the opposite side, so the two bromines end up anti. There is no pathway available that would deliver them to the same face.

How do I know whether to expect a rearrangement?

Draw the carbocation and ask whether a hydrogen or an alkyl group on an adjacent carbon could shift to give a more stable cation. Secondary next to a quaternary carbon is the classic setup. If no shift increases stability, no rearrangement occurs.

Does the peroxide effect work with HCl or HI?

No. Only HBr has the right bond energies for both propagation steps to be favourable. This is asked frequently, and the expected answer names the bond strength argument rather than simply stating that it does not happen.

Why does water attack the more substituted carbon in halohydrin formation?

The bromonium ion is unsymmetrical. More positive charge sits on the carbon better able to stabilise it, which is the more substituted one, so that is where the nucleophile bonds. The stereochemistry stays anti regardless.