Alkenes: Chemical Reactions

Why the C=C bond reacts at all. The $\sigma$ bond of the double bond holds the two carbons rigidly in place, but the $\pi$ bond is a diffuse cloud of electron density lying above and below that framework, only loosely bound to the two nuclei. Any species hunting for electrons — an electrophile — is drawn to that cloud, and the reaction that follows is called electrophilic addition: the double bond breaks, and two new $\sigma$ bonds form in its place, one to each carbon.

The general addition reaction. Whatever the electrophile, the overall transformation looks like this:

where: E is the electrophilic (electron-poor) fragment that attacks the $\pi$ cloud first, and Nu is the nucleophilic fragment that completes the bonding afterward — for HBr, E is $\text{H}^{\delta+}$ and Nu is $\text{Br}^-$; for $\text{H}_2\text{O}/\text{H}^+$, E is $\text{H}^+$ and Nu is $\text{H}_2\text{O}$ (giving -OH after loss of a proton). The reaction always proceeds in two bond-forming steps, never one, and that two-step nature is exactly what lets a substituted alkene choose where E and Nu land.

The reaction family built on this idea. Four addition reactions share this same electrophilic pattern, differing only in the reagent and catalyst:

• Hydrogenation: $\text{alkene} + \text{H}_2 \xrightarrow{\text{Ni/Pt/Pd}} \text{alkane}$ — a metal surface, not an ionic mechanism, but still net addition across the bond.
• Halogenation: $\text{alkene} + \text{X}_2 \rightarrow$ a vicinal dihalide; decolourisation of bromine water is the standard laboratory test for a C=C bond.
• Hydration: $\text{alkene} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{alcohol}$, Markovnikov-selective, since it goes through the same carbocation intermediate as HX addition.
• Ozonolysis: alkene $+ \text{O}_3 \rightarrow$ ozonide $\xrightarrow{\text{Zn/H}_2\text{O}}$ two carbonyl compounds — the only reaction here that cleaves the bond rather than just adding across it, which is exactly why it is useful for locating a double bond.

There is no numerical formula behind Markovnikov's rule — it is a consequence of which reaction intermediate is more stable, and that reasoning is the entire content of the topic. Three linked facts explain every addition outcome you will be asked to predict.

First, when an unsymmetrical alkene reacts with an unsymmetrical reagent like HX, the electrophile ($\text{H}^+$) can add to either double-bond carbon, and each choice generates a different carbocation. Alkyl groups on the neighbouring carbons donate electron density toward the positive charge through the inductive and hyperconjugative effects, so a carbocation with more alkyl substituents attached is more stabilised. Reaction always prefers the lower-energy pathway, so whichever protonation step leads to the more substituted (more stable) carbocation happens faster and dominates the product mixture.

Second, once that more stable carbocation exists, the nucleophile (X⁻) has only one electrophilic centre to attack — the carbocation carbon itself — so X necessarily ends up on the carbon that started out with fewer hydrogens. This is exactly Markovnikov's rule, restated mechanistically: it is not an arbitrary memorised pattern but a direct consequence of carbocation stability.

Third, the peroxide (Kharasch) effect changes which intermediate forms, and only for HBr. Peroxides generate a bromine radical, $\text{Br}^\bullet$, which is the chain-carrying species instead of $\text{H}^+$. The radical adds first, and — just as with carbocations — the more substituted (more stable) radical forms preferentially, but this time it is Br that ends up on the less substituted carbon, because Br is now the first atom to attack, not the last. HCl and HI do not show this reversal because their radical chain steps are thermodynamically unfavourable; only the HBr radical cycle is efficient enough to compete with the ionic pathway.

Q. Predict the major product when HBr adds to propene, $\text{CH}_3\text{-CH=CH}_2$, (a) in the absence of peroxide, and (b) in the presence of peroxide.

Step 1. Without peroxide, protonation of the double bond can generate either a secondary carbocation ($\text{CH}_3\text{-CH}^+\text{-CH}_3$, from H adding to the terminal carbon) or a primary carbocation (from H adding to the internal carbon).

Step 2. The secondary carbocation is more stable (more alkyl groups donating electron density), so it forms preferentially; Br⁻ then attacks it.

Step 3. Major product without peroxide: 2-bromopropane, $\text{CH}_3\text{-CHBr-CH}_3$ — Markovnikov addition, Br on the more substituted carbon.

Step 4. With peroxide, $\text{Br}^\bullet$ adds first, to the terminal, less hindered carbon, generating the more stable secondary radical at C2; this radical then abstracts H from another HBr molecule.

Step 5. Major product with peroxide: 1-bromopropane, $\text{BrCH}_2\text{-CH}_2\text{-CH}_3$ — anti-Markovnikov, because the radical mechanism forms the more stable radical, which puts Br on the opposite carbon from the ionic pathway.

Competitive stretch

Now try a more substituted alkene, 2-methylpropene, $(\text{CH}_3)_2\text{C=CH}_2$, with HBr. Without peroxide, protonation at the terminal $\text{CH}_2$ gives a tertiary carbocation $(\text{CH}_3)_2\text{C}^+\text{-CH}_3$ — even more stabilised than a secondary one — so the ionic product is 2-bromo-2-methylpropane (tert-butyl bromide), Br on the fully substituted carbon. With peroxide, the bromine radical adds first to the terminal carbon, generating the more stable tertiary radical, which then abstracts H to give the anti-Markovnikov product 1-bromo-2-methylpropane. The pattern from propene holds exactly: more substituted alkenes give even more decisive Markovnikov selectivity, because the stability gap between the possible carbocations grows with substitution.

• Applying Markovnikov's rule by rote without checking for peroxide — the Kharasch effect flips the regiochemistry, but only for HBr, never for HCl or HI.
• Treating Markovnikov's rule as an arbitrary memorised pattern rather than a consequence of carbocation stability — the mechanistic reasoning also predicts hydration and other addition outcomes, not just HX addition.
• Confusing hydrogenation (metal catalyst, gives an alkane) with hydration (acid catalyst, gives an alcohol) — both are additions across the double bond, but the reagent and product class differ completely.
• Forgetting that reconstructing an alkene from its ozonolysis products means joining the two carbonyl carbons back together with a double bond, not merely listing the two carbonyl compounds separately.

Ask the tutor: Want to work through a few more Markovnikov-versus-peroxide predictions, or practise deducing an alkene's structure from its ozonolysis products?