Addition Reactions: Breaking the Double Bond

See how alkenes react by addition across the C=C bond: bromination of ethene, catalytic hydrogenation with nickel, and Markovnikov's rule for HX addition.

Quick answer

In an addition reaction, two reactant molecules combine across a carbon–carbon double bond to form a single product. The pi bond breaks and two new sigma bonds form, converting the unsaturated alkene into a saturated product.

Example: Ethene adds bromine across its double bond: CH2=CH2 + Br2 → CH2BrCH2Br (1,2-dibromoethane). The orange bromine color disappears as it reacts, which is why this reaction is used as a chemical test for unsaturation.

What you will learn

  • Define addition and contrast it with substitution.
  • Describe electrophilic addition across the C=C bond.
  • Explain bromination and catalytic hydrogenation of alkenes.
  • Apply Markovnikov's rule to predict the major product of HX addition.

What addition means

In an addition reaction, two reactant molecules combine to form a single product across a carbon–carbon double bond. The pi bond breaks, and each carbon forms one new sigma bond to the incoming atoms.

The pattern is beautifully economical: A=B + X–Y → X–A–B–Y. Nothing is lost as a byproduct — every atom of the reactants appears in the product.

Addition is the characteristic reaction of alkenes and alkynes, and the exact reverse of elimination. Together, the two reactions form a reversible pair that chemists exploit constantly.

Electrophilic addition: the core idea

The driving force in alkene addition is the pi bond's exposed, electron-rich cloud. Species that seek electrons — electrophiles — are drawn to it, so most alkene additions are classified as electrophilic addition.

The general picture has two stages. First, the electrophile bonds to one carbon of the double bond using the pi electrons, leaving the other carbon positively charged (a carbocation). Then a nucleophile — an electron-pair donor — bonds to that positive carbon, completing the addition.

You do not need the full arrow-pushing mechanism at this stage; the concept to hold is that the pi electrons initiate the reaction by attacking the electrophile, and everything follows from the resulting intermediate.

Bromination: ethene + bromine

The textbook example is the bromination of ethene:

CH2=CH2 + Br2 → CH2BrCH2Br (1,2-dibromoethane)

Bromine is an orange-brown liquid, and its color vanishes as it adds across the double bond. This visible change makes the reaction a classic test for unsaturation: if an unknown decolorizes bromine, it contains a C=C or C≡C bond.

The reaction needs no catalyst and proceeds at room temperature — a striking contrast with alkane bromination, which demands UV light. The difference is the pi bond: alkenes offer electrons that alkanes simply do not have.

Hydrogenation

Hydrogen adds across the double bond in the presence of a metal catalyst, typically nickel:

CH2=CH2 + H2 → CH3CH3

The catalyst is essential. H2 adsorbs onto the nickel surface, which weakens the H–H bond, while the alkene adsorbs alongside it; the atoms then transfer across the double bond. Without the surface to bring them together, the reaction is negligibly slow.

Industrially, partial hydrogenation of vegetable oils — which contain C=C bonds — produces margarine and shortening by converting some double bonds to single bonds, raising the melting point. The same catalytic principle saturates countless chemical feedstocks.

Addition of hydrogen halides

Hydrogen halides add across double bonds in the same electrophilic pattern. With ethene the outcome is unambiguous:

CH2=CH2 + HBr → CH3CH2Br (bromoethane)

With an unsymmetrical alkene such as propene, two products are possible — 1-bromopropane or 2-bromopropane — and they do not form equally. The major product follows Markovnikov's rule, covered next.

Note the atom economy again: every atom of HBr ends up in the product. Addition reactions waste nothing, which is one reason industry favors them.

Markovnikov's rule

Markovnikov's rule predicts the major product when HX adds to an unsymmetrical alkene: the hydrogen attaches to the carbon that already bears more hydrogens. For propene:

CH3CH=CH2 + HBr → CH3CHBrCH3 (2-bromopropane, major) + CH3CH2CH2Br (1-bromopropane, minor)

The reason lies in the carbocation intermediate. Adding H+ to carbon 1 produces a secondary carbocation on carbon 2; adding it to carbon 2 would produce a primary carbocation on carbon 1. Secondary carbocations are more stable than primary ones (tertiary > secondary > primary), so the pathway through the more stable intermediate dominates.

The rule is therefore not arbitrary: it is shorthand for 'the reaction proceeds through the most stable carbocation.'

Addition versus substitution

Addition and substitution are easy to confuse but fundamentally different:

  • Addition: two molecules become one; no byproduct. Requires a pi bond. Example: CH2=CH2 + Br2 → CH2BrCH2Br.
  • Substitution: one group replaces another; a byproduct leaves. Example: CH4 + Cl2 → CH3Cl + HCl.

Elimination is addition's mirror image: where addition consumes a pi bond to make two sigma bonds, elimination breaks two sigma bonds to make a pi bond. The three families — addition, elimination, substitution — interconvert the same functional groups in different directions, forming the core reaction network of organic chemistry.

Common mistakes

✗ Addition and substitution give the same products.

Addition combines two molecules into one with no byproduct, while substitution swaps a group and releases a byproduct. Ethene + Br2 gives only 1,2-dibromoethane; methane + Br2 would give CH3Br + HBr. Atom balance distinguishes them.

✗ Markovnikov's rule is just memorization.

The rule summarizes a real stability trend: the reaction passes through a carbocation, and more substituted carbocations (tertiary > secondary > primary) are more stable, so the pathway through the more stable intermediate dominates. The rule is a shortcut for that reasoning.

✗ Hydrogenation happens without a catalyst.

H2 does not add to alkenes at an appreciable rate on its own — the H–H bond is strong. A metal catalyst such as nickel adsorbs both reactants on its surface, weakening the bonds and bringing the molecules together so addition can occur.

Real-world applications

  • The bromine decolorization test identifies unsaturation in unknown organic samples in teaching and quality-control labs.

  • Catalytic hydrogenation hardens vegetable oils into margarine and produces countless saturated feedstocks.

  • Addition polymerization of ethene and propene manufactures the world's most common plastics.

  • Markovnikov-selective additions are workhorse steps in synthesizing alcohols and haloalkanes from alkenes.

Frequently asked questions

Why do electrophiles attack the double bond?

The pi bond's electrons sit in exposed regions above and below the bond axis, rather than tucked between the nuclei like sigma electrons. This electron-rich, accessible cloud strongly attracts electrophiles — electron-seeking species — making the double bond the natural site of attack.

What does Markovnikov's rule predict for propene + HBr?

It predicts 2-bromopropane as the major product. The hydrogen adds to carbon 1 (which already has two hydrogens), placing the positive charge on carbon 2 to form the more stable secondary carbocation; bromide then bonds there. The alternative 1-bromopropane forms only in minor amounts via the less stable primary carbocation.

Does addition change the hybridization of the carbons?

Yes. Each double-bond carbon starts sp2 hybridized (trigonal planar) and becomes sp3 hybridized (tetrahedral) after addition, since it now forms four single bonds. This geometric change — flat to three-dimensional — accompanies every addition reaction.

Can alkynes undergo addition too?

Yes, and twice over: each of the two pi bonds can accept one addition. Ethyne + HBr first gives bromoethene, and a second equivalent gives 1,1-dibromoethane. Markovnikov's rule applies to the first addition just as it does for alkenes.