Electrophilic Addition to Alkenes

25 min

A rule found by experiment

In 1869 the Russian chemist Vladimir Markovnikov noticed a pattern. When hydrogen bromide adds to propene, two products are possible, 1-bromopropane and 2-bromopropane, yet nearly all of the product is 2-bromopropane. The hydrogen goes to the end carbon, which already carries two hydrogens. His rule worked for every unsymmetrical alkene he tried, but nobody could explain it until chemists learned to follow electrons through a reaction. This lesson does exactly that.

Definition

Electrophilic addition: a reaction in which an electrophile is attracted to the electron-rich C=C bond and two groups add across it, so the double bond becomes a single bond. Example: $\mathrm{CH_2{=}CH_2 + HBr \rightarrow CH_3CH_2Br}$ .

Heterolytic fission: breaking a bond so that one atom takes BOTH electrons, giving a positive and a negative ion. A full-headed curly arrow shows a pair of electrons moving.

Markovnikov's rule: when HX adds to an unsymmetrical alkene, the H atom joins the carbon that already has more hydrogen atoms. The modern form: the reaction goes through the more stable carbocation.

Positive inductive effect: the slight release of electron density by an alkyl group along a σ bond.

The mechanism with hydrogen bromide

H–Br is polar ( $\mathrm{H^{\delta+}{-}Br^{\delta-}}$ ). The π electrons, held above and below the C=C plane, attack the $\mathrm{H^{\delta+}}$ atom. At the same time the H–Br bond breaks heterolytically, both electrons going to bromine. One carbon of the old double bond is now short of electrons: a carbocation. In the fast second step, the bromide ion uses a lone pair to bond to it.

Figure 3.1: Electrophilic addition of HBr to propene. The H adds to the carbon that already has more hydrogens, giving the more stable carbocation.

Figure 3.2: The order of carbocation stability. R stands for an alkyl group.

Each alkyl group pushes a little electron density towards the positive carbon, spreading out its charge, so a carbocation with more alkyl groups is more stable and forms faster. That is Markovnikov's rule explained: in propene, adding H to C1 gives a secondary carbocation (two alkyl groups), while adding H to C2 would give a primary one (one alkyl group). The order of reactivity of the hydrogen halides is HI > HBr > HCl, following the weakening of the H–X bond. (In the presence of organic peroxides, HBr adds the other way round, by a radical mechanism; this "anti-Markovnikov" addition is beyond this unit.)

Addition of halogens, and the bromine-water test

Figure 3.3: Addition of bromine. The approaching π electrons repel the electrons of Br₂, making the near bromine atom δ+.

A Br–Br bond is not polar, but as it nears the π electrons, they repel the bonding pair; the near bromine becomes $\delta+$ and is attacked. In bromine water, the carbocation meets far more water molecules than bromide ions, so the main product is 2-bromoethanol, $\mathrm{CH_2BrCH_2OH}$ . Either way, the orange colour of bromine disappears: this is the standard test for C=C. (Some textbooks draw the intermediate as a three-membered "bromonium" ion; the products are the same.)

Addition of water: hydration

In the laboratory: the alkene is absorbed in cold concentrated sulfuric acid, giving an alkyl hydrogensulfate, which is then warmed with water: $\mathrm{CH_2{=}CH_2 + H_2SO_4 \rightarrow CH_3CH_2OSO_3H}$ ; $\mathrm{CH_3CH_2OSO_3H + H_2O \rightarrow CH_3CH_2OH + H_2SO_4}$ . Markovnikov's rule applies: propene gives propan-2-ol.
In industry: ethene and steam pass over a phosphoric acid catalyst on silica at about 300 °C and 60–70 atm: $\mathrm{CH_2{=}CH_2(g) + H_2O(g) \rightleftharpoons CH_3CH_2OH(g)}$ . Only about 5% reacts on each pass, so unreacted gases are recycled.

Worked example

Problem. Name the major product of: (a) HBr with 2-methylpropene, $\mathrm{(CH_3)_2C{=}CH_2}$ ; (b) HCl with but-1-ene. Explain (a) with the intermediate.

Solution. (a) Adding $\mathrm{H^+}$ to the $\mathrm{CH_2}$ end gives the tertiary carbocation $\mathrm{(CH_3)_3C^+}$ , stabilised by three methyl groups; adding it to the other carbon would give a primary carbocation. Br⁻ then bonds to the tertiary carbon: 2-bromo-2-methylpropane, $\mathrm{(CH_3)_3CBr}$ .

(b) H adds to C1 (the $\mathrm{CH_2}$ end), giving the secondary carbocation on C2: 2-chlorobutane. C2 is chiral, and because the flat carbocation can be attacked equally from either face, a racemic mixture forms.

Worked example

Problem. Ethene is bubbled through bromine water that also contains dissolved sodium chloride. The products are 1,2-dibromoethane, 2-bromoethanol and 1-bromo-2-chloroethane, but no 1,2-dichloroethane. Explain how this supports a two-step mechanism.

Solution. If $\mathrm{Br_2}$ added in one step, both new groups would be Br. The mixture shows that a positive intermediate, $\mathrm{CH_2BrCH_2^+}$ , forms first and is then captured by whichever nucleophile reaches it: $\mathrm{Br^-}$ , $\mathrm{H_2O}$ or $\mathrm{Cl^-}$ . Every product contains at least one Br, because the first step always needs the electrophile $\mathrm{Br^{\delta+}}$ ; $\mathrm{Cl^-}$ is a nucleophile and cannot start the reaction, so no 1,2-dichloroethane forms.

Warning

Curly arrows start at electrons and end where the electrons go. The first arrow starts at the C=C bond (the π electrons), not at the H atom. The arrow in step 2 starts at a lone pair on Br⁻, not at the "−" sign. Draw the dipole on HBr, and put the + charge of the carbocation on the correct carbon.

Tip

To predict the major product, ask: "Which carbocation is more stable?" Put the $\mathrm{H^+}$ on the carbon that makes the positive charge land on the carbon with the most alkyl groups. Then attach X⁻ there.

Chemistry Student: Alkenes: Electrophilic Addition with Hydrogen Bromide (16 min). Watch the curly arrows of both steps, and how the more stable carbocation decides the major product for propene.

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Practice

Define an electrophile, and explain why alkenes react with electrophiles while alkanes do not.
Describe, with curly arrows in words, the mechanism of the reaction of ethene with hydrogen bromide.
Name the major product of: (a) HI with propene; (b) HBr with 2-methylbut-2-ene.
Put these carbocations in order of increasing stability: $\mathrm{(CH_3)_3C^+}$ , $\mathrm{CH_3^+}$ , $\mathrm{CH_3CH_2^+}$ , $\mathrm{(CH_3)_2CH^+}$ .
Propene is absorbed in cold concentrated sulfuric acid and the product is warmed with water. Name the alcohol formed and explain why it is not propan-1-ol.
Give the conditions for the industrial hydration of ethene.
Explain why bromine water reacting with an alkene gives mainly a bromoalcohol, not a dibromide.

Show answersHide answers

An electrophile accepts a pair of electrons to form a new bond. The C=C π bond is a region of high, exposed electron density; alkanes have only σ bonds, held tightly between nuclei.
The π bond (arrow from C=C) attacks $\mathrm{H^{\delta+}}$ of H–Br; the H–Br bond breaks (arrow to Br), giving $\mathrm{CH_3CH_2^+}$ and $\mathrm{Br^-}$ ; a lone pair on $\mathrm{Br^-}$ bonds to the carbocation, giving bromoethane.
(a) 2-iodopropane; (b) 2-bromo-2-methylbutane.
$\mathrm{CH_3^+ < CH_3CH_2^+ < (CH_3)_2CH^+ < (CH_3)_3C^+}$ .
Propan-2-ol: the $\mathrm{H^+}$ adds to C1, giving the more stable secondary carbocation on C2, where the –OSO₃H (then –OH) group ends up.
Steam, phosphoric acid catalyst on silica, about 300 °C and 60–70 atm.
After the first step, the carbocation is surrounded by water molecules, which far outnumber bromide ions, so water usually bonds to it; loss of $\mathrm{H^+}$ then gives the alcohol.

Sources and credits

Written by Ganzaa.
Video: Alkenes: Electrophilic Addition with Hydrogen Bromide, HBr (A-level and IB Chemistry), Chemistry Student, embedded from YouTube.
Industrial conditions are typical textbook values.
The text, mechanisms, diagrams, worked examples and Practice items were written and drawn by Ganzaa.

1A rule found by experiment

2The mechanism with hydrogen bromide

3Addition of halogens, and the bromine-water test

4Addition of water: hydration

A rule found by experiment

The mechanism with hydrogen bromide

Addition of halogens, and the bromine-water test

Addition of water: hydration