Why HBr Adds to Propene Differently When Peroxide Is Present

Ethene is used to ripen bananas and mangoes, links up into polythene, and decolourises bromine water in seconds, while ethane barely reacts at all. The difference is a single pi bond — a weaker, exposed cloud of electrons that invites attack.

This lesson covers naming and preparing alkenes, and the mechanism of electrophilic addition with Markovnikov's and anti-Markovnikov's rules.

**Alkenes are hydrocarbons with a carbon-carbon double bond and the general formula ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$, named with the suffix -ene and the lowest locant for the double bond, and they are prepared by partial reduction of alkynes or by elimination reactions of alkyl halides and alcohols.

Structure.** Each carbon of the double bond is $\mathrm{s}{\mathrm{p}}^2$ hybridised; the double bond has one sigma and one pi bond, is 134 pm long, and prevents free rotation, which gives rise to geometrical isomerism.

Naming:

- $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2$ is ethene and $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2$ is propene
- $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{CHC}{\mathrm{H}}_3$ is but-2-ene, which exists as cis and trans isomers
- $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$ is 2-methylbut-1-ene: the parent chain must contain the double bond and give it the lowest locant

Methods of preparation:

- Partial reduction of alkynes — Lindlar's catalyst, partly poisoned palladium, gives cis alkenes, while sodium in liquid ammonia gives trans alkenes
- Dehydrohalogenation — heating an alkyl halide with alcoholic potassium hydroxide removes HX: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{KOH}\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+\mathrm{KBr}+{\mathrm{H}}_2\mathrm{O}$
- Dehalogenation of vicinal dihalides with zinc dust: $\mathrm{C}{\mathrm{H}}_2\mathrm{BrC}{\mathrm{H}}_2\mathrm{Br}+\mathrm{Zn}\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+\mathrm{ZnB}{\mathrm{r}}_2$
- Acidic dehydration of alcohols — ethanol heated with concentrated sulphuric acid at 443 K: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to \mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

Saytzeff rule. When elimination can give more than one alkene, the more substituted alkene is the major product, so 2-bromobutane gives mainly but-2-ene rather than but-1-ene.

Worked reasoning. Alcoholic KOH brings about elimination, while aqueous KOH brings about substitution to an alcohol — the solvent decides which reaction wins.

An everyday example. Fruit ripening chambers in Indian wholesale markets release small, controlled amounts of ethene gas to ripen bananas and mangoes evenly.

The substance. The double bond locks rotation — which is why the cis and trans forms of but-2-ene are different compounds, unlike the freely interconverting conformations of ethane.

In electrophilic addition, the pi electrons of an alkene attack an electrophile to form a carbocation, which a nucleophile then completes; Markovnikov's rule says the negative part of HX goes to the carbon with fewer hydrogens, while with peroxide present, HBr adds the opposite way.

Mechanism with HBr and propene:

- Step 1 — the pi electrons attack ${\mathrm{H}}^{+}$, forming a carbocation; the more stable secondary carbocation, $\mathrm{C}{\mathrm{H}}_3\stackrel{+}{\mathrm{C}}\mathrm{HC}{\mathrm{H}}_3$, forms rather than the primary one
- Step 2 — $\mathrm{B}{\mathrm{r}}^{-}$ attacks the carbocation, giving 2-bromopropane:

$$\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+\mathrm{HBr}\to \mathrm{C}{\mathrm{H}}_3\mathrm{CHBrC}{\mathrm{H}}_3$$

Markovnikov's rule. When an unsymmetrical reagent adds to an unsymmetrical alkene, the negative part of the reagent attaches to the carbon with fewer hydrogen atoms — because that route passes through the more stable carbocation.

Anti-Markovnikov addition — the peroxide effect. With HBr and an organic peroxide, the reaction follows a free radical mechanism:

- The peroxide produces $\mathrm{B}{\mathrm{r}}^{\bullet }$, which adds to the end carbon to give the more stable secondary radical
- That radical takes a hydrogen atom from HBr, giving 1-bromopropane:

$$\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+\mathrm{HBr}\stackrel{\text{peroxide}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}$$

- The peroxide effect works only with HBr — the H-Cl bond is too strong to be broken by radicals, and iodine radicals combine to form ${\mathrm{I}}_2$ instead of adding

Other additions and reactions:

- Bromine adds across the double bond, decolourising reddish-brown bromine water — a test for unsaturation
- Cold concentrated sulphuric acid adds, and warming with water then gives an alcohol, following Markovnikov's rule
- Ozonolysis splits the double bond into carbonyl compounds, and cold dilute alkaline $\mathrm{KMn}{\mathrm{O}}_4$ is decolourised

Worked example. 2-Methylpropene with HBr gives 2-bromo-2-methylpropane through a tertiary carbocation; with HBr and peroxide it gives 1-bromo-2-methylpropane.

An everyday example. Polythene carry bags and milk pouches are made by addition polymerisation of ethene, as double bonds open up and the units link into long chains.

The substance. Markovnikov's rule is really a rule about carbocation stability — once you know which carbocation is more stable, you can predict the product without memorising the rule.