Predict the major product of HCl adding to 2-methylpropene, (CH_3)_2C=CH_2, and name the carbocation type. [2 marks]

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Why are alkenes reactive, and how does electrophilic addition explain their reactions and product distribution?

Describe the reactions of alkenes including electrophilic addition of hydrogen halides, halogens and water, oxidation, and the mechanism of electrophilic addition including Markovnikov's rule and carbocation stability

A focused answer to the H2 Chemistry learning outcome on alkenes. The reactive C=C pi bond, electrophilic addition of hydrogen halides, halogens and water, oxidation reactions, and the electrophilic addition mechanism with Markovnikov's rule explained by carbocation stability.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

SEAB wants you to describe the reactions of alkenes (electrophilic addition of hydrogen halides, halogens and water, plus oxidation), and to set out the electrophilic addition mechanism, using Markovnikov's rule and carbocation stability to predict the major product. The electrophilic addition mechanism and the bromine test are guaranteed exam content.

The answer

Why alkenes are reactive

An alkene has a carbon-carbon double bond: a sigma bond plus a pi bond. The pi bond is a region of high electron density above and below the plane of the molecule, exposed and relatively weak. It attracts electrophiles (electron-poor species), so the characteristic reaction is electrophilic addition across the double bond.

The key addition reactions

Hydrogen halides (e.g. HBr): $\text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}$ (room temperature).
Halogens (e.g. $\text{Br}_2$ ): $\text{CH}_2\text{=CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}$ . The rapid decolourising of bromine is the standard test for a C=C.
Water (steam, phosphoric acid catalyst, high temperature and pressure): hydration to an alcohol, $\text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ (industrial ethanol).
Hydrogen (nickel catalyst): addition to the alkane.

Oxidation reactions

Cold dilute acidified $\text{KMnO}_4$ : gives a diol (the manganate(VII) is decolourised, purple to colourless), e.g. ethene to ethane-1,2-diol.
Hot concentrated acidified $\text{KMnO}_4$ : cleaves the double bond, giving carbonyl compounds, carboxylic acids, or carbon dioxide depending on the substitution at each carbon. This is used to deduce the position of a double bond from the products.

The electrophilic addition mechanism

Using HBr adding to ethene:

The pi electrons attack the slightly positive H of HBr (the electrophile). The H-Br bond breaks heterolytically, giving a carbocation and a bromide ion.
The bromide ion (a nucleophile) attacks the carbocation, forming the product.

Curly arrows show the movement of electron pairs: from the pi bond to the H, and from the H-Br bond to the Br.

Markovnikov's rule and carbocation stability

When the alkene is unsymmetrical (e.g. propene), two carbocations are possible. The more stable carbocation forms preferentially, giving the major product. Stability order:

\text{tertiary} > \text{secondary} > \text{primary}

because alkyl groups donate electron density (the inductive effect) and stabilise the positive charge. Markovnikov's rule summarises the outcome: the hydrogen adds to the carbon already bearing the most hydrogens (via the more stable carbocation).

Worked example

Predict the major product when hydrogen bromide adds to but-1-ene, $\text{CH}_2\text{=CHCH}_2\text{CH}_3$ , and justify using carbocation stability.

The pi bond attacks the H of HBr. Two carbocations are possible:

H adds to C1, giving a secondary carbocation on C2: $\text{CH}_3\text{-CH}^+\text{-CH}_2\text{CH}_3$ .
H adds to C2, giving a primary carbocation on C1: $^+\text{CH}_2\text{-CH}_2\text{CH}_2\text{CH}_3$ .

The secondary carbocation is more stable (two alkyl groups donate electron density and stabilise the positive charge), so it forms preferentially.

The bromide then attacks C2, giving the major product 2-bromobutane, in line with Markovnikov's rule.

Try it: Electrophilic addition predictor - choose an alkene and electrophile to see the carbocation intermediates and the Markovnikov product.

Examples in context

Example 1. Industrial hydration of ethene. Most industrial ethanol is made by the acid-catalysed hydration of ethene from cracking, an electrophilic addition of water across the double bond. SEAB contrasts this with fermentation, asking candidates to compare the rate, purity and sustainability of the two routes, linking the mechanism to industrial chemistry.

Example 2. Locating a double bond by oxidative cleavage. Treating an alkene with hot concentrated acidified manganate(VII) breaks the C=C and gives carbonyl or carboxylic acid fragments. By identifying the fragments, the original position of the double bond can be deduced, a structure-determination technique that ties the alkene reactions to the analytical content area.

Try this

Q1. Describe a test to distinguish ethene from ethane, with the observation. [2 marks]

Cue. Add bromine water; ethene rapidly decolourises it (orange to colourless), ethane does not (without UV).

Q2. Predict the major product of HCl adding to 2-methylpropene, $(\text{CH}_3)_2\text{C=CH}_2$ , and name the carbocation type. [2 marks]

Cue. Major product 2-chloro-2-methylpropane; via the tertiary carbocation (most stable).

Q3. Write the equation and conditions for the industrial hydration of ethene to ethanol. [2 marks]

Cue. $\text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ ; steam, phosphoric acid catalyst, high temperature and pressure.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Specimen (9729)5 marksWhen hydrogen bromide is added to propene, the major product is 2-bromopropane rather than 1-bromopropane. Outline the mechanism of the reaction and use it to explain why 2-bromopropane is the major product.

Show worked answer →

The reaction is electrophilic addition.

Step 1: the pi electrons of the C=C attack the slightly positive hydrogen of HBr (the electrophile). The H-Br bond breaks heterolytically. This can form either a secondary carbocation (positive on the middle carbon) or a primary carbocation (positive on the end carbon).

The secondary carbocation (CH3-CH+-CH3) is more stable than the primary carbocation (CH3-CH2-CH2+) because the two alkyl groups donate electron density (the inductive effect) and stabilise the positive charge.

Step 2: the bromide ion attacks the more stable secondary carbocation, giving 2-bromopropane as the major product.

This is Markovnikov's rule: the hydrogen adds to the carbon already bearing more hydrogens, via the more stable carbocation.

Markers reward the pi-electron attack on the electrophile, the two possible carbocations, the greater stability of the secondary carbocation with reason, and the bromide attack giving the major product.

2023 (style)4 marksDescribe a chemical test, including reagent, conditions and observation, to distinguish between hexane and hex-1-ene, and write the equation for the positive result.

Show worked answer →

Reagent: bromine (as bromine water or in an inert solvent), at room temperature, no UV needed.

Observation: hex-1-ene (the alkene) rapidly decolourises the orange/brown bromine; hexane (the alkane) does not decolourise it (without UV light).

Equation for the positive result (electrophilic addition):
CH2=CH(CH2)3CH3 + Br2 -> CH2Br-CHBr(CH2)3CH3 (1,2-dibromohexane).

Markers reward bromine as the reagent, the rapid decolourisation by the alkene only, and the addition equation.