# Reactions of Hydrocarbons

> Addition reactions of alkenes with hydrogen, halogens, hydrogen halides and water, Markovnikov's rule, the bromine water test, and substitution reactions of alkanes with halogens in UV light.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/reactions-of-hydrocarbons/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The banana that ripened the rest
Long ago, fruit sellers noticed that one ripe banana in a box made all the others ripen faster. In 1934 scientists found why: ripe fruit gives off a gas called **ethene**, a tiny alkene that tells other fruit to ripen.

The same small molecule is also the most important starting material in the chemical industry. Every year, factories turn hundreds of millions of tonnes of ethene into plastics, antifreeze and ethanol. All of it depends on one weak spot in the molecule: the double bond.

**In short:** Alkenes react by addition: hydrogen, halogens, hydrogen halides or water add across the C=C double bond. Alkenes decolourise bromine water quickly and alkanes do not, which tells them apart, and alkanes react with halogens only by substitution in UV light.

## Two kinds of reaction

| Reaction | Hydrocarbon | What happens |
|---|---|---|
| **Addition** | Alkenes and alkynes (unsaturated) | Atoms add across the double or triple bond. One product. |
| **Substitution** | Alkanes (saturated) | A hydrogen atom is swapped for another atom. Two products. |

The C=C double bond has one strong bond and one weaker bond. The weaker bond breaks easily, so alkenes are **much more reactive** than alkanes.

## Addition reactions of alkenes

The double bond opens and one new atom joins each carbon.

| Adding | Name | Conditions | Product from ethene |
|---|---|---|---|
| $\mathrm{H_2}$ | Hydrogenation | Nickel or Pd/Pt catalyst, heat | Ethane |
| $\mathrm{Br_2}$ or $\mathrm{Cl_2}$ | Halogenation | Room temperature, no UV needed | 1,2-dibromoethane |
| $\mathrm{HCl}$ or $\mathrm{HBr}$ | Hydrohalogenation | Room temperature | Chloroethane |
| $\mathrm{H_2O}$ | Hydration | Dilute $\mathrm{H_2SO_4}$ or $\mathrm{H_3PO_4}$ catalyst, heat | Ethanol |

$$\mathrm{CH_2{=}CH_2} + \mathrm{Br_2} \rightarrow \mathrm{CH_2BrCH_2Br}$$

$$\mathrm{CH_2{=}CH_2} + \mathrm{H_2O} \rightarrow \mathrm{CH_3CH_2OH}$$

## The bromine water test

**Bromine water** is orange-brown.

- Shake it with an **alkene**: it turns **colourless immediately**, because the bromine adds across the double bond.
- Shake it with an **alkane** in the dark: **no change**. (In UV light it fades only slowly, by substitution.)

This is the standard test for **unsaturation**. Potassium permanganate solution (purple) is also decolourised by alkenes.

## Markovnikov's rule

When HX or water adds to an **unsymmetrical** alkene like propene, two products are possible:

$$\mathrm{CH_2{=}CHCH_3} + \mathrm{HBr} \rightarrow \mathrm{CH_3CHBrCH_3} \text{ (major)} \text{ or } \mathrm{CH_2BrCH_2CH_3} \text{ (minor)}$$

**Markovnikov's rule**: the H goes to the carbon of the double bond that **already has more hydrogens**. "The rich get richer." So the Br goes to the middle carbon, and **2-bromopropane** is the main product.

For a **symmetrical** alkene, like but-2-ene, both carbons are the same, so only **one** product forms (2-bromobutane).

Adding $\mathrm{H_2}$ or $\mathrm{Br_2}$ never gives two products, because the same atom goes to each carbon.

## Substitution reactions of alkanes

Alkanes do not react with halogens in the dark. In **ultraviolet (UV) light**, a hydrogen atom is replaced by a halogen atom:

$$\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\text{UV light}} \mathrm{CH_3Cl} + \mathrm{HCl}$$

Notice there are **two products**, one of them a hydrogen halide. With more chlorine, more hydrogens are replaced: $\mathrm{CH_2Cl_2}$, $\mathrm{CHCl_3}$ and $\mathrm{CCl_4}$ form, giving a **mixture**.

Substitution is slow and gives mixtures, so making a haloalkane from an alkene (addition, one product) is usually better.

## Comparing addition and substitution

| | Addition (alkene + $\mathrm{Br_2}$) | Substitution (alkane + $\mathrm{Br_2}$) |
|---|---|---|
| Speed | Fast | Slow |
| Conditions | Room temperature, dark is fine | Needs UV light |
| Products | One | Two (or more), including HBr |

## Reaction pathways

Exam questions often ask you to link reactions in a **flow chart**:

$$\text{ethane} \xrightarrow{\mathrm{Cl_2}\text{, UV}} \text{chloroethane}$$

$$\text{ethene} \xrightarrow{\mathrm{H_2O}\text{, acid}} \text{ethanol} \xrightarrow{\text{oxidise}} \text{ethanoic acid}$$

Identify the reagent, conditions and type of reaction for each arrow.

**Mind map**

```text
Hydrocarbon reactions
  Addition (alkenes)
    H2: Ni catalyst
    Br2 or Cl2
    HX
    H2O: acid catalyst
    One product
  Bromine water test
    Alkene: colourless fast
    Alkane: no change in dark
  Markovnikov
    H to carbon with more H
    Unsymmetrical alkenes only
  Substitution (alkanes)
    Halogen + UV light
    Two products
    Mixture forms
```

### Year 11: Balancing organic equations
Count every atom. In addition reactions, everything joins into one product, so balancing is easy. In substitution, remember the second product.

**Example.** Propane and bromine in UV light, replacing one H:

$\mathrm{C_3H_8} + \mathrm{Br_2} \rightarrow \mathrm{C_3H_7Br} + \mathrm{HBr}$

Left: 3 C, 8 H, 2 Br. Right: 3 C, 7 + 1 = 8 H, 1 + 1 = 2 Br. Balanced.

### Year 12: Harder pathways and isomer counts
**How many products?** But-1-ene + HCl gives two products (1-chlorobutane minor, 2-chlorobutane major). But-2-ene + HCl gives only 2-chlorobutane.

**Alkynes** add twice. Ethyne + 2$\mathrm{H_2}$ (Ni) gives ethane. With one mole of $\mathrm{Br_2}$, ethyne gives 1,2-dibromoethene; with two moles, 1,1,2,2-tetrabromoethane.

**Monosubstitution products.** The number of different monochloro products equals the number of different hydrogen environments. Propane has 2 (1-chloropropane and 2-chloropropane). 2-methylpropane has 2. Butane has 2.

**Working backwards.** A compound adds HCl to give 2-chloro-2-methylpentane as the main product. The Cl is on C2, which had fewer H, so the starting alkene had its double bond at C2: 2-methylpent-2-ene or 2-methylpent-1-ene.

**Markovnikov with water.** 2-methylbut-2-ene, $\mathrm{(CH_3)_2C{=}CHCH_3}$, adds water so the H goes to the CH carbon and the OH to the carbon with no H: the main product is **2-methylbutan-2-ol**.

**Planning two steps.** To make a target, first use **addition** to put atoms on an alkene (one product), then **substitution** in UV light to swap one more H. For $\mathrm{CF_3CHClBr}$: $\mathrm{CF_2{=}CHCl}$ adds HF to give $\mathrm{CF_3CH_2Cl}$, then $\mathrm{Br_2}$ in UV light replaces one H.

## Quick summary

- Alkenes undergo **addition**: $\mathrm{H_2}$ (catalyst), halogens, hydrogen halides, water (acid catalyst).
- Bromine water: decolourised fast by alkenes, not by alkanes in the dark.
- Markovnikov: H adds to the carbon that already has more H.
- Alkanes undergo **substitution** with halogens in UV light, giving two products.

## Quiz

Try these questions from the chapter quiz.

1. Alkenes mainly undergo:
   - A. Addition reactions
   - B. Substitution reactions
   - C. Condensation reactions
   - D. No reactions
   - Answer: A. Addition reactions. The double bond opens to add atoms.

2. Alkanes react with chlorine by:
   - A. Addition
   - B. Substitution
   - C. Hydration
   - D. Polymerisation
   - Answer: B. Substitution. A hydrogen is swapped for a chlorine.

3. What do alkanes need to react with halogens?
   - A. Dark conditions
   - B. UV light
   - C. Water
   - D. An acid catalyst
   - Answer: B. UV light. UV light provides the energy to start the reaction.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/reactions-of-hydrocarbons/#quiz