UNIT 2 CHEMISTRY · PEARSON AAQ · E1.7.1
Addition Reactions of Alkenes
FREE CHEMISTRY REVISION · TOPIC 56
Addition Reactions of Alkenes
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.7.1
What you will learn
Predict the products of alkene addition with hydrogen, halogens, hydrogen halides and steam. Write balanced equations and suitable structural formulae, name products and state the conditions. Pearson E1.7.1 requires product prediction and conditions for symmetric and asymmetric alkenes; reaction mechanisms and practical techniques are not required.
1. The common pattern
An alkene contains a C=C double bond: one σ and one π contribution. In a simple addition across this bond, the C–C σ framework remains, the π contribution is replaced and a new single bond forms from each original double-bond carbon to an incoming atom or group. The product has C–C rather than C=C at that position. No small molecule is eliminated.
| Added reagent | Groups added across C=C | Product family | Conditions to recall |
|---|---|---|---|
| Hydrogen, H₂ | H and H | Alkane | Nickel catalyst and heat; a common teaching example is about 150 °C |
| Halogen, especially Br₂ or Cl₂ | X and X | Dihalogenoalkane with halogens on adjacent carbons | Room temperature in a suitable inert, non-aqueous solvent; no UV needed |
| Hydrogen halide, e.g. HCl or HBr | H and X | Halogenoalkane | Usually room temperature; no UV or metal catalyst needed for these simple examples |
| Steam, H₂O(g) | H and OH | Alcohol | Acid catalyst; ethene example: H₃PO₄, about 300 °C and 60–70 atm |
These are representative conditions, not one universal recipe for every substrate and catalyst. Platinum or palladium can also catalyse hydrogenation, often under milder conditions. Chlorine and bromine are the standard halogen examples: do not assume iodine adds readily or that fluorine is safely interchangeable. Halogen-in-water chemistry can differ from simple addition in an inert solvent; use the conditions stated in the question.
2. Hydrogenation
Worked reaction 1 — ethene with hydrogen
CH₂=CH₂ + H₂ → CH₃CH₃
Ethene + hydrogen → ethane. Use nickel catalyst and heat, for example about 150 °C. One H attaches to each original double-bond carbon, changing both CH₂ groups to CH₃. Atom check: C₂H₄+H₂=C₂H₆. Only one H₂ molecule is needed per C=C.
Worked reaction 2 — propene with hydrogen
CH₂=CHCH₃ + H₂ → CH₃CH₂CH₃
Propene → propane, with H₂/Ni and heat. The original CH₂ end becomes CH₃; the original CH becomes CH₂. The existing terminal CH₃ remains attached. C₃H₆+H₂=C₃H₈. Adding identical H atoms makes reversal of the addition direction give the same connectivity.
3. Addition of halogens
Worked reaction 3 — bromine and ethene
CH₂=CH₂ + Br₂ → BrCH₂CH₂Br
Product: 1,2-dibromoethane. Use bromine at room temperature in a suitable inert, non-aqueous medium; UV is not required. One Br attaches to each original double-bond carbon. Both carbons remain CH₂, so C₂H₄+Br₂=C₂H₄Br₂. Do not replace an H with Br or place both Br atoms on the same carbon.
Worked reaction 4 — chlorine and propene
CH₂=CHCH₃ + Cl₂ → CH₂ClCHClCH₃
Product: 1,2-dichloropropane, under room-temperature alkene halogen-addition conditions. Carbon 1 gains Cl and keeps two H; carbon 2 gains Cl and keeps one H. C₃H₆+Cl₂=C₃H₆Cl₂. The alkene is asymmetric, but identical Cl atoms added at either end give the same product connectivity.
Worked reaction 5 — bromine and but-2-ene
CH₃CH=CHCH₃ + Br₂ → CH₃CHBrCHBrCH₃
Product connectivity: 2,3-dibromobutane. Use room temperature and a suitable inert solvent. Each central carbon gains one Br. C₄H₈+Br₂=C₄H₈Br₂. This structural equation does not specify the product stereochemistry; do not infer a stereoisomer ratio from it.
4. Addition of hydrogen halides
Worked reaction 6 — ethene with hydrogen chloride
CH₂=CH₂ + HCl → CH₃CH₂Cl
Ethene gives chloroethane at room temperature. H adds to one carbon and Cl to the other; because ethene is symmetric, reversing these placements gives the same connectivity. C₂H₄+HCl=C₂H₅Cl. This is addition, not replacement of hydrogen.
For an asymmetric alkene, adding two different groups in opposite directions can give different connectivities called regioisomers. Recognise the possible structures before deciding which is favoured. For simple ordinary HX addition without peroxide-induced alternative chemistry, the usual Markovnikov prediction is that H adds preferentially to the double-bond carbon already bearing more H, while X adds to the other. This is a product-prediction guide, not a requirement to draw a mechanism. It does not justify asserting the same exclusivity or product ratio for every reaction.
Worked reaction 7 — propene with HBr
CH₂=CHCH₃ + HBr → CH₃CHBrCH₃ (usual major product).
At room temperature under ordinary addition conditions without peroxides, predict 2-bromopropane as the principal product. The CH₂ end initially has more H, so gains H; the middle carbon gains Br. Atom check: C₃H₆+HBr=C₃H₇Br. Opposite orientation has the connectivity CH₂BrCH₂CH₃, 1-bromopropane; its formation and relative amount depend on conditions. Do not claim an exclusive product from atom balance alone.
Worked reaction 8 — pent-2-ene with HBr
CH₃CH=CHCH₂CH₃ + HBr → CH₃CHBrCH₂CH₂CH₃
or → CH₃CH₂CHBrCH₂CH₃
These are alternative equations, each balanced separately; do not put a full mole of both products on the right of one 1:1 equation. Products: 2-bromopentane and 3-bromopentane. Both original double-bond carbons have one H and the carbon ends are not equivalent. Ordinary HX addition can give a mixture; neither symmetry nor the simple hydrogen-count rule establishes an exclusive product or a 50:50 ratio. Each product is C₅H₁₁Br.
5. Addition of steam: hydration
Worked reaction 9 — ethene to ethanol
CH₂=CH₂(g) + H₂O(g) ⇌ CH₃CH₂OH(g)
Steam adds H and OH across C=C. Product: ethanol. Representative conditions: phosphoric acid (H₃PO₄) catalyst, about 300 °C and 60–70 atm. The high-temperature reaction mixture is gaseous; ethanol can subsequently be cooled and condensed. C₂H₄+H₂O=C₂H₆O. The OH oxygen is bonded to carbon and hydrogen: its two-bond valency is preserved.
Hydration is reversible, so do not imply complete conversion merely because the equation is balanced. A catalyst increases the rate of reaching equilibrium; it does not change the equilibrium yield at a fixed temperature. Hydrogenation adds H₂ and makes an alkane; hydration adds H₂O and makes an alcohol.
Worked reaction 10 — asymmetric hydration prediction
CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃ (usual major alcohol under acid-catalysed hydration conditions).
Propene gives propan-2-ol as the usual principal product. H adds to the CH₂ end and OH to the more substituted carbon. State steam and an acid catalyst with heat/pressure appropriate to the specified process; the quoted 300 °C and 60–70 atm values above describe the ethene example, not a universal operating recipe for all alkenes. C₃H₆+H₂O=C₃H₈O. Do not silently assume every asymmetric alkene gives only one product under every set of conditions.
6. A reliable equation-writing method
- Locate and number the two carbons in C=C. Retain all carbon–carbon connections and branches.
- Replace C=C with C–C and attach one incoming atom/group to each of those two carbons.
- Retain the original H atoms, then add the reagent’s atoms; check carbon total bond order four, oxygen two and halogens one.
- Name the product from its actual structure, including halogen/OH positions. Consider alternative orientations when the incoming groups differ and the alkene is asymmetric.
- Check each element balances, then add conditions above the arrow or alongside the equation. If a displayed formula is requested, show every atom and every bond.
| Exam-style application | Answer with reasoning |
|---|---|
| 0.120 mol propene undergoes complete hydrogenation. How much H₂ reacts? | 0.120 mol H₂: one mole H₂ per mole of a monoalkene. Product is 0.120 mol propane if conversion is complete. |
| What reagent converts but-2-ene to 2,3-dibromobutane? | Br₂, room temperature in a suitable inert solvent. Put one Br on each former double-bond carbon. |
| Why is CH₃CH₂Br not the product of ethene plus Br₂? | It contains only one Br and one extra H. Br₂ addition gives BrCH₂CH₂Br. |
| Can a balanced formula alone choose a regioisomer? | No. Atom balance checks totals, while connectivity and reaction conditions determine product identity and distribution. |
| Write ethanol formation without hiding the OH group | CH₂=CH₂ + H₂O ⇌ CH₃CH₂OH, with steam/H₃PO₄, about 300 °C and 60–70 atm for the ethene process. |
Common misconceptions
- Hydrogenation and hydration are the same: H₂ gives an alkane; H₂O gives an alcohol.
- Both atoms from Br₂ attach to one carbon: simple alkene addition places them on adjacent former double-bond carbons.
- A hydrogen is removed in addition: all reagent atoms are incorporated in these simple equations.
- Every asymmetric alkene yields two equally abundant products: symmetry, orientation and conditions matter.
- Every asymmetric alkene yields exactly one product: pent-2-ene/HBr provides a counterexample.
- Water at room temperature alone is the stated steam-hydration process: the required catalyst and operating conditions must be included.
- The acid catalyst changes equilibrium yield: it changes rate, not the equilibrium constant at fixed temperature.
Examination tips
- Write structural rather than molecular formulae when the question tests product position.
- Use one equivalent of reagent per C=C for these monoalkene examples.
- Differentiate halogen X₂ from hydrogen halide HX.
- Give the reagent, catalyst and temperature/pressure where needed; do not substitute a mechanism for conditions.
- For asymmetric addition, state possible regioisomers and qualify any major-product prediction.
- Check balanced atoms and valencies independently of naming.
Reference checks: OpenStax hydrogenation; halogen addition; HX orientation and pent-2-ene product mixture. Representative ethene hydration conditions are also stated in Pearson’s published chemistry specification; the required AAQ lesson scope is linked below. Mechanistic material in reference chapters is not required here.
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