Worked example 1
Write the balanced equation for the complete combustion of butane, C₄H₁₀.
Try it first: Balance the carbons and hydrogens before you touch the oxygens.
0 of 4 steps revealed.
What you'll be able to do: Classify an organic reaction as substitution, addition, elimination, condensation or combustion and predict its organic product.
Most introductory organic reactions fall into a short list of types. Once you can name the type from what happens to the bonds, predicting the product becomes routine.
These are recommended, not required. You can start this lesson at any time.
One atom or group is swapped for another while the carbon skeleton stays intact. Alkanes undergo free-radical halogenation with a halogen under UV light, and haloalkanes are substituted by nucleophiles such as OH⁻ to give alcohols.
For methane chlorination under UV light, CH₄ + Cl₂ → CH₃Cl + HCl shows substitution: one C–H is replaced by C–Cl. Further substitution can occur, so the first equation is not a promise of one pure product. In aqueous OH⁻ substitution of a simple haloalkane, R–Br + OH⁻ → R–OH + Br⁻, the leaving group is replaced. Reagent, substrate and solvent matter; calling a molecule saturated does not mean substitution is its only possible reaction.
CH₄ + Cl₂ → CH₃Cl + HCl (UV light)
A pi bond opens and two fragments add across it, so a double bond becomes single. Hydrogenation adds H₂ over a metal catalyst; halogenation adds Br₂; hydration adds water with an acid catalyst; hydrohalogenation adds HX. Addition across a multiple bond is characteristic of unsaturated substrates; the specific bond and conditions determine the reaction.
| Transformation of propene | Illustrative conditions | Main introductory product |
|---|---|---|
| Hydrogenation, + H₂ | Suitable metal catalyst | Propane |
| Halogen addition, + Br₂ | Inert nonaqueous solvent | 1,2-dibromopropane |
| Hydration, + H₂O | Acid catalysis | Propan-2-ol |
| Hydrohalogenation, + HBr | Ordinary ionic conditions, no peroxides | 2-bromopropane |
For the unsymmetrical additions of H₂O and HBr, the usual major-product pattern places H on the carbon already bearing more H, and OH or Br on the more substituted carbon. This Markovnikov pattern has conditions and exceptions: peroxide-initiated HBr addition follows a different radical pathway, and rearrangements can matter for other substrates. Bromine water can produce a bromohydrin instead of the simple dibromide, so a product prediction should include its reaction medium. Alkynes can add successive equivalents, giving different products after one versus two additions.
CH₂=CH₂ + Br₂ → CH₂Br–CH₂Br
The reverse of addition: a small molecule such as H₂O or HX leaves and a pi bond forms. Dehydrating an alcohol with concentrated sulfuric acid gives an alkene.
CH₃CH₂OH → CH₂=CH₂ + H₂O (conc. H₂SO₄, heat)
A condensation joins species while eliminating a small molecule. In acid-catalyzed esterification, CH₃COOH + CH₃OH ⇌ CH₃COOCH₃ + H₂O forms methyl ethanoate. It is reversible: excess reactant or removal of water can favor ester formation. Acid-catalyzed hydrolysis uses water to regenerate acid and alcohol; base hydrolysis instead gives carboxylate plus alcohol under the usual conditions.
Amide bonds can be made by suitable condensation or activated-acyl chemistry. Simply mixing a carboxylic acid with an amine usually first gives an ammonium carboxylate by proton transfer; a direct high-yield amide is not guaranteed without appropriate activation or dehydration conditions. Peptide bonds are amide linkages, but their controlled formation requires more than identifying N and COOH groups.
CH₃COOH + CH₃CH₂OH ⇌ CH₃COOCH₂CH₃ + H₂O
Complete combustion in plenty of oxygen gives CO₂ and H₂O and releases a lot of energy. Limited oxygen gives incomplete combustion, producing CO or soot along with water. Balance combustion by carbons first, then hydrogens, then oxygens last.
For hydrocarbon CxHy, complete combustion is CxHy + (x + y/4)O₂ → x CO₂ + (y/2)H₂O. Start by counting C and H, then oxygen, and multiply the whole equation if integer coefficients are wanted. For ethane: 2 C₂H₆ + 7 O₂ → 4 CO₂ + 6 H₂O.
Incomplete combustion is not one unique reaction because product proportions depend on oxygen supply and conditions. If the problem specifies CO only as the carbon product, balance 2 C₂H₆ + 5 O₂ → 4 CO + 6 H₂O. If it specifies soot, 2 C₂H₆ + 3 O₂ → 4 C + 6 H₂O. A mixture of CO₂, CO and soot requires its stated composition; do not assume one from the phrase “limited oxygen” alone. State water's phase when using a combustion enthalpy.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
Elimination removes atoms or groups from adjacent carbons and creates a π bond; addition consumes a π bond while installing groups across it.
Tracking both the carbon skeleton and the small molecule lost or added clarifies how hydration of an alkene relates to dehydration of an alcohol. Reaction conditions determine which direction is favored.
An atom or group replaces another; the number of bonds to carbon does not change.
Two fragments add across a multiple bond, reducing unsaturation.
Two molecules join with the loss of a small molecule, usually water.
Alkanes substitute; alkenes and alkynes add.
Balance C, then H, then O, and use fractional O₂ coefficients before doubling if needed.
CₙH₂ₙ₊₂ + ((3n + 1)/2) O₂ → n CO₂ + (n + 1) H₂O
R–CH=CH–R′ + H₂ → R–CH₂–CH₂–R′ (Ni or Pt)
R–COOH + R′–OH ⇌ R–COO–R′ + H₂O
Write the balanced equation for the complete combustion of butane, C₄H₁₀.
Try it first: Balance the carbons and hydrogens before you touch the oxygens.
0 of 4 steps revealed.
Propene reacts with bromine. Name the reaction type and give the product.
Try it first: What kind of bond does propene have that propane does not?
0 of 4 steps revealed.
Classify CH₃CH₂OH → CH₂=CH₂ + H₂O and the reverse hydration CH₂=CH₂ + H₂O → CH₃CH₂OH.
Try it first: Look for the C=C bond appearing or disappearing.
0 of 3 steps revealed.
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Expecting an alkene to undergo substitution with Br₂. | The pi bond is more reactive than any C–H bond, so bromine adds across it instead of replacing hydrogen. | Check for unsaturation first: unsaturated means addition, saturated means substitution. |
| Balancing oxygen before carbon and hydrogen in a combustion equation. | Oxygen appears in two products, so its count keeps changing until the others are fixed. | Balance C, then H, then O, allowing a fractional O₂ and doubling at the end. |
| Forgetting the water in esterification. | Condensation always expels a small molecule, so the equation will not balance without it. | Count atoms on both sides and add H₂O to close the gap. |
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After the lesson · Question bank
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19 topic questions across 2 difficulty levels.
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Classify reactions by comparing bonds: substitution replaces a group, addition consumes a multiple bond, elimination creates one, and condensation joins species with loss of a small molecule. Product prediction requires substrate and conditions, including solvent, catalyst and regiochemical pathway. Esterification is reversible and amide synthesis requires suitable chemistry beyond simple acid–amine mixing. Balance complete or specified incomplete combustion from the actual carbon products, then hydrogen and oxygen.
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