UNIT 2 CHEMISTRY · PEARSON AAQ · E1.1
Key Terms in Organic Chemistry
FREE CHEMISTRY REVISION · TOPIC 50
Key Terms in Organic Chemistry
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.1
What you will learn
Classify organic compounds by their elements, carbon arrangement, bonds and functional groups. Use general formulae with their stated restrictions. This lesson introduces the language of Section E; detailed representation techniques belong to E1.2 and systematic naming rules to E1.3. You do not need those later techniques to understand these classifications.
1. Hydrocarbons, saturation and unsaturation
A hydrocarbon contains carbon and hydrogen only. A compound containing oxygen, nitrogen or a halogen is not a hydrocarbon, even if it contains a long carbon chain.
A saturated hydrocarbon has only single bonds between carbon atoms. Its carbon atoms carry as many hydrogens as their carbon–carbon connections allow. An unsaturated hydrocarbon contains a carbon–carbon multiple bond, such as C=C or C≡C. The alkene family in this lesson contains a C=C double bond.
A carbon–carbon double bond counts as two bonds when checking carbon’s usual valency of four; a triple bond counts as three. Hydrogen normally forms one bond in these neutral molecules. Do not count a double bond as one when checking how many hydrogens a carbon can carry.
| Description | Hydrocarbon? | Saturation classification |
|---|---|---|
| CH₃–CH₃ | Yes: C and H only | Saturated; C–C is single |
| CH₂=CH₂ | Yes: C and H only | Unsaturated; contains C=C |
| CH₃–CH₂–OH | No: contains oxygen | A saturated alcohol; it is not a saturated hydrocarbon |
| CH₃–CH₂–Cl | No: contains chlorine | A saturated halogenoalkane; it is not a hydrocarbon |
Saturated/unsaturated describes carbon bonding, whereas hydrocarbon describes which elements are present. Keep these questions separate. In this introductory context, an O–H group does not itself make a carbon chain unsaturated.
2. Straight-chain, branched-chain and cyclic
A straight-chain compound has an unbranched, open carbon chain. “Straight” describes connectivity: a bent or zigzag drawing can still show an unbranched chain. A branched-chain compound has a carbon branch attached to the main chain. A cyclic compound has atoms connected in a ring; the examples here have carbon rings.
A ring does not automatically contain a double bond. A carbon ring with only single carbon–carbon bonds is saturated. Cyclic compounds can also contain C=C bonds and be unsaturated. Count bonds from the actual structure rather than guessing from the presence of a ring.
Worked classification 1 — same formula, different arrangement
The first two diagrams both contain four C atoms and ten H atoms. In the straight-chain example the carbons form one unbranched sequence. In the branched example, the central CH carbon is bonded to three other carbon atoms and one H; this gives a branch. Both are saturated hydrocarbons with formula C₄H₁₀.
Counting the branched diagram: three CH₃ groups contribute 9 H, and the central CH contributes 1 H; total 10. A bend in a drawing alone would not prove branching.
Worked classification 2 — formula alone can be ambiguous
The three-carbon ring contains three CH₂ groups: C₃H₆. All its carbon–carbon bonds are single, so it is saturated. The open-chain structure CH₃–CH=CH₂ also has formula C₃H₆, but contains a double bond and is unsaturated.
Therefore C₃H₆ alone does not prove that a molecule is an alkene. You need bonding or structural information. The acyclic-alkene formula has a restricted scope.
3. Homologous series and functional groups
A homologous series is a family of organic compounds sharing a general formula and the same characteristic functional group or bonding pattern. Successive members differ by CH₂. They have similar chemical behaviour and show systematic physical-property trends as chain length changes; members do not have identical properties.
A functional group is an atom or group of atoms responsible for characteristic chemical reactions. For example, C=C identifies the alkene functional group and –OH is the hydroxyl group of an alcohol. Alkanes contain only C–C and C–H single bonds; they do not have a separate characteristic group such as –OH.
| Family | Characteristic feature | Example |
|---|---|---|
| Alkanes | Saturated acyclic hydrocarbon skeleton | CH₃–CH₃ |
| Alkenes | Carbon–carbon double bond, C=C | CH₂=CH₂ |
| Halogenoalkanes | C–X, where X is a halogen | CH₃–CH₂–Cl |
| Alcohols | –OH attached to a saturated carbon atom | CH₃–CH₂–OH |
Halogens commonly represented by X are F, Cl, Br and I. In a halogenoalkane, a halogen replaces a hydrogen of an alkane skeleton. In an alcohol, an –OH group replaces a hydrogen. The hydrogen within –OH still counts in the molecular formula.
Worked classification 3 — an alcohol is not a hydrocarbon
CH₃–CH₂–OH contains an –OH group attached to a saturated carbon atom, so it is an alcohol. Count atoms: C=2; H=3+2+1=6; O=1. Molecular formula: C₂H₆O.
It is not a hydrocarbon because oxygen is present. The C–C bond is single, so the carbon skeleton is saturated. These statements are compatible.
4. General formulae: state the restrictions
A general formula expresses the relationship between atom counts for a specified family. Here n is the number of carbon atoms in one molecule, and X represents one halogen atom. The formulae below apply to the simple families stated; rings, extra double bonds or multiple functional groups can change them.
| Family and scope | General formula | Minimum n | For n=3 |
|---|---|---|---|
| Saturated acyclic alkanes | CₙH₂ₙ₊₂ | 1 | C₃H₈ |
| Acyclic alkenes with exactly one C=C | CₙH₂ₙ | 2 | C₃H₆ |
| Saturated acyclic monohalogenoalkanes | CₙH₂ₙ₊₁X | 1 | C₃H₇X |
| Saturated acyclic monohydric alcohols | CₙH₂ₙ₊₁OH, equivalent to CₙH₂ₙ₊₂O | 1 | C₃H₇OH = C₃H₈O |
Monohalogeno means one halogen substituent. Monohydric means one alcohol –OH group. The alcohol formula has 2n+1 hydrogens attached to the carbon skeleton plus one in –OH: total 2n+2. These are neutral closed-shell molecules in the simple structures considered here.
The general formula is a useful check, but not always a unique identification. Saturated single-ring hydrocarbons can also have CₙH₂ₙ. Different oxygen-containing families can share CₙH₂ₙ₊₂O; an alcohol requires the appropriate –OH connectivity.
Worked classification 4 — one halogen replaces one H
For a saturated acyclic monobromo compound with n=3: H=2(3)+1=7, giving C₃H₇Br. The corresponding alkane would be C₃H₈. Replacing one H by Br reduces the hydrogen count by one; it does not add Br while retaining all eight hydrogens.
A carbon still needs four bonds. In CH₃–CH₂–CH₂–Br, terminal CH₂ next to Br has two C/H-independent bonds: one to C and one to Br; its two H atoms complete four.
Worked classification 5 — substitute n carefully
For five-carbon members within the table’s restrictions:
| Family | Substitution | Formula |
|---|---|---|
| Alkane | H=2(5)+2=12 | C₅H₁₂ |
| Alkene | H=2(5)=10 | C₅H₁₀ |
| Monochloroalkane | H=2(5)+1=11 | C₅H₁₁Cl |
| Monohydric alcohol | Carbon-skeleton H=11; OH adds 1 H | C₅H₁₁OH = C₅H₁₂O |
Do not read C₅H₁₁OH as having only eleven hydrogens. The separate H in OH raises the total to twelve.
5. Apply family relationships without overclaiming
Worked classification 6 — successive homologues
CH₃OH and C₂H₅OH are both saturated acyclic monohydric alcohols. Written as molecular formulae, they are CH₄O and C₂H₆O. The difference is C₁H₂ = CH₂, so they are successive members of this homologous series.
Using Ar(C)=12.0 and Ar(H)=1.0, the relative molecular mass difference is 12.0+2(1.0)=14.0. The same –OH group explains similar characteristic chemistry, but the compounds need not have the same boiling point or solubility.
Worked classification 7 — the functional group must be present
Two supplied structures are CH₃–CH₂–OH and CH₃–O–CH₃. Both have molecular formula C₂H₆O. The first has an O–H bond and is an alcohol. The second has oxygen between two carbons and has no –OH group, so it is not an alcohol.
The second structure is shown only to demonstrate why a molecular formula alone is insufficient. You do not need naming or reaction rules for that other family here. Identify the connectivity actually supplied.
Classification order: identify the elements → inspect the carbon connections → look for multiple bonds and functional groups → count atoms → compare with a general formula only within its stated scope.
Common misconceptions
- Every carbon compound is a hydrocarbon: a hydrocarbon contains C and H only.
- A ring is always unsaturated: inspect whether carbon–carbon multiple bonds are present.
- CₙH₂ₙ always means alkene: a saturated single ring can have the same formula.
- A bent chain is branched: branching concerns carbon connections, not drawing shape.
- CₙH₂ₙ₊₁OH has 2n+1 total H atoms: include the H in OH.
- Any molecule containing O is an alcohol: the appropriate –OH connectivity is needed.
- All homologues have identical properties: similar chemistry and physical trends do not mean identical properties.
- A double bond counts as one for valency: it contributes two towards carbon’s usual four.
Examination tips
- State both evidence and classification: “contains only C/H and a C=C, so it is an unsaturated hydrocarbon.”
- Read atom groups carefully and include H attached to oxygen.
- State acyclic and one double bond when using the simple alkene formula.
- Use the supplied structure when a formula matches more than one family.
- Check C has four bonds, H and halogens one, and O two in these neutral examples.
- Leave detailed drawing conventions and systematic naming rules for the following topics.
Check your understanding
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