Formulas, functional groups and the naming of organic compounds

1. Overview

Organic chemistry is the chemistry of carbon compounds, and there are millions of them. This section gives you its language: the family a compound belongs to, the formulas used to show its structure, and the rules that turn a structure into a name and back again. Every later organic topic assumes you can read a skeletal formula, count its atoms and name the compound.

Key Definitions

• Hydrocarbon: a compound made up of carbon and hydrogen atoms only.
• Alkane: a hydrocarbon whose molecules contain only C-C and C-H single bonds and no functional group; alkanes that are not cyclic fit the general formula CₙH₂ₙ₊₂.
• Functional group: an atom or group of atoms in an organic molecule that is responsible for the characteristic chemical reactions of the compound and has a large effect on its physical properties.
• Alkyl group: the group of atoms left when one hydrogen atom is removed from an alkane, such as methyl, CH₃-, or ethyl, CH₃CH₂-; in formulas such as R-OH it is shown as R.
• General formula: a formula written in terms of n that fits every member of a homologous series, such as CₙH₂ₙ₊₂ for the alkanes.
• Molecular formula: the actual number of atoms of each element present in one molecule of a compound.
• Empirical formula: the simplest whole-number ratio of the atoms of each element present in a compound.
• Structural formula: the minimum detail needed to show how the atoms in a molecule are arranged without any ambiguity, written with conventional groups, such as CH₃CH(OH)CH₃ for propan-2-ol.
• Displayed formula: a drawing of a molecule that shows every atom and every bond, so it shows how the atoms are placed relative to each other and how many bonds join them.
• Skeletal formula: a simplified drawing in which carbon atoms, and the hydrogen atoms bonded to them, are not written; lines show the bonds, each unlabelled line end or corner is a carbon atom, and all other atoms are written in.

Content

Hydrocarbons and alkanes

A hydrocarbon is a compound of carbon and hydrogen only. Hexane, C₆H₁₄, and ethene, C₂H₄, are hydrocarbons; ethanol, CH₃CH₂OH, is not, because it also contains oxygen.

Alkanes are the simplest hydrocarbons. They contain only C-C and C-H single bonds, so they have no functional group. These bonds are strong and almost non-polar, so alkanes react with very few reagents (see 14.1).

The stem of a name gives the number of carbon atoms in the main chain: meth- (1), eth- (2), prop- (3), but- (4), pent- (5), hex- (6). Take one H off an alkane and you have an alkyl group: methyl, CH₃-, ethyl, CH₃CH₂-, and so on. In formulas such as R-OH, R and R′ stand for alkyl groups (in a few, R can be H, as in methanal, HCHO), and X stands for a halogen atom.

Functional groups

Apart from the alkanes, every AS class is defined by its functional group:

Class	Functional group	Name of the group
alkene	C=C	carbon-carbon double bond
halogenoalkane	R-X	halogen
alcohol	R-OH	hydroxyl
aldehyde	R-CHO	carbonyl
ketone	R-CO-R′	carbonyl
carboxylic acid	R-COOH	carboxyl
ester	R-COO-R′	ester
primary amine	R-NH₂	amine
nitrile	R-C≡N	nitrile

In an aldehyde the C=O carbon is at the end of the chain and carries an H atom. In a ketone it is inside the chain, bonded to two other carbon atoms.

The functional group controls how a compound behaves:

• Chemical properties. Most reactions happen at the functional group, because the C-C and C-H bonds around it are so unreactive. So members of a class have similar reactions, whatever their chain length: every alkene decolourises aqueous bromine, and every carboxylic acid gives carbon dioxide with sodium carbonate.
• Physical properties. The group changes the intermolecular forces. Butane, CH₃CH₂CH₂CH₃, and propan-1-ol, CH₃CH₂CH₂OH, both have 34 electrons per molecule, so their instantaneous dipole-induced dipole (id-id) forces are similar. Yet butane is a gas at room temperature and propan-1-ol is a liquid, because its O-H group lets its molecules form hydrogen bonds with each other.

Primary, secondary and tertiary alcohols and halogenoalkanes have one, two or three carbon atoms bonded to the carbon that carries the -OH or -X, as in butan-1-ol, butan-2-ol and 2-methylpropan-2-ol, (CH₃)₃COH. You will use these types in 15.1 and 16.1. For amines, "primary" counts the carbons bonded to the nitrogen: RNH₂ has one, so even CH₃CH(NH₂)CH₃ is a primary amine. Classifying amines is not tested at AS.

Six kinds of formula

Hexanoic acid has molecular formula C₆H₁₂O₂, empirical formula C₃H₆O (divide by 2), general formula CₙH₂ₙO₂ with n = 6, and structural formula CH₃CH₂CH₂CH₂CH₂COOH. Its displayed formula shows all 20 atoms and every bond. Its skeletal formula is a zigzag of five lines ending in a carbon with a double line to O and a single line to OH.

Structural formulas. Write the chain in order, each carbon followed by its H atoms, with any branch in brackets straight after its carbon, as in CH₃CH(CH₃)CH₂OH. Identical groups on one carbon can be grouped: (CH₃)₃COH. Use the standard groups -CHO (not -COH, which reads as C-O-H), -CO-, -COOH, -COO- (usually acid part first: CH₃COOCH₃), -CN and -NH₂. The = of a C=C may be written (CH₃CH=CHCH₃) or left out (CH₃CHCHCH₃).

Displayed formulas. Write every atom and draw every bond as a line (two for a double bond, three for a triple bond), including each C-H of a CH₃ group and the O-H of an -OH group. The angles you draw need not be the real ones: a displayed formula shows which atoms are bonded, not the shape (13.3).

Skeletal formulas.

1. Each line is a bond. The carbon chain is a zigzag, and every line end and every corner is a carbon atom, unless another symbol is written there.
2. Carbon atoms, and the H atoms on them, are not written.
3. C=C is two parallel lines. C≡N is three lines ending at N; the carbon where they start is part of the chain.
4. Every other atom is written, with any H on it: OH, NH₂, Cl, O. The H of -CHO is on carbon, so it is not drawn: a chain-end carbon with just a double line to O is a CHO group, and a corner with a double line to O is a ketone C=O. (Writing CHO at the chain end is also clear.)

Carbon forms four bonds, so the H atoms on each carbon are 4 minus the bonds drawn to it (a double line counts as two, a triple line as three): a line end with one bond is CH₃, two bonds make CH₂, three make CH and four make C.

General formulas

For molecules with one functional group and no ring, with n as the total number of carbon atoms:

Class	General formula	n = 4
alkane	CₙH₂ₙ₊₂	C₄H₁₀
alkene	CₙH₂ₙ	C₄H₈
halogenoalkane	CₙH₂ₙ₊₁X	C₄H₉Br
alcohol	CₙH₂ₙ₊₁OH	C₄H₉OH
aldehyde, ketone	CₙH₂ₙO	C₄H₈O
acid, ester	CₙH₂ₙO₂	C₄H₈O₂
primary amine	CₙH₂ₙ₊₁NH₂	C₄H₉NH₂
nitrile	CₙH₂ₙ₋₁N	C₄H₇N

Some books write acids as CₙH₂ₙ₊₁COOH, where n counts only the alkyl carbons, so check what n counts. Classes can share a general formula (aldehydes and ketones; acids and esters; alkenes and cycloalkanes), so a molecular formula alone does not fix the class (13.4). Each C=C or ring takes two H atoms off: an alcohol with one C=C is CₙH₂ₙO.

Using a general formula. An unbranched nitrile has $M_{\text{r}}$ = 83.0. From CₙH₂ₙ₋₁N, $M_{\text{r}}$ = 12.0n + 1.0(2n − 1) + 14.0 = 14.0n + 13.0, so n = 5: C₅H₉N. The C≡N is always at a chain end, so this can only be CH₃CH₂CH₂CH₂CN, pentanenitrile.

Molecular and empirical formulas from a structure

Count the carbons, then the H atoms carbon by carbon, then the other atoms (including H on O or N). Write C first, then H, then the other elements in alphabetical order (C₄H₇ClO), and check the H count against the general formula. For the empirical formula, divide by the highest common factor; if it is 1, the two formulas are the same. (Empirical formulas from masses: see 2.3.)

Worked example 1: every formula for one compound

Question: For 2-methylpentanal, write the structural formula, describe the displayed and skeletal formulas, and give the molecular and empirical formulas.

Step 1: decode the name. "-al": an aldehyde, so C1 is the CHO carbon. "pent": a five-carbon chain. "2-methyl": a CH₃ branch on C2.

Step 2: structural formula. From C5 to C1: CH₃CH₂CH₂CH(CH₃)CHO.

Step 3: displayed formula. Five chain carbons in a row. C5 has three H, C4 and C3 two H each, C2 one H and a line to the branch carbon (three H), and C1 a double line to O and a single line to H.

Step 4: skeletal formula. A zigzag of four lines from C5 to C1, a short branch line at C2 (so three lines meet there), and a double line from C1 to O.

Step 5: count. C: 5 + 1 = 6. H: 3 + 2 + 2 + 1 + 3 (branch) + 1 (on C1) = 12. O: 1. This fits CₙH₂ₙO with n = 6. The highest common factor of 6, 12 and 1 is 1.

Answer: CH₃CH₂CH₂CH(CH₃)CHO; molecular formula C₆H₁₂O; empirical formula C₆H₁₂O.

Naming: the method

A name has prefixes (branches and halogens), a stem (chain length) and an ending (the main functional group), as in 4-chloropentan-2-one: prefix "4-chloro", stem "pentan", ending "-2-one".

1. Find the main group. It sets the ending. Halogens and alkyl branches are always prefixes.
2. Find the longest continuous carbon chain that contains the main group (for an alkene, both C=C carbons), following bonds into brackets.
3. Number the chain from the end that gives the main group the lowest number. If that is a tie, give the lowest number to any C=C, then to the prefixes.
4. Add prefixes in alphabetical order, each with its number. For two or three of the same, use di- or tri- and give every number: 2,2-dimethylbutane. Ignore di- and tri- when ordering.
5. Punctuate: commas between numbers, hyphens between numbers and letters, and no spaces except in acids and esters (ethanoic acid, ethyl methanoate).

Class	Ending or prefix: example
alkane	-ane: 2-methylbutane
alkene	-ene: but-1-ene
halogenoalkane	fluoro-, chloro-, bromo-, iodo-: 2-iodopropane
alcohol	-ol: butan-2-ol
aldehyde	-al: pentanal
ketone	-one: pentan-3-one
carboxylic acid	-oic acid: 2-methylpropanoic acid
ester	alkyl ...-oate: ethyl methanoate
primary amine	-ylamine: ethylamine
nitrile	-nitrile: butanenitrile

Rules for particular classes

• The "e" of the stem is dropped before a vowel and kept before a consonant: propan-1-ol, propanal, propanone, but propanenitrile and ethane-1,2-diol.
• Numbers. The carbon of -CHO, -COOH and -C≡N is always C1, so it needs no number: butanal, not butan-1-al. Other groups need one when more than one position is possible (pentan-2-one), but not in ethanol, propene, propanone or butanone.
• Nitriles. The C≡N carbon is counted in the chain. At AS, nitriles are unbranched.
• Esters. RCOOR′ is named from its alcohol and acid: the alkyl group R′ on the single-bonded O comes first, as a separate word, and the part containing C=O (counting the C=O carbon) gives the "-oate" word. Both parts are unbranched, up to six carbons each.
• Amines. CH₃CH₂CH₂NH₂ is propylamine (also 1-aminopropane or propan-1-amine). The amino- form suits NH₂ on a middle carbon: CH₃CH(NH₂)CH₃ is 2-aminopropane.
• OH with another group. With a nitrile, aldehyde, ketone or acid group, that group sets the ending and the OH becomes the prefix hydroxy-: HOCH₂CH₂CN is 3-hydroxypropanenitrile. A C=C stays in the stem as "-en-", and the group that sets the ending gets the lower number (Worked example 4).

Worked example 2: naming from structural formulas

Question: Name (a) CH₃CH(CH₃)CH₂CH(OH)CH₃, (b) CH₃CH₂CH₂CH₂CH₂CN, (c) CH₃CH=C(CH₃)CH₂CH₃ and (d) CH₃CHClCH₂CHO.

(a) Main group OH, so -ol. The longest chain through the C-OH carbon has five carbons. Numbering from the right puts OH on C2 (from the left, C4), and then the methyl branch is on C4.

(b) A nitrile: 5 + 1 (the CN carbon) = 6 carbons. The CN carbon is C1, so no number, and the "e" stays before "nitrile".

(c) The longest chain containing the C=C has five carbons. From the left the C=C starts at C2 (from the right, C3), and the methyl branch is then on C3.

(d) An aldehyde, so CHO is C1. Four carbons: CHO, CH₂, CHCl, CH₃, so Cl is on C3.

Answer: (a) 4-methylpentan-2-ol; (b) hexanenitrile; (c) 3-methylpent-2-ene; (d) 3-chlorobutanal.

Worked example 3: an ester written the other way round

Question: An ester has the structural formula CH₃CH₂OOCCH₂CH₂CH₂CH₂CH₃. Name it, give its molecular and empirical formulas, and explain why "hexyl ethanoate" is wrong.

Step 1: find the C=O carbon. "OOC" is the ester group written backwards: CH₃CH₂-O-C(=O)-. The C=O carbon is bonded to the five-carbon chain on the right.

Step 2: name it. The acid part has 1 + 5 = 6 carbons: hexanoate. The group on the single-bonded O is ethyl. Written acid part first: CH₃CH₂CH₂CH₂CH₂COOCH₂CH₃.

Step 3: count. C: 2 + 6 = 8. H: 5 + 11 = 16. O: 2. C₈H₁₆O₂ fits CₙH₂ₙO₂. The highest common factor of 8, 16 and 2 is 2, giving C₄H₈O.

Step 4: the wrong name. Hexyl ethanoate, CH₃COOCH₂CH₂CH₂CH₂CH₂CH₃, has the six-carbon chain on the single-bonded O. It is a different compound with the same molecular formula, so the name must follow the C=O.

Answer: ethyl hexanoate; molecular formula C₈H₁₆O₂; empirical formula C₄H₈O.

Worked example 4: formulas from skeletal formulas

Question: Give the molecular formula, empirical formula and name of each compound.

• (a) A zigzag of five bonds, so six carbons in a row. Numbering from the left, a line from C2 ends in OH, and the C5 to C6 bond is a double line.
• (b) A zigzag of five lines whose two end lines lead to Cl atoms.

(a) H on each carbon. C1: 1 bond, 3 H. C2: 3 bonds (C1, C3, O), 1 H. C3 and C4: 2 bonds, 2 H each. C5: 3 bonds (single plus double line), 1 H. C6: 2 bonds (double line), 2 H.

(a) Total. H: 3 + 1 + 2 + 2 + 1 + 2 = 11, plus 1 in OH = 12, so C₆H₁₂O: two H fewer than C₆H₁₄O because of the C=C. The highest common factor is 1. The OH gets the lower number: hex-5-en-2-ol.

(b) The Cl atoms are written in, so the carbons are only the four corners between them. Each has 2 bonds drawn (the end ones: one C-C and one C-Cl), so each is CH₂: C₄H₈Cl₂. Dividing by 2 gives C₂H₄Cl.

Answer: (a) C₆H₁₂O, empirical formula C₆H₁₂O, hex-5-en-2-ol; (b) C₄H₈Cl₂, empirical formula C₂H₄Cl, 1,4-dichlorobutane.

Key Equations

General formulas (n = total number of carbon atoms; one functional group, no ring): $$\text{alkane: } \text{C}_n\text{H}_{2n+2} \qquad \text{alkene: } \text{C}_n\text{H}_{2n}$$ $$\text{alcohol: } \text{C}_n\text{H}_{2n+1}\text{OH}$$ $$\text{aldehyde, ketone: } \text{C}_n\text{H}_{2n}\text{O}$$ $$\text{acid, ester: } \text{C}_n\text{H}_{2n}\text{O}_2 \qquad \text{nitrile: } \text{C}_n\text{H}_{2n-1}\text{N}$$

H atoms on a carbon in a skeletal formula (a double line counts as two bonds, a triple line as three): $$\text{H on a carbon} = 4 - \text{bonds drawn to it}$$

Empirical formula (HCF = highest common factor of the numbers of atoms in the molecular formula): divide each number of atoms by the HCF.

$M_{\text{r}}$ from a general formula (for example a nitrile; $M_{\text{r}}$ has no unit): $$M_{\text{r}}(\text{C}_n\text{H}_{2n-1}\text{N}) = 14.0n + 13.0$$

Common Mistakes to Avoid

• Mistake: writing -OH or -CH₃ as groups inside a displayed formula. Fix: draw every bond. In propan-2-ol, each end carbon needs three separate C-H lines, and the OH must be drawn as O with a line to H.
• Mistake: drawing a carbon with three or five bonds. Fix: check each carbon for exactly four, a double bond counting as two. A CHO carbon has a double bond to O, one bond to H and one to the chain: no room for another H.
• Mistake: miscounting carbons in a skeletal formula. Fix: count line ends and corners, not lines. A plain zigzag of four lines has five carbons (pentane), and in a nitrile the corner where the triple line starts is a carbon too.
• Mistake: giving a molecular formula when a structural formula is asked for. Fix: C₄H₉OH could be any of four alcohols. Write CH₃CH₂CH₂CH₂OH for butan-1-ol or CH₃CH(OH)CH₂CH₃ for butan-2-ol.
• Mistake: naming from the chain as written, or leaving out a needed number. Fix: CH₃CH(CH₂CH₃)CH₂CH₃ is 3-methylpentane, not 2-ethylbutane, because the longest chain runs through the ethyl group. "Pentanol" is incomplete: write pentan-1-ol, pentan-2-ol or pentan-3-ol.
• Mistake: leaving the CN carbon out of a nitrile name, or garbling amine names. Fix: CH₃CH₂CH₂CN has four carbons, so it is butanenitrile, not propanenitrile or "butannitrile". CH₃CH₂CH₂CH₂NH₂ is butylamine, not "butamine".
• Mistake: naming an ester the wrong way round. Fix: the alkyl group on the single-bonded O comes first, and the C=O carbon belongs to the "-oate" part: CH₃CH₂COOCH₂CH₂CH₂CH₂CH₃ is pentyl propanoate.

Exam Tips

• Name: check whether the question wants the compound's systematic name (propanoic acid), its class (carboxylic acid) or its functional group (carboxyl, -COOH). The -OH of an alcohol is a hydroxyl group; hydroxide is the OH⁻ ion.
• Draw the displayed formula: every atom and every bond, including each C-H and O-H. Draw the skeletal formula: no C and no H on carbon, but every other atom written in.
• Give the structural formula: unambiguous, with branches in brackets, such as (CH₃)₂CHCH₂OH. Before you finish any structure, check the bonds: C 4, N 3, O 2, H and halogens 1.
• Deduce the molecular formula: give the full count, checked against the general formula. Simplify only if the empirical formula is asked for.