Formulae, Functional Groups and Naming

Organic chemistry is the chemistry of carbon compounds, and there are millions of them. You can only make sense of that number by learning a small set of rules: how to write the structure of a molecule, how to spot its functional group (the part that reacts), and how to name it so that every chemist draws the same thing. This note covers the six kinds of formula, homologous series, the functional groups on the AS syllabus, how to classify carbon atoms, alcohols and halogenoalkanes as primary, secondary or tertiary, and the IUPAC naming rules. Almost every organic question in Papers 1 and 2 asks you to name, draw or identify something, so these marks are among the most reliable you can earn.

Why carbon makes so many compounds

A carbon atom has four electrons in its outer shell and forms four covalent bonds. It bonds strongly to other carbon atoms, so it can form long chains, branched chains and rings, and it bonds strongly to hydrogen, oxygen, nitrogen and the halogens. The C–C and C–H bonds are strong and non-polar, which is why the hydrocarbon "skeleton" of a molecule is unreactive and the reactions happen at the functional group.

Two other atoms you will draw all the time have fixed bonding too: hydrogen and the halogens form one bond, oxygen forms two, and nitrogen forms three. Checking that every atom in a structure you have drawn has the right number of bonds catches most drawing errors.

The six kinds of formula

Each type of formula gives a different amount of information. You must be able to read and write all of them.

Here are all six for butan-1-ol and ethanoic acid.

type	butan-1-ol	ethanoic acid
empirical	$\ce{C4H10O}$	$\ce{CH2O}$
molecular	$\ce{C4H10O}$	$\ce{C2H4O2}$
general (series)	$\ce{C_{n}H_{2n+1}OH}$	$\ce{C_{n}H_{2n+1}COOH}$
structural	$\ce{CH3CH2CH2CH2OH}$	$\ce{CH3COOH}$
displayed	every C–H, C–C, C–O and O–H bond drawn	every bond drawn, with C=O as a double line
skeletal	a four-carbon zigzag ending in OH	a two-carbon line with =O and OH on the end carbon

For butan-1-ol the empirical and molecular formulae are the same because 4, 10 and 1 have no common factor. For ethanoic acid they differ: $\ce{C2H4O2}$ simplifies to $\ce{CH2O}$.

Writing structural formulae

A structural formula must be unambiguous: someone reading it must be able to draw only one molecule. The conventions:

• Write each carbon with the atoms attached to it: $\ce{CH3CH2CH3}$ for propane.
• Put branches and side groups in brackets after the carbon they are attached to: $\ce{CH3CH(CH3)CH3}$ is methylpropane; $\ce{CH3CH(OH)CH3}$ is propan-2-ol.
• Show double bonds between carbons explicitly: $\ce{CH2=CHCH3}$ for propene. Writing $\ce{CH2CHCH3}$ is ambiguous and loses the mark.
• Standard group abbreviations: $\ce{-CHO}$ for an aldehyde group, $\ce{-COOH}$ for a carboxyl group, $\ce{-COO-}$ for an ester link, $\ce{-CN}$ for a nitrile, $\ce{-NH2}$ for an amine.

Skeletal formulae

In a skeletal formula every line end and every vertex is a carbon atom, and each carbon carries enough hydrogen atoms to make up four bonds. Atoms other than carbon and hydrogen are written in, along with any hydrogen atoms attached to them (so an alcohol shows $\ce{OH}$, not just O).

To read a skeletal formula, count the vertices and line ends to get the carbons, then fill in hydrogens. In 2-methylbutane above there are five carbon atoms: four along the zigzag and one on the branch. The carbon at the junction has three bonds to carbon, so it carries one hydrogen.

Homologous series and functional groups

The trend in physical properties is easy to explain. Each extra $\ce{CH2}$ adds electrons, so the instantaneous dipole–induced dipole forces between molecules get stronger and boiling points rise steadily along a series. The chemical properties stay similar because the functional group, which does the reacting, stays the same.

The carbon skeleton itself can be straight-chained (no branches, every carbon bonded to at most two others), branched (at least one carbon bonded to three or four other carbons) or cyclic (the carbons form a ring, as in cyclohexane, $\ce{C6H12}$).

The AS functional groups

Some of these general formulae overlap. Aldehydes and ketones share $\ce{C_{n}H_{2n}O}$; carboxylic acids and esters share $\ce{C_{n}H_{2n}O2}$; alkenes and cycloalkanes share $\ce{C_{n}H_{2n}}$. That overlap is the source of functional group isomerism (see Isomerism).

Primary, secondary and tertiary

Many reactions depend on how many carbon atoms are attached to a particular carbon, so you need to classify carbon atoms, alcohols and halogenoalkanes.

class	alcohol example	halogenoalkane example
primary	$\ce{CH3CH2CH2OH}$ propan-1-ol	$\ce{CH3CH2CH2Br}$ 1-bromopropane
secondary	$\ce{CH3CH(OH)CH3}$ propan-2-ol	$\ce{CH3CHBrCH3}$ 2-bromopropane
tertiary	$\ce{(CH3)3COH}$ 2-methylpropan-2-ol	$\ce{(CH3)3CBr}$ 2-bromo-2-methylpropane

A quick way to classify: count the hydrogen atoms on the carbon carrying the functional group. Two hydrogens (as in $\ce{-CH2OH}$) means primary; one hydrogen ($\ce{>CHOH}$) means secondary; none means tertiary. Methanol, $\ce{CH3OH}$, is counted as primary.

Naming organic compounds

IUPAC names are built from three parts: prefixes (side chains and some functional groups), a stem (the number of carbons in the main chain) and a suffix (the main functional group).

carbons	1	2	3	4	5	6
stem	meth-	eth-	prop-	but-	pent-	hex-
alkyl group	methyl	ethyl	propyl	butyl	pentyl	hexyl

Rules for each series

• Alkanes: stem + -ane. $\ce{CH3CH(CH3)CH2CH3}$ is 2-methylbutane.
• Alkenes: number the position of the first carbon of the C=C. $\ce{CH3CH=CHCH3}$ is but-2-ene; $\ce{CH2=CHCH2CH3}$ is but-1-ene. Ethene and propene need no number because there is only one possible position.
• Halogenoalkanes: prefix. $\ce{CH3CHClCH3}$ is 2-chloropropane; $\ce{CH2BrCH2Br}$ is 1,2-dibromoethane.
• Alcohols: the "e" of the alkane becomes "-ol" with a position number: propan-1-ol, propan-2-ol, butane-1,2-diol (the "e" is kept before a consonant, as in "diol").
• Aldehydes: -al. The CHO carbon is always carbon 1, so no number is needed: $\ce{CH3CH2CHO}$ is propanal.
• Ketones: -one with a number when needed: $\ce{CH3COCH2CH2CH3}$ is pentan-2-one. Propanone and butanone need no number.
• Carboxylic acids: -oic acid. The COOH carbon is carbon 1 and is counted in the stem: $\ce{CH3CH2COOH}$ is propanoic acid, not ethanoic acid.
• Esters: named "alkyl alkanoate". The alkyl part comes from the alcohol and is written first; the alkanoate part comes from the acid and includes the C=O carbon. $\ce{CH3COOCH2CH3}$ is ethyl ethanoate; $\ce{HCOOCH3}$ is methyl methanoate; $\ce{CH3CH2COOCH3}$ is methyl propanoate.
• Amines: alkyl + amine. $\ce{CH3NH2}$ is methylamine; $\ce{CH3CH2NH2}$ is ethylamine. (The IUPAC alternatives methanamine and ethanamine are also accepted.)
• Nitriles: -nitrile, and the carbon of the $\ce{C#N}$ group is counted in the stem. $\ce{CH3CN}$ is ethanenitrile; $\ce{CH3CH2CN}$ is propanenitrile.
• Hydroxynitriles: the nitrile takes the suffix and the OH becomes "hydroxy". $\ce{CH3CH(OH)CN}$ is 2-hydroxypropanenitrile.

Worked examples

Practice

Answers

1. (a) $\ce{C_{n}H_{2n}}$ (b) $\ce{C_{n}H_{2n+1}OH}$ (or $\ce{C_{n}H_{2n+2}O}$) (c) $\ce{C_{n}H_{2n+1}COOH}$ (or $\ce{C_{n}H_{2n}O2}$).
2. (a) Four-carbon chain with methyls on C2 and C3: 2,3-dimethylbutane. (b) 1-chloropentane. (c) Butanone (no number needed, the C=O can only be on C2). (d) The C=O side has four carbons (butanoate), the alkyl on O is methyl: methyl butanoate.
3. (a) $\ce{CH2=CHCH(CH3)CH3}$ (b) $\ce{CH3CH2COCH2CH2CH3}$ (c) $\ce{(CH3)2CHCHO}$ (d) $\ce{CH3CH2CH2NH2}$.
4. (a) Secondary: the OH carbon is bonded to two carbons. (b) Primary: $\ce{(CH3)2CHCH2OH}$, the OH carbon is bonded to one carbon. (c) Tertiary: $\ce{CH3CH2CBr(CH3)2}$, the C–Br carbon is bonded to three carbons. (d) Primary: $\ce{(CH3)2CHCH2Br}$.
5. (a) $\ce{C2H4}$ → $\ce{CH2}$ (b) $\ce{C6H14}$ → $\ce{C3H7}$ (c) $\ce{C4H8O2}$ → $\ce{C2H4O}$.
6. "2-ethylbutane" would be $\ce{CH3CH(CH2CH3)CH2CH3}$. The longest chain runs through the ethyl branch and has five carbons, with a methyl on C3. The correct name is 3-methylpentane. Any name with an ethyl group on C2 has failed to find the longest chain.
7. Each successive alcohol has one more $\ce{CH2}$ group, so more electrons; the instantaneous dipole–induced dipole forces between molecules become stronger, and more energy is needed to separate the molecules, so boiling point rises. (Hydrogen bonding through the OH group is present in all of them.) The chemical reactions depend on the functional group, the OH, which is the same in each, so the reactions are similar.
8. $\ce{CH3CH2CH2CH2OH}$ butan-1-ol, primary. $\ce{CH3CH2CH(OH)CH3}$ butan-2-ol, secondary. $\ce{(CH3)2CHCH2OH}$ 2-methylpropan-1-ol, primary. $\ce{(CH3)3COH}$ 2-methylpropan-2-ol, tertiary.
9. $n(\ce{CO2}) = 0.880 / 44.0 = 0.0200\ \text{mol}$, so $n(\ce{C}) = 0.0200\ \text{mol}$ and mass of C $= 0.240\ \text{g}$. $n(\ce{H2O}) = 0.315 / 18.0 = 0.0175\ \text{mol}$, so $n(\ce{H}) = 0.0350\ \text{mol}$ and mass of H $= 0.035\ \text{g}$. Mass of N $= 0.345 - 0.240 - 0.035 = 0.070\ \text{g}$, so $n(\ce{N}) = 0.070 / 14.0 = 0.00500\ \text{mol}$. Ratio C : H : N $= 0.0200 : 0.0350 : 0.00500 = 4 : 7 : 1$, empirical formula $\ce{C4H7N}$ ($M_r = 69.0$). Since $M_r = 69.0$, the molecular formula is also $\ce{C4H7N}$. Straight-chain nitrile with four carbons (including the $\ce{C#N}$ carbon): $\ce{CH3CH2CH2CN}$, butanenitrile. The branched isomer is $\ce{(CH3)2CHCN}$, 2-methylpropanenitrile.
10. (a) The aldehyde takes the suffix and is C1; OH is on C3: 3-hydroxybutanal. (b) Aldehyde ($\ce{-CHO}$) and hydroxy ($\ce{-OH}$). (c) The OH carbon (C3) is bonded to C2 and C4: secondary. (d) $\ce{C4H8O2}$; empirical $\ce{C2H4O}$. (e) A four-carbon zigzag: the right-hand end carbon is the CHO, drawn as a line to "=O" (that carbon has one H, hidden); the next carbon is a $\ce{CH2}$ vertex; the next carries "OH" (one H on that carbon); the left end is a $\ce{CH3}$ line end.