Stems and position numbers

1. The stem of a name gives the number of carbon atoms in the longest unbranched chain.
   1. Meth- is 1, eth- is 2, prop- is 3 and but- is 4.
   2. Longer chains carry on with pent- for 5, hex- for 6, hept- for 7, oct- for 8, non- for 9 and dec- for 10.
2. The ending gives the homologous series: -ane for alkanes, -ene for alkenes, -ol for alcohols and -oic acid for carboxylic acids.
3. In longer alkenes and alcohols the functional group can sit in more than one place, so a position number is added.
   1. Number the carbon atoms from the end of the chain that gives the functional group the lowest number.
   2. For an alkene, the number is the first carbon of the $C=C$ bond.
   3. For an alcohol, the number is the carbon atom joined to the $-OH$ group.
4. The number goes between the stem and the ending, with a hyphen on each side, as in but-1-ene and propan-2-ol.
5. A number is left out when only one position is possible, so you write propene, ethanol and ethanoic acid.

Unbranched alkanes

1. An unbranched alkane has all its carbon atoms in one chain joined by single bonds, and its name ends in -ane.

2. In a structural formula each end carbon is written ${CH}_3$ and each carbon inside the chain is written ${CH}_2$.

   1. Methane is ${CH}_4$.
   2. Ethane is ${CH}_3{CH}_3$.
   3. Propane is ${CH}_3{CH}_2{CH}_3$.
   4. Butane is ${CH}_3{CH}_2{CH}_2{CH}_3$.
   5. Pentane is ${CH}_3{CH}_2{CH}_2{CH}_2{CH}_3$, and longer chains follow the same pattern.

3. The displayed formula of butane shows four carbon atoms in a row, with three hydrogen atoms on each end carbon and two on each middle carbon.

   $$\begin{array}{ccccccccccc}& & H& & H& & H& & H& & \\ & & |& & |& & |& & |& & \\ H& -& C& -& C& -& C& -& C& -& H\\ & & |& & |& & |& & |& & \\ & & H& & H& & H& & H& & \end{array}$$

Unbranched alkenes

1. An unbranched alkene has one $C=C$ bond in its chain, and its name ends in -ene.
2. Ethene, ${CH}_2={CH}_2$, and propene, ${CH}_2={CHCH}_3$, need no position number.
   1. In propene the double bond must touch an end carbon, and numbering from that end always gives position 1.
3. A four-carbon chain allows two different positions, which gives two compounds.
   1. But-1-ene, ${CH}_2={CHCH}_2{CH}_3$, has the double bond between carbon 1 and carbon 2.
   2. But-2-ene, ${CH}_{3}CH={CHCH}_3$, has the double bond between carbon 2 and carbon 3.
4. A double bond between carbons 3 and 4 is the same as but-1-ene counted from the other end, so but-3-ene is never a correct name.

Unbranched alcohols

1. An unbranched alcohol has an $-OH$ group on its chain, and its name ends in -ol.
2. The one- and two-carbon alcohols need no position number.
   1. Methanol is ${CH}_{3}OH$.
   2. Ethanol is ${CH}_3{CH}_{2}OH$.
3. A three-carbon chain gives two alcohols.
   1. Propan-1-ol, ${CH}_3{CH}_2{CH}_{2}OH$, has the $-OH$ on an end carbon.
   2. Propan-2-ol, ${CH}_{3}CH(OH){CH}_3$, has the $-OH$ on the middle carbon.
4. A four-carbon chain also gives two unbranched alcohols.
   1. Butan-1-ol, ${CH}_3{CH}_2{CH}_2{CH}_{2}OH$, has the $-OH$ on an end carbon.
   2. Butan-2-ol, ${CH}_{3}CH(OH){CH}_2{CH}_3$, has the $-OH$ on the second carbon.
5. In a structural formula, an $-OH$ on a middle carbon is written in brackets after that carbon, as in $CH(OH)$.

Unbranched carboxylic acids

1. A carboxylic acid has a $-COOH$ group at the end of its chain, and its name ends in -oic acid.

2. The carbon of the $-COOH$ group counts as part of the chain and is always carbon 1, so no position number is needed.

   1. Methanoic acid is $HCOOH$, with one carbon atom.
   2. Ethanoic acid is ${CH}_{3}COOH$, with two carbon atoms.
   3. Propanoic acid is ${CH}_3{CH}_{2}COOH$, with three carbon atoms.
   4. Butanoic acid is ${CH}_3{CH}_2{CH}_{2}COOH$, with four carbon atoms.

3. In the displayed formula the last carbon has a double bond to one oxygen and a single bond to an $O-H$ group, as in propanoic acid.

   $$\begin{array}{ccccccccccc}& & H& & H& & O& & & & \\ & & |& & |& & \Vert & & & & \\ H& -& C& -& C& -& C& -& O& -& H\\ & & |& & |& & & & & & \\ & & H& & H& & & & & & \end{array}$$

Naming esters

1. An ester is made from an alcohol and a carboxylic acid, and contains the $-COO-$ group joining two carbon chains.
2. The name of an ester has two words.
   1. The first word comes from the alcohol, with -anol changed to -yl.
   2. The second word comes from the acid, with -anoic acid changed to -anoate.
3. The four unbranched alcohols up to four carbons give four first words.
   1. Methanol gives methyl, ethanol gives ethyl, propan-1-ol gives propyl and butan-1-ol gives butyl.
4. The four carboxylic acids up to four carbons give four second words.
   1. Methanoic acid gives methanoate, ethanoic acid gives ethanoate, propanoic acid gives propanoate and butanoic acid gives butanoate.
5. Any first word can pair with any second word, so these alcohols and acids make 16 unbranched esters, from methyl methanoate to butyl butanoate.
6. The structural formula is written the other way round from the name, with the acid part first.
   1. Ethyl ethanoate is ${CH}_3{COOCH}_2{CH}_3$, where ${CH}_{3}COO$ comes from ethanoic acid and ${CH}_2{CH}_3$ comes from ethanol.
   2. Methyl propanoate is ${CH}_3{CH}_2{COOCH}_3$, where ${CH}_3{CH}_{2}COO$ comes from propanoic acid and ${CH}_3$ comes from methanol.

Drawing esters

1. Draw the displayed formula of an ester in four steps.
   1. Draw the acid part, ending in a carbon atom with a double bond to an oxygen atom.
   2. Join that carbon by a single bond to a second oxygen atom.
   3. Join the second oxygen to the first carbon of the chain from the alcohol.
   4. Add hydrogen atoms until every carbon has four bonds, and check that each oxygen has two.
2. The second oxygen sits between two carbon atoms, so the carbon chain is broken at that point and you count carbons on each side separately.