Unit 2: Naming organic molecules

Emma Harrage

Matter and Materials: Identify and apply knowledge of organic molecules in everyday life and industrial context

Unit 2: Naming organic molecules

Emma Harrage

By the end of this unit you will be able to:

Understand isomers.
Name (IUPAC) organic molecules and write their structural formulae.

What you should know

Before you start this unit, make sure you can:

Understand hydrocarbons and functional groups. Refer to level 4 subject outcome 5.3 unit 1 if you need help with this.

Introduction

Parts of the text in this unit were sourced from Siyavula Physical Science Gr 12 Learner’s Book, Chapter 4, released under a CC-BY licence.

In this unit you will learn about isomers of hydrocarbons and how to name organic compounds according to IUPAC naming conventions.

Isomers

It is possible for two organic compounds to have the same molecular formula but a different structural formula. Look at the two organic compounds that are shown in figure 1 for example.

Both butane and 2-methylpropane (isobutane) are used in camping stoves and lighters.

If you were to count the number of carbon and hydrogen atoms in each [latex]\scriptsize \displaystyle {{\text{C}}_{\text{4}}}{{\text{H}}_{{\text{10}}}}[/latex] compound, you would find that they are the same. They both have the same molecular formula, but their structure is different and so are their properties. Such compounds are called .

Isomers are molecules with the same molecular formula and often (though not always) with the same kinds of chemical bonds between atoms, but with the atoms arranged differently.

The isomers shown in figure 1 differ only in the location of the carbon atoms. The functional groups are the same, but butane has all four carbons in one chain, while 2-methylpropane has three carbons in the longest chain and a methyl group attached to the second carbon in the chain.

It is also possible to have positional isomers (see figure 2). In this case the −OH functional group can be on different carbon atoms, for example carbon 1 for pentan-1-ol, on carbon 2 for pentan-2-ol or on carbon 3 for pentan-3-ol.

Figure 2: The structural and condensed structural representations of the isomers (a) pentan-1-ol, (b) pentan-2-ol and (c) pentan-3-ol

It is important to note that molecules need not have the same functional groups to be isomers. For example, ethanol and methoxymethane (in figure 3) have the same molecular formula ([latex]\scriptsize \displaystyle {{\text{C}}_{2}}{{\text{H}}_{\text{6}}}\text{O}[/latex]) but different functional groups and properties. These types of isomers are .

Figure 3: Functional isomers of [latex]\scriptsize \displaystyle {{\text{C}}_{2}}{{\text{H}}_{\text{6}}}\text{O}[/latex]

Match the organic compound in Column A with its isomer in Column B:

The full solutions are at the end of the unit.

Naming organic molecules

To give compounds a name, certain rules must be followed. When naming organic compounds, the International Union of Pure and Applied Chemistry IUPAC (IUPAC) nomenclature (naming scheme) is used. This is to give consistency to the names. It also enables every compound to have a unique name, which is not possible with the common names used (for example in industry). We will first look at some of the steps that need to be followed when naming a compound, and then try to apply these rules to some specific examples.

A good general rule to follow is to start at the end (the suffix) and work backwards (from right to left) in the name.

Molecules can contain both double or triple bonds and other functional groups (e.g. an alkene and an alcohol functional group in one molecule – propenol). However, all molecules explored in this book will contain only single carbon-carbon bonds when combined with other functional groups.

Recognise the functional group in the compound. This will determine the suffix of the name.

Table 1: The suffix associated with various functional groups

Functional group Suffix

Alkane -ane

Alkene -ene

Alkyne -yne

Alcohol -ol

Find the longest continuous carbon chain that contains the functional group (it won’t always be a straight chain) and count the number of carbon atoms in this chain. This number will determine the prefix (the beginning) of the compound’s name.

Table 2: The prefix of a compound’s name is determined by the number of carbon atoms in the longest chain that contains the functional group
Number of carbon atoms	Prefix
[latex]\scriptsize 1[/latex]	Meth-
[latex]\scriptsize 2[/latex]	Eth-
[latex]\scriptsize 3[/latex]	Prop-
[latex]\scriptsize 4[/latex]	But-
[latex]\scriptsize 5[/latex]	Pent-
[latex]\scriptsize 6[/latex]	Hex-
[latex]\scriptsize 7[/latex]	Hept-
[latex]\scriptsize 8[/latex]	Oct-
[latex]\scriptsize 9[/latex]	Non-
[latex]\scriptsize 10[/latex]	Dec-

Number the carbons in the longest carbon chain containing the functional group. (Important: if the molecule is not an alkane i.e. has a functional group, you need to start numbering so that the functional group is on the carbon with the lowest possible number). Start with the carbon at the end closest to the functional group.

Look for any branched groups:
.
Name them by counting the number of carbon atoms in the branched group and referring to Table 2, these groups will all end in -yl.
.
Note the position of the group on the main carbon chain. If there is more than one of the same types of branched group, then both numbers must be listed (e.g. 2,4 -) and one of the prefixes listed in Table 3 below must be used. (Important: if the molecule is an alkane the branched group must be on the carbon with the lowest possible number.)
.
The branched groups must be listed before the name of the main chain in alphabetical order (ignoring di/tri/tetra).
.
If there are no branched groups, this step can be ignored.

Table 3: Prefixes for multiple substituents with the same name; these apply to multiple functional groups as well
Number	Prefix
[latex]\scriptsize 2[/latex]	Di-
[latex]\scriptsize 3[/latex]	Tri-
[latex]\scriptsize 4[/latex]	Tetra-

For the alkyl halides the halogen atom is treated in much the same way as branched groups:
.
To name them, take the name of the halogen atom (e.g. iodine) and replace the -ine with -o (e.g. iodo).

Table 4: Naming halogen atoms in organic molecules
Halogen	Name
Fluorine	Fluoro
Chlorine	Chloro
Bromine	Bromo
Iodine	Iodo

Give the halogen atom a number to show its position on the carbon chain. If there is more than one halogen atom the numbers should be listed and a prefix should be used (e.g. 3,4-diiodo- or 1,2,2-trichloro-). See Table 4 for a list of the prefixes.
.
The halogen atoms must be listed before the name of the main chain in alphabetical order (ignore di/tri/tetra).
.
If there are no halogen atoms this step can be ignored.

Combine the elements of the name into a single word in the following order:
- branched groups/halogen atoms in alphabetical order (ignoring prefixes)
- prefix of main chain
- finally, name the ending according to the functional group and its position on the longest carbon chain.

Name this alkane:

Solution

Identify the functional group
.
The compound is a hydrocarbon with single bonds between the carbon atoms. It is an alkane and will have a suffix of -ane.
Find the longest carbon chain
.
There are four carbon atoms in the longest chain. The prefix of the compound will be but-.
Number the carbon atoms in the longest chain
.
The numbering has been done for you here.
Look for any branched group, name them and give their position on the carbon chain
.
There are no branched groups in this compound.
Combine the elements of the name into a single word
.
The name of the compound is butane.

Name this alkanes:

Solution

Identify the functional group
.
The compound is a hydrocarbon with single bonds between the carbon atoms. It is an alkane and will have the suffix -ane.
Find the longest carbon chain
.
There are three carbon atoms in the longest chain. The prefix for this compound is prop-.
Number the carbons in the carbon chain
.
If we start at the carbon on the left, we can number the atoms as shown in red (below left). If we start at the carbon on the right, we can number the atoms as shown in blue (below right).
Look for any branched groups, name them and give their position on the carbon chain
There is a branched group attached to the second carbon atom. In this case the methyl group is on carbon 2 regardless of which side you number the longest chain from.
.
This group has the formula [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}[/latex], which is methane without a hydrogen atom. However, because it is not part of the main chain, it is given the suffix -yl (i.e. methyl). The position of the methyl group comes just before its name (see the next step).
Combine the elements of the compound’s name into a single word in the order of branched group; prefix; name ending according to the functional group
.
The compound’s name is 2-methylpropane.

Give the IUPAC name for the following compound:

[latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{3}}}[/latex]

Solution

Remember that the side groups are shown in brackets after the carbon atom to which they are attached.

Draw the structural formula from its condensed structural formula:
Identify the functional group
.
The compound is a hydrocarbon with single bonds between the carbon atoms. It is an alkane and will have the suffix -ane.
Find the longest carbon chain
.
There are four carbon atoms in the longest chain. The prefix for this compound is but-.
Number the carbons in the carbon chain
.
If we start at the carbon on the left, we can number the atoms as shown in red (below left). If we start at the carbon on the right, we can number the atoms as shown in blue (below right).
Look for any branched groups, name them and give their position on the carbon chain
.
There are two methyl groups attached to the main chain. The first one is attached to the second carbon atom and the second methyl group is attached to the third carbon atom. Notice that in this example it does not matter how you have chosen to number the carbons in the main chain; the methyl groups are still attached to the second and third carbon atoms and so the naming of the compound is not affected.
.
This group will be 2,3-dimethyl-
Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group
.
The compound’s name is 2,3-dimethylbutane.

Give the IUPAC name for the following compound:

Solution

Identify the functional group.
.
The compound is a hydrocarbon with single bonds between the carbon atoms. It is an alkane and will have the suffix -ane.
Find the longest carbon chain and number the carbons in the longest chain.

There are six carbons in the longest chain if they are numbered as shown in red (on the left). There are only five carbon atoms if they are numbered as shown in blue (right). Therefore, the red numbering (on the left) is correct and the prefix for the compound is hex-.
Look for any branched groups, name them and give their position on the carbon chain.

There is one methyl group attached to the main chain. If we number as shown in red (on the left) the methyl is attached to the fourth carbon atom. If we number as shown in blue (on the right) the methyl is attached to the third carbon atom.
After functional groups, the branched groups should have the lowest numbers possible. Therefore the blue numbering (on the right) is correct. The methyl is attached to the third carbon atom (3-methyl).
Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group
.
The compound’s name is 3-methylhexane.

Draw the semi-structural, structural and condensed structural formula for the organic compound 2,2,4-trimethylhexane.

Solution

Identify the functional group
.
The name ends in -ane therefore the compound is an alkane.
Determine the number of carbon atoms in the longest chain
.
The longest chain has the prefix hex-. There are therefore [latex]\scriptsize \displaystyle 6[/latex] carbon atoms in the longest chain.
.
Structural formula:
Look for any branched groups and place them on the structure.The compound is 2,2,4-trimethylhexane. Therefore there are three branched groups. Two on carbon [latex]\scriptsize \displaystyle 2[/latex] and one on carbon [latex]\scriptsize \displaystyle 4[/latex].
.
Semi-structural formula:
Combine this information and add the hydrogen atoms
.
Carbon atoms can have four single bonds. Therefore wherever a carbon atom has less than four bonds draw in hydrogen atoms until there are four bonds.
Condense the structural formula
.
First condense the main chain: [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CC}{{\text{H}}_{\text{2}}}\text{CHC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]
.
Then add the side chains (in brackets) on the relevant carbon atoms:
.
Condensed structural formula: [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]

Give the structural formula for each of the following alkanes:
1. octane
2. 3-ethylpentane
Give the IUPAC name for each of the following alkanes:
1. [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]
2. [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{2}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{3}}}[/latex]
3. .
4. .

The full solutions are at the end of the unit.

Naming alkenes

Let’s look at some examples of how to name alkenes.

Give the IUPAC name for the following compound:

Solution

Identify the functional group
.
The compound has a double carbon-carbon bond and is an alkene. It will have the suffix -ene.
Find the longest carbon chain containing the functional group
.
The functional group is a double bond, so the longest chain must contain the double bond. There are four carbon atoms in the longest chain and so the prefix for this compound will be but-.
Number the carbon atoms
.
Remember that the carbon atoms must be numbered so that the functional group is at the lowest numbered carbon atom possible. In this case, it doesn’t matter whether we number the carbons from the left to right, or from the right to left. The double bond will still fall between the second and third carbon atoms.
Look for any branched groups, name them and give their position on the carbon chain
.
There are no branched groups in this molecule.
Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain
.
The name of this compound is but-2-ene or 2-butene.

Draw the structural and molecular formula for the organic compound 3-methylbut-1-ene.

Solution

Identify the functional group
.
The suffix -ene means that this compound is an alkene and there must be a double bond in the molecule. The number [latex]\scriptsize \displaystyle 1[/latex] immediately before the suffix means that the double bond must be at the first carbon in the chain (but-1-ene).
Determine the number of carbon atoms in the longest chain containing the functional group
.
The prefix for the compound is but- so there must be four carbons in the longest chain containing the double bond.
Look for any branched groups
.
There is a methyl group at the third carbon atom in the chain. Count from the left so that the double bond carbon is the first carbon atom.
Combine this information and add the hydrogen atoms.
Reduce the structural formula to the molecular formula
.
There are [latex]\scriptsize \displaystyle 5[/latex] carbon atoms and [latex]\scriptsize \displaystyle 10[/latex] hydrogen atoms so the molecular formula is [latex]\scriptsize \displaystyle {{\text{C}}_{\text{5}}}{{\text{H}}_{{\text{10}}}}[/latex].
(Remember that there is no structural information given by the molecular formula.)

Give the IUPAC name for the following compound:

Solution

Identify the functional group
The compound is an alkene and will have the suffix -ene. There is a double bond between the first and second carbons and between the third and fourth carbons. The organic compound therefore contains ‘1,3-diene’.
Find the longest carbon chain containing the functional group, and number the carbon atoms
Remember that the main carbon chain must contain both the double bonds. There are four carbon atoms in the longest chain containing the double bonds and so the prefix for this compound will be but-. The carbon atoms are already numbered [latex]\scriptsize \displaystyle 1[/latex] to [latex]\scriptsize \displaystyle 4[/latex] in the diagram.
Look for any branched groups, name them and give their position on the carbon chain
There is an ethyl group on the second carbon.
Note that if we had numbered from the right to left the suffix would still have been 1,3-diene, however the ethyl group would have been on the third carbon. So we have to number from left to right.
Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain
The name of this compound is 2-ethylbut-1,3-diene.

Give the IUPAC name for each of the following alkenes:
1. [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{2}}}\text{CHC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]
2. .
Give the structural formula for each of the following alkenes:
1. hex-1-ene
2. 4-ethyloct-3-ene

The full solutions are at the end of the unit.

Naming alkynes

The suffix for an alkyne is -yne. Let’s look at some examples of how to name alkynes.

Give the IUPAC name for the following compound:

Solution

Identify the functional group
There is a triple bond between two of the carbon atoms, so this compound is an alkyne. The suffix will be -yne.
Find the longest carbon chain containing the functional group
The functional group is a triple bond, so the longest chain must contain the triple bond. There are six carbon atoms in the longest chain. The prefix of the compound’s name will be hex-.
Number the carbons in the longest chain
In this example, you will need to number the carbons from right to left so that the triple bond is between carbon atoms with the lowest numbers (the suffix for the compound will therefore be -2-yne).
Look for any branched groups, name them and assign the number of the carbon atom to which the group is attached.
There is a methyl ([latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}[/latex]) group attached to the fifth carbon (remember we have numbered the carbon atoms from right to left).
Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain
If we follow this order, the name of the compound is 5-methylhex-2-yne.

Give the IUPAC name for the following compound:

Solution

Identify the functional group
There are two triple bonds. The suffix will therefore be -diyne.
Find the longest carbon chain containing the functional group
The functional group is a triple bond, so the longest chain must contain all triple bonds. The longest carbon chain contains seven carbon atoms; therefore the prefix will be hept-.
Number the carbons in the longest chain:
Numbering from left to right (shown in red) the first triple bond is on carbon 1 and the second is on carbon 5. The suffix will therefore be -1,5-diyne.
(Numbering from right to left (shown in blue) will give the suffix -2,6-diyne, and is incorrect).
Look for any branched groups
There are no branched groups for this molecule.
Combine the elements of the name into a single word in the following order: branched groups; prefix; name ending according to the functional group and its position along the longest carbon chain
The name of the compound is hept-1,5-diyne.

Draw the structural and condensed structural formula for the organic compound 6-methylhept-3-yne.

Solution

Identify the functional group
The suffix -3-yne means that this compound is an alkyne and there must be a triple bond located on carbon number [latex]\scriptsize \displaystyle 3[/latex].
Determine the number of carbon atoms in the longest chain containing the functional group
The prefix for the compound is hept- so there must be seven carbons in the longest chain.
Look for any branched groups
There is a methyl group located on carbon number [latex]\scriptsize 6[/latex].
Combine this information and add the hydrogen atoms
Condense the structural formula
First condense the main chain:[latex]\scriptsize \displaystyle ~\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CCC}{{\text{H}}_{\text{2}}}\text{CHC}{{\text{H}}_{\text{3}}}[/latex]
Then add the side chains (in brackets) on the relevant carbon atoms:[latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CCC}{{\text{H}}_{\text{2}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{3}}}[/latex]

Give the structural formula for each of the following alkynes:
1. pent-1-yne
2. 5-methylhept-3-yne
Give the IUPAC names for the following alkynes:
1. [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CCC}{{\text{H}}_{\text{3}}}[/latex]
2. .

The full solutions are at the end of the unit.

Naming alkyl halides

All the same rules apply when naming the alkyl halides as for naming the hydrocarbons. The halogen atom is treated in the same way as a branched group.

Give the IUPAC name for the following compound:

Solution

Identify the functional group
There is a halogen atom and no other functional group. This compound is therefore a haloalkane and will have the suffix -ane.
Find the longest carbon chain containing the functional group
There are three carbons in the longest chain containing the halogen atom. The prefix is prop-.
Number the carbon atoms in the longest chain
You need to number the carbon atoms so that the halogen atom is on the carbon atom with the lowest number. In this case you can number from either side.
Name the halogen atom and assign the number for the carbon atom to which it is attached
The halogen is a chlorine atom. It is attached to carbon number [latex]\scriptsize \displaystyle 2[/latex]and so will have the name 2-chloro.
Look for branched groups
There are no branched groups in this compound.
Combine the elements of the name into a single word in the following order: halogen atoms; prefix; name ending according to functional group
The name of the compound is 2-chloropropane.

Give the IUPAC name for the following compound:

Solution

Identify the functional group
There are three halogen atoms and no other functional groups. This compound is therefore a haloalkane and will have the suffix -ane.
Find the longest carbon chain containing the functional group
There are four carbons in the longest chain containing all the halogen atoms. The prefix for this compound will be but-.
Number the carbon atoms in the longest chain
You need to number the carbon atoms so that the halogen atoms are on the carbon atoms with the lowest numbers. You must number from left to right here so that one halogen atom is on carbon [latex]\scriptsize \displaystyle ~1[/latex] and two halogen atoms are on carbon [latex]\scriptsize \displaystyle ~3[/latex].
Name the halogen atoms and assign the number for the carbon atom attached to it
There are two halogen atoms that are bromine atoms and one that is fluorine. One bromine is attached to carbon [latex]\scriptsize \displaystyle ~1[/latex] and one is attached to carbon [latex]\scriptsize \displaystyle ~3[/latex]. The fluorine atom is attached to carbon [latex]\scriptsize \displaystyle ~3[/latex]. So you have 1,3-dibromo- and 3-fluoro.
Look for branched groups
There are no branched groups in this compound.
Combine the elements of the name into a single word in the following order: halogen atoms in alphabetical order; prefix; name ending according to functional group
The name of the compound is 1,3-dibromo-3-fluorobutane.
Note that we place the halogens in alphabetical order: bromo (ignore the di/tri/tetra) is before fluoro.

Draw the structural and condensed structural formula for the organic compound 2-iodo-3-methylpentane.

Solution

Identify the functional group
This compound has the suffix -ane, but also contains a halogen atom. It is therefore a haloalkane. Note that the methyl and iodo are written in alphabetical order.
Find the longest carbon chain containing the functional group
The prefix is pent- therefore there are [latex]\scriptsize \displaystyle 5[/latex] carbons in the longest chain.
Place the halogen atom(s) and any branched groups
There is an iodine atom on the second carbon atom, and a methyl branched group on the third carbon atom.
Combine this information and add the hydrogen atoms
Condense the structural formula
First condense the main chain: [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CHCHC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]
Then add the side chains and halogen atoms (in brackets) on the relevant carbon atoms: [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CH}\left( \text{I} \right)\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]

Naming alcohols

The rules used to name the alcohols are like those already discussed for the hydrocarbons. The suffix of an alcohol is –ol.

Give the IUPAC name for the following organic compound:

Solution

Identify the functional group
The compound has an −OH (hydroxyl) functional group and is therefore an alcohol.
Find the longest carbon chain containing the functional group
There are three carbon atoms in the longest chain that contains the functional group. The prefix for this compound will be prop-. As there are only single bonds between the carbon atoms, the prefix includes an- to become propan-.
Number the carbons in the carbon chain
In this case, it doesn’t matter whether you start numbering from the left or right. The hydroxyl group will still be attached to second carbon atom (-2-ol).
Look for branched groups
There are no branched groups in this compound.
Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group
The compound’s name is propan-2-ol or 2-propanol.

Give the IUPAC name for the following compound:

Solution

Identify the functional group
The compound has an −OH (hydroxyl) functional group and is therefore an alcohol. There are two hydroxyl groups in the compound, so the suffix will be -diol.
Find the longest carbon chain that contains the functional group
There are four carbon atoms in the longest chain that contains the functional group (but-) and only single bonds (an-). The prefix for this compound will be butan-.
Number the carbons in the carbon chain:

.
There are two hydroxyl groups attached to the main chain. If we number as shown in red (on the left) they are attached to the first and second carbon atoms. If we number as shown in blue (on the right) they are attached to the third and fourth carbon atoms.
The functional groups should have the lowest numbers possible. Therefore the red numbering is correct. The hydroxyl groups are attached to the first and second carbon atoms (1,2-diol).
Look for branched groups
There are no branched groups in this compound.
Combine the elements of the compound’s name into a single word in the order of branched groups; prefix; name ending according to the functional group
The compound’s name is butan-1,2-diol.

Draw the structural and condensed structural representations for the organic compound 4-ethyloctan-2,5-diol.

Solution

Identify the functional group
The compound has the suffix -ol. It is therefore an alcohol.
Find the longest carbon chain that contains the functional group
The prefix is oct- therefore there are 8 carbons in the longest chain containing the functional group.
Place the functional group as well as any branched groups
There is one −OH attached to carbon [latex]\scriptsize 2[/latex] and one attached to carbon [latex]\scriptsize \displaystyle 5[/latex]. There is also an ethyl ([latex]\scriptsize \displaystyle \text{-C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]) branched group attached to carbon [latex]\scriptsize \displaystyle ~4[/latex].
Combine this information and add the hydrogen atoms:
Condense the structural formula
First condense the main chain: [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CHC}{{\text{H}}_{\text{2}}}\text{CHCHC}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]
Then add the side chains and alcohol functional groups (in brackets) on the relevant carbon atoms: [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CH}\left( {\text{OH}} \right)\text{C}{{\text{H}}_{\text{2}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}} \right)\text{CH}\left( {\text{OH}} \right)\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]

Summary

In this unit you have learnt the following:

If two compounds are isomers, it means that they have the same molecular formula but different structural formulae.
Organic compounds are named according to their functional group and its position in the molecule, the number of carbon atoms in the molecule and the position of any double and triple bonds. The IUPAC rules for nomenclature are used in the naming of organic molecules.

Unit 2: Assessment

Suggested time to complete: 30 minutes

Give the structural formula for each of the following compounds:
1. 3-methylhexane
2. 4-ethyl-4-methylhept-2-yne
3. 2,3-difluoropentane
4. 2-methylpropan-1-ol
Give the IUPAC name for the following haloalkanes:
1. .
2. .
3. [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{2}}}\left( \text{F} \right)\text{C}{{\left( \text{I} \right)}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}[/latex]
Give the IUPAC name for each of the following:
1. [latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CH}\left( {\text{OH}} \right)\text{C}{{\text{H}}_{\text{3}}}[/latex]
2. .

The full solutions are at the end of the unit.

Unit 2: Solutions

Exercise 2.1

A1 ([latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\left( {\text{OH}} \right)[/latex]) and B3 [latex]\scriptsize \displaystyle \left( {\text{C}{{\text{H}}_{\text{3}}}\text{CH}\left( {\text{OH}} \right)\text{C}{{\text{H}}_{\text{3}}}} \right)[/latex] are isomers.
B1 ([latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{CH}\left( {\text{C}{{\text{H}}_{\text{3}}}} \right)\text{C}{{\text{H}}_{\text{3}}}[/latex]) and A2 are isomers.
B2 and A3 are isomers

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Exercise 2.2

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1. The suffix -ane tells us that this is an alkane. The prefix oct- tells us that there are eight carbon atoms in the longest chain.
2. The suffix -ane tells us that this is an alkane. The prefix pent- tells us that there are five carbon atoms in the longest chain. 3-ethyl tells us that there is an ethyl branched group ([latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{-}[/latex]) group attached to the third carbon atom.
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1. This is hard to do unless you draw the structural formula of the molecule. It is recommended that you do this in exams.
  .
  There are five carbons in the longest chain, so the prefix is pent-. There are only single carbon-carbon bonds and no other functional group, so the compound is an alkane, and the suffix is -ane. There is one methyl group at position [latex]\scriptsize \displaystyle 3[/latex] (you can number from either end of the chain for this example). So the compound is 3-methylpentane.
2. There are five carbons in the longest chain, so the prefix is pent-. There are only single carbon-carbon bonds, so the compound is an alkane, and the suffix is -ane. There is one methyl group at position [latex]\scriptsize \displaystyle 2[/latex] and one at position [latex]\scriptsize \displaystyle 4[/latex] (once again you can number from either end). So the compound is 2,4-dimethylpentane.
3. There are four carbon atoms in the longest chain, so the prefix is but-. There are two methyl branches at positions [latex]\scriptsize \displaystyle 2[/latex] and [latex]\scriptsize \displaystyle 3[/latex]. The functional group is an alkane, so the suffix is -ane. Combining all this information we get: 2,3-dimethylbutane. Note that in this example it does not matter which end you start numbering from.
4. There are four carbon atoms in the longest chain, so the prefix is but-. The functional group is an alkane, so the suffix is -ane. There is one branched group which is a methyl group, and this is at position [latex]\scriptsize \displaystyle 2[/latex]. The molecule is 2-methylbutane.
  .
  Note that in this example it does matter which way your number the chain as the branched group needs to have the lowest number possible and so the compound is not 3-methylbutane.

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Exercise 2.3

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1. Draw the structural representation:
  
  The molecule contains a double carbon-carbon bond. It is an alkene and so the suffix is -ene. There are five carbons in the longest chain, so the prefix is pent-. There are no branched groups. The double bond occurs between carbons [latex]\scriptsize 1[/latex] and [latex]\scriptsize 2[/latex]. So the molecule is 1-pentene or pent-1-ene.
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  Note that the way you number the carbon atoms matters here, the molecule is not pent-4-ene.
2. The compound contains two double carbon-carbon bonds. It is an alkene and so the suffix is -diene. There are four carbons in the longest chain containing the double bonds, so the prefix is but- there are no branched chains. The first double bond occurs between carbons one and two. The second double bond occurs between carbons three and four. The compound is but-1,3-diene.
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1. The prefix hex- tells us there are six carbon atoms in the chain. The suffix -1-ene tells us there is a double bond between the first and second carbon atoms.
2. The prefix oct- tells us there are eight carbon atoms in the longest chain containing the functional group. The suffix -3-ene tells us there is a double bond between the third and fourth carbon atoms. 4-ethyl tells us that there is an ethyl ([latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{-}[/latex]) branched group attached to the fourth carbon atom.

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Exercise 2.4

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1. The prefix pent- tells us there are five carbon atoms in the longest chain. The suffix -1-yne tells us there is a triple bond between the first and second carbon atoms.
2. The prefix hept- tells us there are seven carbon atoms in the longest chain. The suffix -3-yne tells us there is a triple bond between the third and fourth carbon atoms. 5-methyl tells us there is a methyl branched chain on the fifth carbon atom.
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1. There is a triple carbon-carbon bond. This compound is an alkyne and will have the suffix -yne. There are four carbon atoms in the longest chain, therefore the prefix will be but-. The triple bond is between the second and third carbon atoms regardless of how you number the chain (-2-yne). There are no branched groups. This molecule is but-2-yne or 2-butyne.
2. There is a triple carbon-carbon bond. This compound is an alkyne and will have the suffix -yne. There are six carbon atoms in the longest chain, therefore the prefix will be hex-. The triple bond is between the third and fourth carbon atoms regardless of how you number the chain (-3-yne). There are two branched methyl groups. Depending on the order of numbering they are either on the fourth carbon atom (left to right) or the second carbon atom (right to left). The lower numbering is correct (right to left) and as there are two methyl groups this is 2,2-dimethyl. The molecule is 2,2-dimethylhex-3-yne.

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Unit 2: Assessment

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1. The prefix hex- tells us there are six carbon atoms in the longest chain. The suffix -ane tells us this is an alkane and that there are only single carbon-carbon bonds and no other functional groups. 3-methyl tells us there is a branched methyl group on the third carbon atom.
2. The prefix hept- tells us there are seven carbon atoms in the longest chain. The suffix -2-yne tells us this is an alkyne and there is a triple bond between the second and third carbon atoms. The 4-ethyl tells us there is an ethyl ([latex]\scriptsize \displaystyle \text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{-}[/latex]) branched group on the fourth carbon atom. The 4-methyl tells us there is a methyl (CH3−) branched group on the fourth carbon atom.
3. The pent- tells us there are five carbon atoms in the longest chain. The -ane tells us there are only single carbon-carbon bonds. The 2,3-difluoro means there are two fluorine atoms, one attached to carbon [latex]\scriptsize \displaystyle 2[/latex] and the other attached to carbon [latex]\scriptsize \displaystyle 3[/latex].
4. The prefix propan- tells us there are three carbon atoms in the longest chain and only single carbon-carbon bonds. The suffix -1-ol tells us there is a hydroxyl group attached to the first carbon atom. 2-methyl tells us there is a methyl branched group attached to the second carbon atom.
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1. There is a halogen atom and only single carbon-carbon bonds, therefore this is a haloalkane and the suffix is -ane. There are three carbon atoms in the longest chain, therefore the prefix is prop-. There is an iodine atom attached to the second carbon atom (2-iodo). There are no branched groups. Therefore this molecule is 2-iodopropane.
2. There are halogen atoms and only single carbon-carbon bonds, therefore this is a haloalkane and the suffix is -ane. There are eight carbon atoms in the longest chain, therefore the prefix is oct-. There are two chlorine atoms, one attached to the first carbon atom, and one attached to the fourth carbon atom (1,4-dichloro). There are no branched groups. This molecule is 1,4-dichlorooctane.
3. .
  
  There are halogen atoms and only single carbon-carbon bonds, therefore this is a haloalkane and the suffix is -ane. There are four carbon atoms in the longest chain containing all the halogen atoms, therefore the prefix is but-. There is a fluorine atom attached to the first carbon atom (1-fluoro). There are two iodine atoms attached to the second carbon atom (2,2-diiodo). These must be put in alphabetical order, ignoring any prefixes on the halogen atoms. F comes before i (1-fluoro-2,2-diiodo). This molecule is 1-fluoro-2,2-diiodobutane.
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1. .
  
  There is a hydroxyl group, therefore the compound is an alcohol, and the suffix is -ol. There are four carbon atoms in the longest chain, so the prefix is but-. There are only single carbon-carbon bonds, therefore the prefix becomes butan-. There are no branched groups. The hydroxyl group is attached to the second carbon atom. The molecule is 2-butanol or butan-2-ol.
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  Note that the way we number the carbon atoms matters. The hydroxyl group is given the lowest possible number and so this compound is not butan-3-ol.
2. There are two hydroxyl groups, therefore the compound is an alcohol, and the suffix is -diol. There are seven carbon atoms in the longest chain, so the prefix is hept-. There are only single carbon-carbon bonds, therefore the prefix becomes heptan-. The hydroxyl groups are both attached to the third carbon atom (-3,3-diol). The branched methyl chain is attached to the fourth carbon atom. The molecule is 4-methylheptan-3,3-diol.
  .
  Note that the way we number the carbon atoms matters. The hydroxyl groups are given the lowest possible numbers and so this compound is not 4-methylheptan-5,5-diol.

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National Curriculum (Vocational) Physical Science Level 4 by Emma Harrage is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

Functional group	Suffix
Alkane	-ane
Alkene	-ene
Alkyne	-yne
Alcohol	-ol

Unit outcomes

What you should know

Introduction