How to Name Haloalkanes with More Than One Halogen

Imagine a six-carbon chain with two different halogens and a methyl branch:

\(\ce{CH3-CH(Br)-CH2-CH(Cl)-CH(CH3)-CH3}\)

You could number the chain from the left or the right. Both directions are chemically valid, but they produce different locants. Which numbering gives the correct IUPAC name?

The correct name is 5-bromo-3-chloro-2-methylhexane.

The reason is that numbering from the right gives the lowest set of locants: 2, 3, 5 rather than 2, 4, 5.

This is the central rule for naming haloalkanes with multiple halogens:

1. Choose the correct parent carbon chain.
2. Number it to give the lowest possible set of locants to the substituents.
3. List the halogen prefixes in alphabetical order.
4. Use prefixes such as di-, tri-, and tetra- when the same halogen appears more than once.

Let’s work through why each step matters, including the situations where two numbering directions appear equally good.

01The four halogen prefixes

A halogen attached to an alkane is named using a prefix. The four halogens commonly encountered in organic nomenclature have these names:

These prefixes are used whether the molecule contains one halogen or several.

For example:

• One chlorine: chloropropane
• Two chlorine atoms: dichloropropane
• Three chlorine atoms: trichloropropane
• One bromine and one chlorine: bromochloroalkane, with the appropriate locants included in the full name.

The important distinction is that halogens are treated as substituents, not as the principal functional group.

They do not change the carbon-chain root. A four-carbon parent remains butane, whether it contains one bromine atom or three chlorine atoms.

02A reliable method for naming multiple halogens

Step 1: Choose the parent carbon chain

Start by finding the longest eligible continuous chain of carbon atoms.

The parent chain determines the main name:

Only carbon atoms count towards the chain length. Halogen atoms do not count as additional carbons.

For example, consider:

\(\ce{CH3-CH(Br)-CH2-CH(Cl)-CH3}\)

There are five carbon atoms connected in one continuous chain. The parent is therefore pentane.

The bromine and chlorine atoms are attached to this chain, so they will become prefixes in the final name.

If the molecule also contains a principal functional group, such as an alcohol group, you must account for that before choosing the parent chain. The parent must be suitable for naming the principal group.

For a more detailed explanation of competing carbon chains, see How to Choose the Parent Chain in Organic Compound Naming.

Step 2: Number the parent chain

Once you have selected the parent, number its carbon atoms from one end to the other.

For a simple saturated haloalkane, choose the direction that gives the lowest possible set of locants to all substituents together.

This includes:

• Halogens, such as bromine and chlorine.
• Alkyl branches, such as methyl and ethyl groups.

Do not give halogens automatic priority over alkyl branches.

Consider a six-carbon chain with:

• Bromine on carbon 2.
• Chlorine on carbon 4.
• A methyl branch on carbon 5.

Numbering from the left produces:

\[ 2,4,5 \]

Numbering from the right produces:

\[ 2,3,5 \]

The second set is lower, so numbering from the right is correct.

How do you compare two sets of locants?

Compare the numbers in ascending order, starting with the first number. If they match, move to the next number. The set with the lower number at the first point of difference wins.

You do not add the locants together.

For example, compare:

• Set A: 2, 4, 5
• Set B: 2, 3, 5

Both start with 2. The next numbers are 4 and 3. Since 3 is lower, Set B wins.

This is the lowest-set-of-locants principle used in IUPAC nomenclature.

Step 3: Resolve a tie using alphabetical order

Sometimes numbering from either end produces exactly the same set of locants.

When this happens, alphabetical order decides which substituent receives the lower number.

For example, consider:

\[ \ce{BrCH2-CH2Cl} \]

Numbering from the bromine end gives:

• Bromine at carbon 1.
• Chlorine at carbon 2.

Numbering from the chlorine end gives:

• Chlorine at carbon 1.
• Bromine at carbon 2.

Both directions produce the same locant set: 1, 2.

So which direction should you choose?

Bromo comes before chloro alphabetically. Therefore, bromine receives the lower number.

The correct name is 1-bromo-2-chloroethane.

It is not 2-bromo-1-chloroethane.

Remember the alphabetical order of the four halogen prefixes:

\[ \text{bromo} < \text{chloro} < \text{fluoro} < \text{iodo} \]

This is alphabetical order, not the order of the halogens in the periodic table.

Step 4: Write the prefixes in alphabetical order

After numbering the parent chain, write each substituent before the parent name.

When different halogens are present, list their prefixes alphabetically.

For example:

If the same halogen appears more than once, use a multiplicative prefix:

For example, a molecule with chlorine atoms on carbons 1 and 3, and a bromine atom on carbon 2, is named:

2-bromo-1,3-dichloropropane

Notice that bromo comes before dichloro.

The prefix di- does not count when determining alphabetical order. You alphabetise dichloro under C, not D.

The same principle applies when alkyl branches are present. All substituent prefixes are arranged alphabetically.

Step 5: Check the punctuation

The final name follows a consistent format.

• Commas separate multiple numbers.
• Hyphens separate numbers from words.
• There are no spaces between the different parts of the name.

For example:

1,2-dibromo-3-chlorobutane

Not:

1-2 dibromo 3 chloro butane

The first form is the correct systematic format.

03What if the molecule contains another functional group?

This is an important exception to the simple haloalkane naming method.

A halogen is a substituent prefix. It does not automatically determine the suffix or the parent chain.

Suppose a molecule contains both an alcohol group, \(\ce{-OH}\), and two different halogens.

The alcohol may be the principal functional group, so the compound is named using the suffix -ol. The halogens remain prefixes.

Consider:

\[ \ce{HOCH2-CH(Br)-CH2Cl} \]

The parent chain contains three carbon atoms. Numbering must begin at the end nearest the hydroxyl group because the alcohol determines the suffix.

The groups are therefore located at:

• Carbon 1: hydroxyl group.
• Carbon 2: bromine.
• Carbon 3: chlorine.

The correct name is:

2-bromo-3-chloropropan-1-ol

Notice what happened: the molecule contains two halogens, but neither gets priority over the alcohol suffix.

If a higher-priority functional group is present, its naming rules must be considered before optimising the locants of the halogen prefixes.

This is why you should identify functional groups before numbering a complicated organic structure.

04Worked examples

05Common mistakes to avoid

A useful final check is to read your name backwards: can you reconstruct the correct number of carbon atoms, the position of every halogen, and every branch?

If not, revisit your numbering.

06Questions and solutions

Try each question before reading its solution. The questions become progressively more demanding, particularly when numbering directions compete or another functional group changes the naming priority.

Question 1

Give the IUPAC name of:

\[ \ce{CH3-CH(Cl)-CH2-CH(Cl)-CH3} \]

Solution 1

The correct name is 2,4-dichloropentane.

The parent chain contains five carbon atoms, so the parent is pentane.

Numbering from either end gives chlorine atoms at carbons 2 and 4. Both directions produce the same locant set.

There are two chlorine atoms, so use the prefix dichloro.

The final name is:

2,4-dichloropentane

The key detail is that both chlorine positions must be included. Writing only 2-chloropentane would describe a different molecule.

Question 2

Give the IUPAC name of:

\[ \ce{CH3-CH(Br)-CH(CH3)-CH(Cl)-CH3} \]

Explain how alphabetical order determines the numbering direction.

Solution 2

The correct name is 2-bromo-4-chloro-3-methylpentane.

The parent chain contains five carbon atoms, so the parent is pentane.

Numbering from the left produces:

• Bromine at carbon 2.
• Methyl at carbon 3.
• Chlorine at carbon 4.

The locant set is 2, 3, 4.

Numbering from the right produces:

• Chlorine at carbon 2.
• Methyl at carbon 3.
• Bromine at carbon 4.

The locant set is also 2, 3, 4.

Since the locant sets are identical, alphabetical order determines the direction.

The substituent names are arranged as:

\[ \text{bromo} < \text{chloro} < \text{methyl} \]

Bromo comes first, so it must receive the lower locant.

Numbering from the left is therefore correct.

The final name is:

2-bromo-4-chloro-3-methylpentane

A common mistake is to number from the chlorine end because chlorine appears at carbon 2 in that direction. But the alphabetical tie-break gives bromine priority.

Question 3

A seven-carbon saturated chain has the following structure:

\[ \ce{CH3-CH(Br)-CH2-CH2-CH(Cl)-CH(CH3)-CH3} \]

Determine its IUPAC name.

Show the locant sets obtained by numbering from both ends. Explain why comparing the sums of the locants is not a reliable method.

Solution 3

The correct name is 6-bromo-3-chloro-2-methylheptane.

The parent chain contains seven carbon atoms, so the parent is heptane.

First, number from the left.

The substituents are:

• Bromine at carbon 2.
• Chlorine at carbon 5.
• Methyl at carbon 6.

The locant set is:

\[ 2,5,6 \]

Now number from the right.

The substituents become:

• Bromine at carbon 6.
• Chlorine at carbon 3.
• Methyl at carbon 2.

The locant set is:

\[ 2,3,6 \]

Compare the sets term by term.

Both begin with 2. The second numbers are 5 and 3.

Since 3 is lower, the second numbering direction is correct.

Interestingly, both sets have the same sum:

\[ 2+5+6=13 \]

\[ 2+3+6=11 \]

In this example the sums actually differ, but adding locants is not the IUPAC selection rule. A different pair of sets could have equal sums while still having a clear lowest set.

The correct method is always to compare the numbers in ascending order and find the first point of difference.

Finally, arrange the prefixes alphabetically:

• 6-bromo
• 3-chloro
• 2-methyl

The final name is:

6-bromo-3-chloro-2-methylheptane

The locants appear in the name alongside their respective substituents, so they do not have to appear in increasing numerical order across the entire name.

Question 4

A molecule has the following structure:

\[ \ce{HOCH2-CH(Cl)-CH2-CH(Br)-CH3} \]

Give its IUPAC name and explain why the parent chain must be numbered from the hydroxyl end rather than the end nearest a halogen.

Solution 4

The correct name is 4-bromo-2-chloropentan-1-ol.

The parent chain contains five carbon atoms, so the parent is pentane.

However, this molecule contains an alcohol group as well as two halogens.

The alcohol is the principal functional group and determines the suffix -ol.

Therefore, numbering must give the hydroxyl group the lowest possible locant.

Starting from the hydroxyl end gives:

• Hydroxyl group at carbon 1.
• Chlorine at carbon 2.
• Bromine at carbon 4.

The suffix becomes pentan-1-ol.

The halogen prefixes are arranged alphabetically, so bromo comes before chloro.

The final name is:

4-bromo-2-chloropentan-1-ol

The tempting mistake is to start numbering from the opposite end because that gives bromine carbon 2 rather than carbon 4.

But that would place the hydroxyl group at carbon 5. The principal functional group takes priority over the halogen prefixes when assigning locants.

This is why identifying functional groups must happen before numbering.

Question 5

Consider the following branched organic molecule:

\[ \ce{HOCH2-C(CH3)(CH2CH3)-CH(Cl)-CH(Br)-CH2-CH3} \]

Determine its correct IUPAC name.

Your explanation must address all of the following:

• The longest continuous carbon chain in the molecule.
• The longest eligible parent chain containing the hydroxyl group.
• The correct numbering direction.
• The alphabetical order of the substituents.

Assume the molecule is acyclic, contains no other functional groups, and has no stereochemical information specified.

Solution 5

The correct name is 4-bromo-3-chloro-2-ethyl-2-methylhexan-1-ol.

The important complication is that the longest continuous carbon chain in the entire structure does not contain the hydroxyl group.

A student might initially choose the seven-carbon path running from the terminal carbon of the ethyl branch, through the central carbon, and along the right-hand chain.

That path does contain seven carbon atoms.

However, it does not contain the carbon bonded to the hydroxyl group.

Because the alcohol is the principal functional group, the parent chain must include the carbon attached to the hydroxyl group. A longer chain that leaves the principal functional group behind is not an acceptable parent chain for this name.

Step 1: Identify the longest eligible parent chain.

Start at the hydroxyl-bearing carbon:

\[ \ce{HOCH2} \]

Continue through the central carbon and along the right-hand chain:

\[ \ce{HOCH2-C-CH(Cl)-CH(Br)-CH2-CH3} \]

This chain contains six carbon atoms.

The parent is therefore hexane, modified to hexan-1-ol.

Step 2: Number from the hydroxyl end.

The hydroxyl group must receive carbon 1.

The parent-chain positions are:

The carbon chain is numbered from the hydroxyl end because the alcohol determines the suffix.

Step 3: Identify every prefix.

The molecule contains:

• Bromine at carbon 4: 4-bromo.
• Chlorine at carbon 3: 3-chloro.
• Ethyl at carbon 2: 2-ethyl.
• Methyl at carbon 2: 2-methyl.

Step 4: Arrange the prefixes alphabetically.

The correct order is:

\[ \text{bromo} < \text{chloro} < \text{ethyl} < \text{methyl} \]

The parent name is hexan-1-ol.

Combining everything gives:

4-bromo-3-chloro-2-ethyl-2-methylhexan-1-ol

The important lesson is that parent-chain selection is not always a simple competition to find the greatest number of carbon atoms.

You must first identify the principal functional group. Only then can you choose the longest eligible parent chain and number it correctly.

07Where to go next

Once you can name a haloalkane correctly, the next useful step is understanding how its halogen affects its chemical reactions.

For example, in a substitution reaction, a halogen attached to a carbon chain can be replaced by a hydroxyl group to produce an alcohol. The carbon skeleton may remain unchanged, but the functional group and therefore the compound’s naming class change.

Continue with Alcohol Production from Haloalkanes: HSC Chemistry Guide to connect the structures you’ve just named with their chemical reactions.

IUPAC and NSW course context

The naming conventions in this guide follow the principles of IUPAC substitutive nomenclature, including lowest sets of locants, alphabetical ordering of prefixes, and the use of multiplicative prefixes. See the IUPAC Blue Book recommendations.

This guide supports the organic naming conventions covered in the NSW Chemistry Stage 6 Syllabus (2017). The Chemistry 11-12 Syllabus (2025) is scheduled to begin implementation for Year 11 in 2028. This article does not claim that the specific multiple-halogen naming rule is a separately stated dot point in the 2025 syllabus. See the NSW Department of Education’s Module 7 organic chemistry guide and NESA’s syllabus implementation information.