Skip to main content
Chemistry LibreTexts

16.3: Branched Hydrocarbons

  • Page ID
    64109
    • Anonymous
    • LibreTexts

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)
     Learning Objectives
    • Name a branched hydrocarbon from its structure.
    • Draw the structural formula of a branched hydrocarbon from its name.

    Not all hydrocarbons are straight chains. Many hydrocarbons have branches of C atoms attached to a chain. These branched alkanes are isomers of straight-chain alkanes having the same number of C atoms. However, they are different compounds with different physical and chemical properties. As such, they need different names. How do we name branched hydrocarbons?

    There are a series of rules for naming branched alkanes (and, ultimately, for all organic compounds). These rules make up the system of nomenclature for naming organic molecules. Worldwide, the International Union of Pure and Applied Chemistry (IUPAC) has developed the system of nomenclature for organic compounds. So these rules are sometimes called the IUPAC rules of nomenclature. By learning and applying these rules, you can name any organic compound when given its structure or determine the unique structure of a molecule from its name. You have already learned the basics of nomenclature—the names of the first 10 normal hydrocarbons. Here, we will add some steps to the procedure so you can name branched hydrocarbons.

    First, given the structure of an alkane, identify the longest continuous chain of C atoms. Note that the longest chain may not be drawn in a straight line. The longest chain determines the parent name of the hydrocarbon. For example, in the molecule

    Structural formula of hexane with a methyl group attached to carbon 2.

    the longest chain of carbons has six C atoms. Therefore, it will be named as a hexane. However, in the molecule

    Structural formula of heptane with a methyl group attached to carbon 3.

    the longest chain of C atoms is not six, but seven, as shown. So this molecule will be named as a heptane.

    The next step is to identify the branches, or substituents, on the main chain. The names of the substituents, or alkyl groups, are derived from the names of the parent hydrocarbons; however, rather than having the ending -ane, the substituent name has the ending -yl. Table \(\PageIndex{1}\) - Substituent Names, lists the substituent names for the five smallest substituents.

    Table \(\PageIndex{1}\) Substituent Names
    Substituent Formula Number of C Atoms Name of Substituent
    CH3 1 methyl-
    CH3CH2 2 ethyl-
    CH3CH2CH2 3 propyl-
    CH3CH2CH2CH2 4 butyl-
    CH3CH2CH2CH2CH2 5 pentyl-
    and so forth and so forth and so forth

    In naming the branched hydrocarbon, the name of the substituent is combined with the parent name of the hydrocarbon without spaces. However, there is likely one more step. The longest chain of the hydrocarbon must be numbered, and the numerical position of the substituent must be included to account for possible isomers. As with double and triple bonds, the main chain is numbered to give the substituent the lowest possible number. For example, in this alkane

    Structural formula of pentane with a methyl group attached to carbon 3.

    the longest chain is five C atoms long, so it is a pentane. There is a one-carbon substituent on the third C atom, so there is a methyl group at position 3. We indicate the position using the number, which is followed by a hyphen, the substituent name, and the parent hydrocarbon name—in this case, 3-methylpentane. That name is specific to that particular hydrocarbon and no other molecule. Organic chemistry nomenclature is very specific!

    It is common to write the structural formula of a hydrocarbon without the H atoms, for clarity. So we can also represent 3-methylpentane as

    3-methylpentane displayed without any hydrogen atoms.

    where it is understood that any unwritten covalent bonds are bonds with H atoms. With this understanding, we recognize that the structural formula for 3-methylpentane refers to a molecule with the formula of C6H14.

    Example \(\PageIndex{1}\)

    Name this molecule.

    The molecule has a chain of 7 carbons with an ethyl group coming off of the 5th carbon from the left.

    Solution

    The longest continuous carbon chain has seven C atoms, so this molecule will be named as a heptane. There is a two-carbon substituent on the main chain, which is an ethyl group. To give the substituent the lowest numbering, we number the chain from the right side and see that the substituent is on the third C atom. So this hydrocarbon is 3-ethylheptane.

    Exercise \(\PageIndex{1}\)

    Name this molecule.

    A chain of five carbons with a methyl group on the 4th carbon from the left.

    Answer

    2-methylpentane

    Branched hydrocarbons may have more than one substituent. If the substituents are different, then give each substituent a number (using the smallest possible numbers) and list the substituents in alphabetical order, with the numbers separated by hyphens and with no spaces in the name. So the molecule

    A 5 carbon chain with a methyl group on the second carbon & an ethyl group on the third carbon.

    is 3-ethyl-2-methylpentane.

    If the substituents are the same, then use the name of the substituent only once, but use more than one number, separated by a comma. Also, put a numerical prefix before the substituent name that indicates the number of substituents of that type. The numerical prefixes are listed in Table \(\PageIndex{2}\) - Numerical Prefixes to Use for Multiple Substituents. The number of the position values must agree with the numerical prefix before the substituent.

    Table \(\PageIndex{2}\): Numerical Prefixes to Use for Multiple Substituents
    Number of Same Substituent Numerical Prefix
    2 di-
    3 tri-
    4 tetra-
    5 penta-
    and so forth and so forth

    Consider this molecule:

    butane.png

    The longest chain has four C atoms, so it is a butane. There are two substituents, each of which consists of a single C atom; they are methyl groups. The methyl groups are on the second and third C atoms in the chain (no matter which end the numbering starts from), so we would name this molecule 2,3-dimethylbutane. Note the comma between the numbers, the hyphen between the numbers and the substituent name, and the presence of the prefix di- before the methyl. Other molecules—even with larger numbers of substituents—can be named similarly.

    Example \(\PageIndex{2}\)

    Name this molecule.

    A seven carbon chain with two methyl groups on the second carbon & an ethyl group on the third carbon.

    Solution

    The longest chain has seven C atoms, so we name this molecule as a heptane. We find two one-carbon substituents on the second C atom and a two-carbon substituent on the third C atom. So this molecule is named 3-ethyl-2,2-dimethylheptane.

    Exercise \(\PageIndex{2}\)

    Name this molecule.

    An eight carbon chain with two propyl groups on the fourth carbon from the left & another propyl group on the carbon fifth from the left.

    Answer

    4,4,5-tripropyloctane

    Alkenes and alkynes are named in a similar fashion. The biggest difference is that when identifying the longest carbon chain, it must contain the C–C double or triple bond. Furthermore, when numbering the main chain, the double or triple bond gets the lowest possible number. This means that there may be longer or higher-numbered substituents than may be allowed if the molecule were an alkane. For example, this molecule

    2,4-dimethyl-3-heptene .png

    is 2,4-dimethyl-3-heptene (note the number and the hyphens that indicate the position of the double bond).

    Example \(\PageIndex{3}\)

    Name this molecule.

    A 6 carbon chain with a triple bond between the 3rd and 4th carbons with 2 methyl groups on the carbon 2nd from the left.

    Solution

    The longest chain that contains the C–C triple bond has six C atoms, so this is a hexyne molecule. The triple bond starts at the third C atom, so this is a 3-hexyne. Finally, there are two methyl groups on the chain; to give them the lowest possible number, we number the chain from the left side, giving the methyl groups the second position. So the name of this molecule is 2,2-dimethyl-3-hexyne.

    Exercise \(\PageIndex{3}\)

    Name this molecule.

    A five carbon chain with a double bond between the second and third carbons from the left and a methyl group on each of the three middle carbons is shown.

    Answer

    2,3,4-trimethyl-2-pentene

    Once you master naming hydrocarbons from their given structures, it is rather easy to draw a structure from a given name. Just draw the parent chain with the correct number of C atoms (putting the double or triple bond in the right position, as necessary) and add the substituents in the proper positions. If you start by drawing the C atom backbone, you can go back and complete the structure by adding H atoms to give each C atom four covalent bonds. From the name 2,3-dimethyl-4-propyl-2-heptene, we start by drawing the seven-carbon parent chain with a double bond starting at the third carbon:

    A seven carbon chain with a double bond between the second and third carbons from the left.

    We add two one-carbon substituents to this structure on the second and third C atoms:

    Two methyl groups are added to the carbons (one to each of them) that are involved in the double bond.

    We finish the carbon backbone by adding a three-carbon propyl group to the fourth C atom in the parent chain:

    The propyl group is added to the fourth carbon from the left in the seven member chain.

    If we so choose, we can add H atoms to each C atom to give each carbon four covalent bonds, being careful to note that the C atoms in the double bond already have an additional covalent bond. (How many H atoms do you think are required? There will need to be 24 H atoms to complete the molecule.)

    Example \(\PageIndex{4}\)

    Draw the carbon backbone for 2,3,4-trimethylpentane.

    Solution

    First, we draw the five-carbon backbone that represents the pentane chain:

    A straight chain of 5 carbons.

    According to the name, there are three one-carbon methyl groups attached to the second, the third, and the fourth C atoms in the chain. We finish the carbon backbone by putting the three methyl groups on the pentane main chain:

    The chain of five carbons now has three methyl groups, one each going on to the second, third, and fourth carbons.

    Exercise \(\PageIndex{4}\)

    Draw the carbon backbone for 3-ethyl-6,7-dimethyl-2-octene.

    Answer
    An eight member chain of carbons is shown. There is a double bond between the second and third carbons from the left. An ethyl group is on the third carbon. The sixth and seventh carbons both have a methyl group.

    Naming substituted benzene molecules is straightforward. If there is only one substituent, the substituent is named as a side chain on a benzene molecule, like this:

    The structures of chlorobenzene and ethylbenzene are shown.

    If there are two or more substituents on a benzene molecule, the relative positions must be numbered, just as an aliphatic chain of C atoms is numbered. The substituent that is first alphabetically is assigned position 1, and the ring is numbered in a circle to give the other substituents the lowest possible number(s).

    The structures of 1,3-dichlorobenzene and 1-bromo-2-ethylbenzene are shown.

    If a benzene ring is treated as a substituent, it is given the name phenyl-. The following molecule is 3-phenylpentane:

    A five carbon chain with a phenyl group on the third carbon.

    where the H atoms have been omitted for clarity.

    Summary

    A unique name can be given to branched hydrocarbons. A unique structure can be drawn for the name of a hydrocarbon.

    Exercise \(\PageIndex{5}\)
    1. How does a branched hydrocarbon differ from a normal hydrocarbon?
    2. How does a substituent get its unique name?
    3. Name this molecule. A six carbon chain with a double bond between the second and third carbons and a methyl group on the third carbon.
    4. Name this molecule. A five carbon chain with two methyl groups on the second carbon and another methyl group on the third carbon.
    5. Name this molecule. A five carbon chain with a double bond between the fourth and fifth carbons and two methyl groups on the second carbon.
    6. Name this molecule. Two chains of five carbons each intersect at the second carbon of each of them.
    7. Name this molecule. A five carbon chain with a double bond between the second and third carbons and two methyl groups, one on the second carbon and one on the fourth carbon.
    8. Name this molecule. A five carbon chain with a methyl group on the first carbon and a triple bond between the second and third carbons.
    9. Name this molecule. An eight carbon chain with an ethyl group on the third carbon and an ethyl group on the fourth carbon.
    10. Name this molecule. A seven carbon chain with a double bond between the fourth and fifth carbons. There is a methyl group on the third and fourth carbons.
    11. Name this molecule. An aromatic 6 membered ring with a chlorine on the top carbon and a bromine on the bottom carbon.
    12. Name this molecule. A six membered aromatic ring with an ethyl group on the top carbon and each of the two carbons on the right have a methyl group.
    13. Draw the carbon backbone for each molecule.
      1. 3,4-diethyloctane
      2. 2,2-dimethyl-4-propylnonane
    14. Draw the carbon backbone for each molecule.
      1. 3-ethyl-4-methyl-3-heptene
      2. 3,3-diethyl-1-pentyne
    15. Draw the carbon backbone for each molecule.
      1. 4-ethyl-4-propyl-2-octyne
      2. 5-butyl-2,2-dimethyldecane
    16. Draw the carbon backbone for each molecule.
      1. 3,4-diethyl-1-hexyne
      2. 4-propyl-3-ethyl-2-methyloctane
    17. The name 2-ethylhexane is incorrect. Draw the carbon backbone and write the correct name for this molecule.
    18. The name 3-butyl-7-methyloctane is incorrect. Draw the carbon backbone and write the correct name for this molecule.
    Answers
    1. A branched hydrocarbon does not have all of its C atoms in a single row.
    2.  
    3. 3-methyl-2-hexene
    4.  
    5. 4,4-dimethyl-1-pentene
    6.  
    7. 2,4-dimethyl-2-pentene
    8.  
    9. 3,4-diethyloctane
    10.  
    11. 1-bromo-4-chlorobenzene
    12.  
      1. An eight carbon chain with two ethyl groups, one each on the third and fourth carbons.
      2. A nine carbon chain with two methyl groups on the second carbon and a propyl group on the fourth carbon.
    13.  
      1. A nine carbon chain with a triple bond between the second and third carbon and an ethyl group and propyl group on the fourth carbon.
      2. A ten carbon chain with two methyl groups on the second carbon and a butyl group on the fifth carbon.
    14.  
    15. A seven carbon chain with a an ethyl group on the second carbon.

    This page titled 16.3: Branched Hydrocarbons is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform.