Alkanes, Alkenes, and Alkynes

Alkane molecules have only single bonds (\chemfig{C-[,.6]C}) between carbon atoms.

Alkene molecules have a double bond (\chemfig{C=[,.6]C}).

Alkyne molecules have a triple bond (\chemfig{C~[,.6]C}).

The following examples show the 4 carbon alkane, alkene, and alkyne.

Notice that the alkene only has one double bond and the alkyne only has one triple bond.

Name Formula Structure
butane C4H10 \definesubmol{a}{-C(-[2,.6]H)(-[6,.6]H)-[,.2]} \definesubmol{e}{-C(-[6,.6]H)=[,.6]C(-[6,.6]H)-[,.2]} \definesubmol{y}{-C~[,.6]C-[,.2]} \chemfig{[,.4]H-[,.2]!a!a!a!a-[,.4]H}
butene C4H8 \definesubmol{a}{-C(-[2,.6]H)(-[6,.6]H)-[,.2]} \definesubmol{e}{-C(-[6,.6]H)=[,.6]C(-[6,.6]H)-[,.2]} \definesubmol{y}{-C~[,.6]C-[,.2]} \chemfig{[,.4]H-[,.2]!e!a!a-[,.4]H}
butyne C4H6 \definesubmol{a}{-C(-[2,.6]H)(-[6,.6]H)-[,.2]} \definesubmol{e}{-C(-[6,.6]H)=[,.6]C(-[6,.6]H)-[,.2]} \definesubmol{y}{-C~[,.6]C-[,.2]} \chemfig{[,.4]H-[,.2]!y!a!a-[,.4]H}

All of the carbon atoms above have exactly 4 bonds.

A multiple-bonded C atom has less bonds available for H’s. That’s why butene (C4H8) has two less H atoms than butane (C4H10), and butyne(C4H6) has four less H’s than butane.

Generic Formula for Alkane / Alkene / Alkyne

Type Generic
Formula
Example
n = # of C’s
Example
2n+2 = # of H’s
Example
Formula
alkane CnH2n+2 n = 5 2n + 2 = 12 C5H12
alkene CnH2n n = 5 2n = 10 C5H10
alkyne CnH2n-2 n = 5 2n – 2 = 8 C5H8

Example: If n=5, then 2n + 2 = 2(5) + 2 = 10 + 2 = 12

You can use these generic formulas to predict the formula for any alkane, alkene, or alkyne.

Example:
What is the molecular formula of a 32 carbon alkyne?

Solution:
•Generic formula is CnH2n-2
•# of C’s = n = 32
•# of H’s = 2n – 2 = 2(32) – 2 = 62
•Formula is C32H62

Naming Alkanes / Alkenes / Alkynes

Use a prefix (found here) to indicate the number of C’s.

Use the appropriate ending to show which type of molecule it is:

Type Functional
Group
Name
Ending
alkane \chemfig{C-[,.6]C}
(all single bonds)
-ane
alkene \chemfig{C=[,.6]C}
(one double bond)
-ene
alkyne \chemfig{C~[,.6]C}
(one triple bond)
-yne

 

Example 1: Name this molecule: \definesubmol{a}{-C(-[2,.6]H)(-[6,.6]H)-[,.2]} \definesubmol{e}{-C(-[6,.6]H)=[,.6]C(-[6,.6]H)-[,.2]} \definesubmol{y}{-C~[,.6]C-[,.2]} \chemfig{[,.4]H-[,.2]!a!a!e!a!a-[,.4]H}


Solution 1:
 It has 6 carbons, so the prefix is “hex-.” It has one double bond, so it’s an alkene and it ends with “-ene”. Its name is hexene.

Example 2: Draw pentyne.

Solution 2: It has 5 carbons since the prefix is “pent-.” It has one triple bond since the ending is “-yne.” There are multiple ways to draw this one depending on where you want to put the triple bond:

\definesubmol{a}{-C(-[2,.6]H)(-[6,.6]H)-[,.2]} \definesubmol{e}{-C(-[6,.6]H)=[,.6]C(-[6,.6]H)-[,.2]} \definesubmol{y}{-C~[,.6]C-[,.2]} \chemfig{[,.4]H-[,.2]!y!a!a!a-[,.4]H}

\definesubmol{a}{-C(-[2,.6]H)(-[6,.6]H)-[,.2]} \definesubmol{e}{-C(-[6,.6]H)=[,.6]C(-[6,.6]H)-[,.2]} \definesubmol{y}{-C~[,.6]C-[,.2]} \chemfig{[,.4]H-[,.2]!a!y!a!a-[,.4]H}

The first solution shows a triple bonded carbon on one end, while the second solution shows the triple bond somewhere in the middle. Both are correct structures for pentyne (they are isomers, and branched isomers also exist).