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Homologous Series Chemistry Tutorial

Key Concepts

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Alkane Homologous Series

The alkane homologous series contains compounds that:

Since one carbon will make a total of 4 covalent bonds, the first member (homologue) of the alkane homologous series must be 1 carbon atom bonded to 4 hydrogen atoms as shown below:

  H
|
 
H− C −H
  |
H
 

which has the molecular formula CH4.

The next homologue in the alkane homologous series will contain 2 carbon atoms with a single covalent bond between them, each carbon atom will make up its quota of 4 covalent bonds by bonding to 3 hydrogen atoms as shown below:

  H
|
  H
|
 
H− C C −H
  |
H
  |
H
 

and this molecule has the molecular formula C2H6.

The next homologue in the series will have 3 carbon atoms and 8 hydrogen atoms (C3H8), followed by 4 carbon atoms and 10 hydrogen atoms (C4H10), and so on and so forth...

Some Homologues of the Alkane Homologous Series
molecular structure molecular formula molar mass
(g mol-1)
  H
|
 
H− C −H
  |
H
 
CH4 16
  H
|
  H
|
 
H− C C −H
  |
H
  |
H
 
C2H6 30
  H
|
  H
|
  H
|
 
H− C C C −H
  |
H
  |
H
  |
H
 
C3H8 44
  H
|
  H
|
  H
|
  H
|
 
H− C C C C −H
  |
H
  |
H
  |
H
  |
H
 
C4H10 58
  H
|
  H
|
  H
|
  H
|
  H
|
 
H− C C C C C −H
  |
H
  |
H
  |
H
  |
H
  |
H
 
C5H12 72

Notice a pattern? For each successive homologue we add CH2 to the chain.

Start with CH4 (the first homologue)

then add CH2 to make C2H6 (the next homologue in the series)

then add CH2 to make C3H8 (the next homologue in the series)

then add CH2 to make C4H10 (the next homologue in the series)

then add CH2 to make C5H12 (the next homologue in the series)

Which means the difference in the molar mass between adjacent homologues in the series is 14 g mol-1:

C has a molar mass of 12 g mol-1
H has a molar mass of 1 g mol-1
Therefore CH2 has a molar mass of 12 + (2 × 1) = 12 + 2 = 14 g mol-1

So, each member of the alkane series, each homologue, contains a chain that is n lots of carbon atoms long, Cn.

Each homologue will also contain 2 × n hydrogen atoms (from each CH2 unit), plus there is an extra hydrogen atom at each end of the chain that is, an extra 2 H atoms, so the total number of hydrogen atoms in the chain is 2n+2, H2n+2.

Putting it all together allows us to write a general formula for the homologous series of alkanes: CnH2n+2

For the first member of the alkane series (the first homologue), n=1 so the molecular formula will be C1H2×1+2 which is CH4

For the second member of the alkane series (the second homologue), n=2 so the molecular formula will be C2H2×2+2 which is C2H6

Click to investigate the properties of alkanes.

Alkene Homologous Series

Alkene homologous series contains compounds that:

The first member, the first homologue, of the alkene homologous series must contain 2 carbon atoms which are joined by a double bond, and, because each carbon atom must make 4 bonds in total, each of the carbon atoms must be bonded to 2 hydrogen atoms as shown below:

H           H
  \       /  
    C = C    
  /       \  
H           H
C2H4

(M = 28 g mol-1)

For the next homologue we add 1 more C atom, and it will need to make 4 bonds in total as shown in the structure below:

H       H   H  
  \     |   |  
    C = C C −H
  /         |  
H           H  
C3H6

(M = 42 g mol-1)

For the next homologue we add 1 more C atom, and it will need to make 4 bonds in total as shown in the structure below:

H       H   H   H  
  \     |   |   |  
    C = C C C −H
  /         |   |  
H           H   H  
C4H8

(M = 56 g mol-1)

For the next homologue we add 1 more C atom, and it will need to make 4 bonds in total as shown in the structure below:

H       H   H   H   H  
  \     |   |   |   |  
    C = C C C C −H
  /         |   |   |  
H           H   H   H  
C5H10

(M = 70 g mol-1)

So we can see that we are adding a CH2 unit to one homologue to make the next one in the series.
Since each CH2 unit has a molar mass of 14 g mol-1, the difference in molar mass between adjacent homologues is 14 g mol-1.

And we can describe the homologous series of alkenes by a general formula, CnH2n.

For the first member of the series (the first homologue), n=2, so the molecular formula for this compound will be C2H2×2 which is C2H4

For the second member of the series (the second homologue), n=3, so the molecular formula for this compound will be C3H2×3 which is C3H6

Click to investigate the properties of alkenes.

Alkyne Homologous Series

Alkyne homologous series contains compounds that:

The first member, the first homologue, of the alkyne homologous series must contain 2 carbon atoms which are joined by a triple bond, and, because each carbon atom must make 4 bonds in total, each of the carbon atoms must be bonded to 1 hydrogen atom as shown below:

  H− C≡C −H  
C2H2

(M = 26 g mol-1)

The next member in the series, the next homologue, must contain 3 carbon atoms, and each C atom must make 4 bonds in total:

      H
|
 
H− C≡C C −H
      |
H
 
C3H4

(M = 40 g mol-1)

Which is followed by the next homologue with 4 carbon atoms:

      H
|
  H
|
 
H− C≡C C C −H
      |
H
  |
H
 
C4H6

(M = 54 g mol-1)

So once again we can see that we are adding a CH2 unit to one homologue to make the next one in the series.
Since each CH2 unit has a molar mass of 14 g mol-1, the difference in molar mass between adjacent homologues is 14 g mol-1.

And we can describe the homologous series of alkynes by a general formula, CnH2n−2.

For the first member of the series (the first homologue), n=2, so the molecular formula for this compound will be C2H2×2−2 which is C2H2

For the second member of the series (the second homologue), n=3, so the molecular formula for this compound will be C3H2×3−2 which is C3H4

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Alkanol Homologous Series

Alkanol homologous series contains compounds that:

The first member of the alkanol homologous series contains just 1 carbon atom which needs to make 4 bonds in total, one of these bonds must be to the OH functional group, so the other 3 bonds must be to hydrogen atoms (H), as shown below:

  H
|
 
OH C −H
  |
H
 
CH4O

(M = 32 g mol-1)

The next homologue contains 2 carbon atoms, each carbon atom must make a total of 4 bonds, one of these bonds is to the OH functional group:

  H
|
  H
|
 
OH C C −H
  |
H
  |
H
 
C2H6O

(M = 46 g mol-1)

The next member of the alkanol homologous series has 3 carbon atoms, all of which must make 4 bonds, and one of those bonds must be to the OH functional group:

  H
|
  H
|
  H
|
 
OH C C C −H
  |
H
  |
H
  |
H
 
C3H8O

(M = 60 g mol-1)

The difference between the molecular formulae of adjacent homologues is -CH2, which has a molar mass of 14 g mol-1, so the difference in molar mass of adjacent homologues is 14 g mol-1.

The general formula for members of the alkanol homologous series is CnH2n+2O

The first member of the alkanol homologous series has n=1, so the molecular formula of the first homologue is C1H2×1+2O, that is, CH4O

The second member of the alkanol homologous series has n=2, so the molecular formula of the second homologue is C2H2×2+2O, that is, C2H6O

Click to investigate the properties of alkanols.

Alkanal Homologous Series

Alkanal homologous series contains compounds that:

The first member (homologue) of the alkanal homologous series contains only 1 carbon atom, which must make 4 bonds in total and 2 of those bonds are to an oxygen atom, as shown below:

    H
|
   
  O= C −H  
CH2O

(M = 30 g mol-1)

The second member (homologue) of the alkanal homologous series contains 2 carbon atoms, each of which must make 4 bonds in total, and the end carbon atom must make 2 bonds to an oxygen atom, as shown below:

  H
|
  H
|
 
O= C C −H
      |
H
 
C2H4O

(M = 44 g mol-1)

The third member (homologue) of the alkanal homologous series contains 3 carbon atoms, each of which must make 4 bonds in total, and the end carbon atom must make 2 bonds to an oxygen atom, as shown below:

  H
|
  H
|
  H
|
 
O= C C C −H
      |
H
  |
H
 
C3H6O

(M = 58 g mol-1)

The difference between the molecular formulae of adjacent homologues is -CH2, which has a molar mass of 14 g mol-1, so the difference in molar mass of adjacent homologues is 14 g mol-1.

The general formula for members of the alkanal homologous series is CnH2nO

The first member of the alkanal homologous series has n=1, so the molecular formula of the first homologue is C1H2×1O, that is, CH2O

The second member of the alkanal homologous series has n=2, so the molecular formula of the second homologue is C2H2×2O, that is, C2H4O

Click to investigate the properties of alkanals.

Alkanone Homologous Series

Alkanone homologous series contains compounds that:

The first member (homologue) of the alkanone homologous series contains 3 carbon atoms, each of which must make 4 bonds in total, and the carbon atom in the middle of the chain must make 2 bonds to an oxygen atom, as shown below:

  H
|
  O
||
  H
|
 
H− C C C −H
  |
H
      |
H
 
C3H6O

(M = 58 g mol-1)

The next homologue contains 4 carbon atoms, each of which must make 4 bonds in total, and the carbon atom in the middle of the chain must make 2 bonds to an oxygen atom, as shown below:

  H
|
  O
||
  H
|
  H
|
 
H− C C C C −H
  |
H
      |
H
  |
H
 
C4H8O

(M = 72 g mol-1)

The next homologue contains 5 carbon atoms, each of which must make 4 bonds in total, and the carbon atom in the middle of the chain must make 2 bonds to an oxygen atom, as shown below:

  H
|
  O
||
  H
|
  H
|
  H
|
 
H− C C C C C −H
  |
H
      |
H
  |
H
  |
H
 
C5H10O

(M = 86 g mol-1)

The difference between the molecular formulae of adjacent homologues is -CH2, which has a molar mass of 14 g mol-1, so the difference in molar mass of adjacent homologues is 14 g mol-1.

The general formula for members of the alkanone homologous series is CnH2nO

The first member of the alkanone homologous series has n=3, so the molecular formula of the first homologue is C3H2×3O, that is, C3H6O

The second member of the alkanone homologous series has n=4, so the molecular formula of the second homologue is C4H2×4O, that is, C4H8O

Click to investigate the properties of alkanones.

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Alkanoic Acids Homologous Series

Alkanoic acid homologous series contains compounds that:

The first member (homologue) of the alkanoic acid homologous series contains only 1 carbon atom, which must make 4 bonds in total and 2 of those bonds are to an oxygen atom, one of those bonds is to an OH group, as shown below:

    O
||
   
  HO- C −H  
CH2O2

(M = 46 g mol-1)

The second member (homologue) of the alkanoic acid homologous series contains 2 carbon atoms, which must make 4 bonds in total, the first carbon atom must have 2 of those bonds to an oxygen atom and 1 of those bonds is to an OH group, as shown below:

  O
||
  H
|
 
HO- C C −H
      |
H
 
C2H4O2

(M = 60 g mol-1)

The third member (homologue) of the alkanoic acid homologous series contains 3 carbon atoms, which must make 4 bonds in total, the first carbon atom must have 2 of those bonds to an oxygen atom and 1 of those bonds is to an OH group, as shown below:

  O
||
  H
|
  H
|
 
HO- C C C −H
      |
H
  |
H
 
C3H6O2

(M = 74 g mol-1)

The difference between the molecular formulae of adjacent homologues is -CH2, which has a molar mass of 14 g mol-1, so the difference in molar mass of adjacent homologues is 14 g mol-1.

The general formula for members of the alkanoic acid homologous series is CnH2nO2

The first member of the alkanoic acid homologous series has n=1, so the molecular formula of the first homologue is C1H2×1O2, that is, CH2O2

The second member of the alkanoic acid homologous series has n=2, so the molecular formula of the second homologue is C2H2×2O2, that is, C2H4O2

Click to investigate the properties of alkanoic acids.

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