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Dehydration of Alkanols (alcohols)

Key Concepts

  • Dehydration of alkanols (alcohols) is an elimination reaction.

  • Water is eliminated from the alkanol resulting in an alkene.

  • Any concentrated strong acid can cause the dehydration of an alkanol, but hot, concentrated sulfuric acid is the most commonly used.

  • The reagent used to dehydrate the alkanol is known as a dehydrating agent.

  • The dehydration of a primary alkanol (alcohol) produces the alk-1-ene (1-alkene)

      H
    |
    H
    |
     
    R- C- C- OH
      |
    H
    |
    H
     
    hot conc. H2SO4
      H
    |
    H
    |
     
    R- C= C  
        |
    H
     
    + H-O-H

  • The dehydration of a secondary alkanol or a tertiary alkanol (alcohol) may produce a mixture of alkene structural isomers1.

Dehydration of a Primary Alkanol

When hot, concentrated sulfuric acid is added to a primary alkanol a water molecule is eliminated from the alkanol molecule.

This eliminated water molecule is made up of the hydroxyl (hydroxy), OH, functional group of the alkanol and an atom of hydrogen, H, from an adjacent carbon atom in the alkanol molecule.

primary alkanol dehydrating agent
alk-1-ene + water
  H
|
H
|
 
R- C- C- OH
  |
H
|
H
 
hot conc. H2SO4
  H
|
H
|
 
R- C= C  
    |
H
 
+ H-O-H

Example: Dehydration of Ethanol

Ethanol, CH3-CH2OH, (ethyl alcohol) is a primary alcohol.
At 180oC, concentrated sulfuric acid will dehydrate ethanol to produce ethene, CH2=CH2, (ethylene) and water.
ethanol
(ethyl alcohol)
hot conc. H2SO4

180oC
ethene
(ethylene)
+ water
CH3-CH2OH hot conc. H2SO4

180oC
CH2=CH2 + H2O
  H
|
H
|
 
H- C- C -OH
  |
H
|
H
 
hot conc. H2SO4

180oC
  H
|
H
|
 
  C= C  
  |
H
|
H
 
+ H-O-H

Dehydration of a Secondary Alkanol

Adding hot, concentrated sulfuric acid to a secondary alkanol eliminates a water molecule from the organic molecule in a dehydration reaction.

This eliminated water molecule is made up of the hydroxyl (hydroxy), OH, functional group of the alkanol and an atom of hydrogen, H, from an adjacent carbon atom in the alkanol molecule.

If the secondary alkanol molecule is symmetrical, only one organic product will result.

  H
|
OH
|
H
|
 
R- C- C- C- R
  |
H
|
H
|
H
 
hot conc. H2SO4
  H
|
  H
|
 
R- C= C- C- R
    |
H
|
H
 
+ H-O-H

This alkanol molecule is symmetrical because both alkyl groups, R, are the same

If the secondary alkanol molecule is not symmetrical, two organic products will be produced.
These two organic products will be structural isomers of the same alkene molecule.

secondary alkanol   structural isomers   water
  H
|
OH
|
H
|
 
R- C- C- C- R'
  |
H
|
H
|
H
 
hot conc. H2SO4
  H
|
  H
|
 
R- C= C- C- R'
    |
H
|
H
 
+
  H
|
  H
|
 
R- C- C= C- R'
  |
H
|
H
   
+ H-O-H

This alkanol molecule is not symmetrical because the two alkyl groups, R and R', are different.

Example: Deydration of a Symmetrical Secondary Alkanol

Propan-2-ol (2-propanol), (CH3)2CHOH, is a symmetrical secondary alcohol.
At 100oC, concentrated sulfuric acid will dehydrate propan-2-ol (2-propanol) to produce propene (propylene) and water.
Since propan-2-ol (2-propanol) is a symmetrical molecule, only one organic product will be produced.
propan-2-ol
(2-propanol)
hot conc. H2SO4

100oC
propene
(propylene)
+ water
(CH3)2CHOH hot conc. H2SO4

100oC
CH3CH=CH2 + H2O
  H
|
OH
|
H
|
 
H- C- C- C- H
  |
H
|
H
|
H
 
hot conc. H2SO4

100oC
  H
|
     
H- C- C= C -H
  |
H
|
H
|
H
 
+ H-O-H

Example: Deydration of an Unsymmetrical Secondary Alkanol

Pentan-2-ol (2-pentanol), CH3-CH2-CH2-CHOH-CH3, is an unsymmetrical secondary alcohol.
Hot concentrated sulfuric acid will dehydrate pentan-2-ol (2-pentanol) to produce water and two possible isomers of pentene, pent-2-ene and pent-1-ene (also named as 2-pentene and 1-pentene).
pentan-2-ol
(2-pentanol)
hot conc. H2SO4
pent-2-ene
(2-pentene)
+ pent-1-ene
(1-pentene)
+ water
CH3-CH2-CH2-CHOH-CH3 hot conc. H2SO4
CH3-CH2-CH=CH-CH3 + CH3-CH2-CH2-CH=CH2 + H2O
  H
|
H
|
H
|
OH
|
H
|
 
H- C- C- C- C- C- H
  |
H
|
H
|
H
|
H
|
H
 
hot conc. H2SO4
  H
|
H
|
    H
|
 
H- C- C- C= C- C -H
  |
H
|
H
|
H
|
H
|
H
 
+
  H
|
H
|
H
|
       
H- C- C- C- C = C -H
  |
H
|
H
|
H
|
H
  |
H
 
+ H-OH
    (major product)   (minor product)

Animated Tutorial


What would you like to do now?

1Saytseff's Rule predicts that the alkene with the greatest number of alkyl groups on the doubly-bonded carbon atoms will be the major product.

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