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Dehydration of Alkanols (dehydration of alcohols) Chemistry Tutorial

Key Concepts

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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 as shown in the reaction given below:

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

The products of the dehydration of a primary alkanol are water and an alk-1-ene (1-alkene).

Example: Dehydration of Ethanol to Produce Ethene (ethylene)

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

The chemical reaction is given below:

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

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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, as shown in the chemical equation below:

  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.

The dehydration of an unsymmetrical secondary alkanol is shown below:

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 100°C, 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 as shown in the chemical equation below:

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), as shown in the chemical equation given below:

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)

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Footnotes:

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