Alcohols can be prepared from alkenes by addition of
concentrated sulfuric acid across the double bond
followed by reaction of the intermediate sulfate with
water.
The overall result is addition of H–OH across the double bond.
The H and the OH become bonded to carbons of the alkene that were doubly-bonded.
There is no change in the arrangement of carbons.
 | 1. conc H2SO4  2. H2O |
A
single alcohol is produced above because the
alkene is symmetrical.
Symmetrical alkenes have the
same groups attached to
both of the doubly-bonded carbons.
In hex-3-ene above each of the doubly-bonded carbons is bonded to an H and a CH2CH3 group.
The alcohol produced is secondary because there are two carbons bonded to the carbon bearing OH.
This reaction may give rise to
two products for
unsymmetrical alkenes that have different groups at the two carbons of the double bond.
The
major product is the
more substituted alcohol (the one with the larger number of carbons bonded to the carbon bearing OH).
Therefore secondary* and tertiary* alcohols can be prepared in good yield, but not primary* alcohols.
This guideline that can be used to predict the major product of the reaction is also known as
Markownikoff's Rule.
 | 1. conc H2SO4  2. H2O |  major product | + |
 major product | + |  minor product |
*Tertiary alcohols have three carbons bonded to the carbon bearing OH.
Secondary alcohols have two carbons bonded to the carbon bearing OH.
Primary alcohols have one carbon bonded to the carbon bearing OH.