* Alcohols undergo dehydration (loss of water molecule) in acidic medium to give olefins. A double bond is formed due to loss of water molecule. It is an elimination reaction. According to Saytzeff's rule (also Zaitsev's rule), during dehydration, more substituted alkene (olefin) is formed as a major product, since greater the substitution of double bond greater is the stability of alkene.
* Sulfuric acid is a strong acid and ionizes to give a proton.
* Thus formed H+ ion attacks the -OH group. The OH2+ group formed is good leaving group due to accumulation of positive charge on oxygen atom. Now the loss of H2O creates positive charge on the carbon atom and thus by forming a tertiary carbocation.
* Finally, one of the hydrogen adjacent to the positively charged carbon is removed as proton, H+ to make a double bond. However there are three adjacent hydrogen atoms (indicated by descriptors I, II & III). Hence three different alkenes are possible.
However according to Saytzeff's rule, highly substituted alkene, as shown below by the loss of H+(I), is formed as a major product.
Note: Since a tertiary carbocation is formed during the reaction, it does not undergo any rearrangement.
1) How do we get alkene if a halogen is present instead of -OH group? Does this reaction too follow Saytzeff's rule?
2) Why do more substituted alkenes are more stable?
3) What are the IUPAC names of alkenes formed in above reaction?