Question:The conversion of 3° alcohols into alkenes under acidic conditions involves two cationic intermediates. For each reaction, draw the complete mechanism using curved arrows.

a)

b)

c)

It is a process by which removal of the hydroxyl group takes place and leads to the formation of an alkene by following the Zaitsev rule and this reaction is followed by the loss of water molecule.

The reaction is an E1 reaction in which protonation of the alcohol oxygen occurs followed by the loss of the water molecule for the formation of the carbocation then protonation of the hydrogen occurs and alkene is formed which is shown below here:

Step followed as:

1. Protonation of the alcohol
2. Breaking of the carbon-oxygen bond and two electrons from bond accepted by an oxygen atom and carbocation formed
3. Stable carbocation do elimination reaction because this lead to the formation of more stable alkene due to the hyperconjugation having nine alpha hydrogen
4. Then formation of alkene with takes place
5. The alkene which have non-terminal double bond is more stable

Step followed as:

1. Protonation of the alcohol
2. Breaking of the carbon-oxygen bond and two electrons from bond accepted by oxygen atom and carbocation formed.
3. These two electron formation of alkene with pi bond and protonation of${\mathrm{H}}^{+}$ occur
4. Stable alkene is formed because of the hyperconjugation having 3 alpha hydrogen