Question

Which one of the following is the slowest step in dehydration of alcohol?

• A. Protonation of Alcohol
  
• B. Loss of leaving group
  $\left(H_{2}O\right)$ molecule
  
• C. Deprotonation to form alkene
  
• D. None of the above

Answer 1

The correct option is B Loss of leaving group

$\left(H_{2}O\right)$ molecule

Dehydration of alcohol is an E1 elimination reaction. The rate of the reaction depends on the stability of the carbocation formed.
Mechanism of dehydration of alcohol:

Step 1: Protonation of alcohol
The $H_{2}S{O}_4$ will protonate the alcohol group to form ${}^{+}O{H}_2$. this step occur very fast.

Step 2: Loss of leaving group (Slowest step)
The formed ${}^{+}O{H}_2$ leaves as $H_{2}O$ leaving behind a carbocation.
Step 3: Loss of proton.
Due to delocalisation of lone pair of electron of oxygen, it carries a positive charge. Then, the proton gets removed to form a carbonyl group.
A is major product as it is more substituted alkene according to saytzeff rule while B is minor product.