Question

In principle, the equilibrium in the dehydration of an alcohol could be shifted to the right be removal of water. Why is this tactic not a good option for the dehydration of 4-methyl-2-pentanol and cyclohexanol

Answer 1

See explanation

Step-by-step explanation:

In this case, we have to remember that if we want to remove water from the reaction vessel we have to heat the vessel. So, we can convert the liquid water into gas water and we can remove it from the vessel. In this case, the products of dehydration for both molecules are (E)-4-methylpent-2-ene and cyclohexene with boiling points of 59.2 ºC and 89 ºC respectively. The boiling point of water is 100 ºC, therefore if we heat the vessel the products and water would leave the system, and the products would be lost.

See figure 1

I hope it helps!