Starting with (Propanol) and
(ethanol). Prepare `pent-2-ene` using `G.R`. Is there more than one way ? If so, which is the best and why ?

Explanation

Preparation of pent-2-ene from ethanol (two `C` atoms) and propanol (three `C` atoms) requires three and two more `C` atoms, respectively, `Pent-2-ene` can be obtained by the dehydration of two alcohols `(I and II)`.

first prepare any one or both `2^@` alcohols by using `G.R`.
(i) Since `2^@` alcohol `(II)` is symmetrical, there is only one way of preparing it from ethanol and propanol.

(ii) Preparation of pent-2-ene z from ethanol and propanol.
(a)
(b)
(iiii) As `2^@` alcohol `(I)` is unsymmetrical, there are two ways of preparing it from ethanol and propanol.

The compounds `(D)` and `( C)` are available, but `( E)` and `(F)` are not available, which requires stepping up of propanol and stepping down of ethanol by one `C` atom. Therefore, it is best to prepare `2^@` alcohol by the available pair `( C)` and `(D)`.

Two alkenes are possible by the dehydration of `2^@` alcohol `(I)`. the major product is formed according to Saytzeff's rule. Thus, the best way of preparing pent-2-ene is via alcohol `(II)`.