Question

Among the following compounds which can be dehydrated very easily:
(A) $C{H_3}C{H_2}C{H_2}C{H_2}C{H_2}OH$
(B) $C{H_3}C{H_2}C{H_2}CHOHC{H_3}$
(C) $C{H_3}C{H_2}CHC{H_3}C{H_2}C{H_2}OH$
(D) $C{H_3}C{H_2}C{H_3}COHC{H_2}C{H_3}$

Answer 1

The stabler the alkene formed as a result of the dehydration of the alcohol, the greater the ease of its dehydration of said alcohol. Since the carbocation is most stable in the case of tertiary alcohols, the rate of dehydration is highest for tertiary alcohols in comparison to secondary and primary alcohols.

Complete answer:
Let’s first discuss what dehydration of alcohol is. Alcohol upon reaction with protic acids tends to lose a molecule of water to form alkenes. It is an example of an elimination reaction. Its rate varies for primary, secondary and tertiary alcohols. This variation of rate can be attributed to the stability of carbocation generated. Since the carbocation is most stable in the case of tertiary alcohols, the rate of dehydration is highest for tertiary alcohols in comparison to secondary and primary alcohols. One way to synthesize alkenes is by dehydration of alcohols, a process in which alcohols undergo ${E_1}$ or ${E_2}$ mechanisms to lose water and form a double bond.
The order of the ease of dehydration of alcohols is: tertiary $>$ secondary $>$ primary. Secondary and tertiary alcohols are best dehydrated by dilute sulfuric acid. The more stable carbocation is generated, the more easily it will be dehydrated.
So, the correct answer is (D) $C{H_3}C{H_2}C{H_3}COHC{H_2}C{H_3}$ .

Note:
If the reaction is not sufficiently heated, the alcohol does not dehydrate to form alkenes, but react with one another to form ethers. Alcohols are amphoteric; they can act as both acid and base. The lone pair of electrons on the oxygen atom makes the $- OH$ group weakly basic.