Question

What is the ease of dehydration of alcohol in primary, tertiary and secondary alcohol?

Answer 1

Hint : Alcohols when reacted with protic acid, it loses a molecule of water in order to form alkenes. This formation of alkene by alcohol with loss of water is termed as dehydration of alcohols. The rate differs depending on primary, tertiary and for secondary alcohol.

Complete Step By Step Answer:
The mechanism involved in general is that the $O{H^{}}$ group in the alcohol donates its two electrons to hydrate ions from the acid reagent, hence forming an alkyl oxonium ion. Dehydration mechanisms differ in different types of alcohol. The alkyloxonium ion acts as a good leaving group and forms a carbocation. The formed base that is deprotonated acid reacts with the hydrogen adjacent to the carbocation and will form the double bond. The primary alcohol dehydrates through the E2 Mechanism. The hydroxyl oxygen donates two electrons to a proton from sulphuric acid ( ${H_2}S{O_4}$ ), forming an alkyloxonium ion. The thus formed conjugate base, that is $HS{O_4}^ -$ reacts with beta hydrogen. The alkyloxonium forms the double bond. This primary alcohol undergoes bi molecular elimination. The secondary and tertiary alcohols dehydrate through the E1 mechanism, similarly to the reaction above. The only difference here is that carbocation will be in intermediate form.
According to the Saytzeff rule, the more substituted alkenes are more stable and will be formed more preferentially than substituted alkenes.
So according to this rule, we clearly know tertiary alcohol is more substituted and least is primary.
So the relative order of dehydration of alcohol will be in the order- tertiary $>$ secondary $>$ primary.

Note :
amongst the trans alkenes are more stable than cis-alkenes and also the major product formed. For example the trans-diastereomer of the $2 -$ butene product is most abundant. Alkenes are also referred to as olefins.