\(R–OH + HX \rightarrow R–X + H_2O\)
In the above reaction, the reactivity of different alcohols is

Tertiary > secondary > primary

The correct answer is option 1. Tertiary > secondary > primary.

The reaction

\(R–OH + HX \rightarrow R–X + H_2O\)

is a nucleophilic substitution reaction that proceeds via the \(S_N1\) mechanism. In the \(S_N1\) mechanism, the rate-determining step is the formation of a carbocation. The stability of the carbocation determines the rate of the reaction. Tertiary carbocations are the most stable, followed by secondary and primary carbocations. Therefore, the reactivity of alcohols in this reaction follows the order tertiary > secondary > primary.

The + I-effect of alkyl groups stabilizes carbocations by donating electrons to the positive charge. This makes tertiary carbocations more stable than secondary carbocations, which are more stable than primary carbocations. As a result, tertiary alcohols are more reactive than secondary alcohols, which are more reactive than primary alcohols.